### Problem 1: Multiples of 3 and 5Published on Friday, 5th October 2001, 06:00 pm; Solved by 798257; Difficulty rating: 5%

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

### Problem 2: Even Fibonacci numbersPublished on Friday, 19th October 2001, 06:00 pm; Solved by 639270; Difficulty rating: 5%

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

### Problem 3: Largest prime factorPublished on Friday, 2nd November 2001, 06:00 pm; Solved by 457201; Difficulty rating: 5%

The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

### Problem 4: Largest palindrome productPublished on Friday, 16th November 2001, 06:00 pm; Solved by 405885; Difficulty rating: 5%

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.

Find the largest palindrome made from the product of two 3-digit numbers.

### Problem 5: Smallest multiplePublished on Friday, 30th November 2001, 06:00 pm; Solved by 413525; Difficulty rating: 5%

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

### Problem 6: Sum square differencePublished on Friday, 14th December 2001, 06:00 pm; Solved by 416073; Difficulty rating: 5%

The sum of the squares of the first ten natural numbers is,

12 + 22 + ... + 102 = 385

The square of the sum of the first ten natural numbers is,

(1 + 2 + ... + 10)2 = 552 = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

### Problem 7: 10001st primePublished on Friday, 28th December 2001, 06:00 pm; Solved by 355881; Difficulty rating: 5%

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

### Problem 8: Largest product in a seriesPublished on Friday, 11th January 2002, 06:00 pm; Solved by 299458; Difficulty rating: 5%

The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?

### Problem 9: Special Pythagorean tripletPublished on Friday, 25th January 2002, 06:00 pm; Solved by 303125; Difficulty rating: 5%

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a2 + b2 = c2

For example, 32 + 42 = 9 + 16 = 25 = 52.

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

### Problem 10: Summation of primesPublished on Friday, 8th February 2002, 06:00 pm; Solved by 277660; Difficulty rating: 5%

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

### Problem 11: Largest product in a gridPublished on Friday, 22nd February 2002, 06:00 pm; Solved by 200567; Difficulty rating: 5%

In the 20×20 grid below, four numbers along a diagonal line have been marked in red.

08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91
22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80
24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50
32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

### Problem 12: Highly divisible triangular numberPublished on Friday, 8th March 2002, 06:00 pm; Solved by 187801; Difficulty rating: 5%

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

### Problem 13: Large sumPublished on Friday, 22nd March 2002, 06:00 pm; Solved by 193337; Difficulty rating: 5%

Work out the first ten digits of the sum of the following one-hundred 50-digit numbers.

37107287533902102798797998220837590246510135740250
46376937677490009712648124896970078050417018260538
74324986199524741059474233309513058123726617309629
91942213363574161572522430563301811072406154908250
23067588207539346171171980310421047513778063246676
89261670696623633820136378418383684178734361726757
28112879812849979408065481931592621691275889832738
44274228917432520321923589422876796487670272189318
47451445736001306439091167216856844588711603153276
70386486105843025439939619828917593665686757934951
62176457141856560629502157223196586755079324193331
64906352462741904929101432445813822663347944758178
92575867718337217661963751590579239728245598838407
58203565325359399008402633568948830189458628227828
80181199384826282014278194139940567587151170094390
35398664372827112653829987240784473053190104293586
86515506006295864861532075273371959191420517255829
71693888707715466499115593487603532921714970056938
54370070576826684624621495650076471787294438377604
53282654108756828443191190634694037855217779295145
36123272525000296071075082563815656710885258350721
45876576172410976447339110607218265236877223636045
17423706905851860660448207621209813287860733969412
81142660418086830619328460811191061556940512689692
51934325451728388641918047049293215058642563049483
62467221648435076201727918039944693004732956340691
15732444386908125794514089057706229429197107928209
55037687525678773091862540744969844508330393682126
18336384825330154686196124348767681297534375946515
80386287592878490201521685554828717201219257766954
78182833757993103614740356856449095527097864797581
16726320100436897842553539920931837441497806860984
48403098129077791799088218795327364475675590848030
87086987551392711854517078544161852424320693150332
59959406895756536782107074926966537676326235447210
69793950679652694742597709739166693763042633987085
41052684708299085211399427365734116182760315001271
65378607361501080857009149939512557028198746004375
35829035317434717326932123578154982629742552737307
94953759765105305946966067683156574377167401875275
88902802571733229619176668713819931811048770190271
25267680276078003013678680992525463401061632866526
36270218540497705585629946580636237993140746255962
24074486908231174977792365466257246923322810917141
91430288197103288597806669760892938638285025333403
34413065578016127815921815005561868836468420090470
23053081172816430487623791969842487255036638784583
11487696932154902810424020138335124462181441773470
63783299490636259666498587618221225225512486764533
67720186971698544312419572409913959008952310058822
95548255300263520781532296796249481641953868218774
76085327132285723110424803456124867697064507995236
37774242535411291684276865538926205024910326572967
23701913275725675285653248258265463092207058596522
29798860272258331913126375147341994889534765745501
18495701454879288984856827726077713721403798879715
38298203783031473527721580348144513491373226651381
34829543829199918180278916522431027392251122869539
40957953066405232632538044100059654939159879593635
29746152185502371307642255121183693803580388584903
41698116222072977186158236678424689157993532961922
62467957194401269043877107275048102390895523597457
23189706772547915061505504953922979530901129967519
86188088225875314529584099251203829009407770775672
11306739708304724483816533873502340845647058077308
82959174767140363198008187129011875491310547126581
97623331044818386269515456334926366572897563400500
42846280183517070527831839425882145521227251250327
55121603546981200581762165212827652751691296897789
32238195734329339946437501907836945765883352399886
75506164965184775180738168837861091527357929701337
62177842752192623401942399639168044983993173312731
32924185707147349566916674687634660915035914677504
99518671430235219628894890102423325116913619626622
73267460800591547471830798392868535206946944540724
76841822524674417161514036427982273348055556214818
97142617910342598647204516893989422179826088076852
87783646182799346313767754307809363333018982642090
10848802521674670883215120185883543223812876952786
71329612474782464538636993009049310363619763878039
62184073572399794223406235393808339651327408011116
66627891981488087797941876876144230030984490851411
60661826293682836764744779239180335110989069790714
85786944089552990653640447425576083659976645795096
66024396409905389607120198219976047599490197230297
64913982680032973156037120041377903785566085089252
16730939319872750275468906903707539413042652315011
94809377245048795150954100921645863754710598436791
78639167021187492431995700641917969777599028300699
15368713711936614952811305876380278410754449733078
40789923115535562561142322423255033685442488917353
44889911501440648020369068063960672322193204149535
41503128880339536053299340368006977710650566631954
81234880673210146739058568557934581403627822703280
82616570773948327592232845941706525094512325230608
22918802058777319719839450180888072429661980811197
77158542502016545090413245809786882778948721859617
72107838435069186155435662884062257473692284509516
20849603980134001723930671666823555245252804609722
53503534226472524250874054075591789781264330331690

### Problem 14: Longest Collatz sequencePublished on Friday, 5th April 2002, 06:00 pm; Solved by 193118; Difficulty rating: 5%

The following iterative sequence is defined for the set of positive integers:

nn/2 (n is even)
n → 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

### Problem 15: Lattice pathsPublished on Friday, 19th April 2002, 06:00 pm; Solved by 159615; Difficulty rating: 5%

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.

How many such routes are there through a 20×20 grid?

### Problem 16: Power digit sumPublished on Friday, 3rd May 2002, 06:00 pm; Solved by 195221; Difficulty rating: 5%

215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 21000?

### Problem 17: Number letter countsPublished on Friday, 17th May 2002, 06:00 pm; Solved by 129567; Difficulty rating: 5%

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.

### Problem 18: Maximum path sum IPublished on Friday, 31st May 2002, 06:00 pm; Solved by 124384; Difficulty rating: 5%

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

### Problem 19: Counting SundaysPublished on Friday, 14th June 2002, 06:00 pm; Solved by 115742; Difficulty rating: 5%

You are given the following information, but you may prefer to do some research for yourself.

• 1 Jan 1900 was a Monday.
• Thirty days has September,
April, June and November.
All the rest have thirty-one,
Saving February alone,
Which has twenty-eight, rain or shine.
And on leap years, twenty-nine.
• A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

### Problem 20: Factorial digit sumPublished on Friday, 21st June 2002, 06:00 pm; Solved by 170536; Difficulty rating: 5%

n! means n × (n − 1) × ... × 3 × 2 × 1

For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

### Problem 21: Amicable numbersPublished on Friday, 5th July 2002, 06:00 pm; Solved by 124908; Difficulty rating: 5%

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where ab, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

### Problem 22: Names scoresPublished on Friday, 19th July 2002, 06:00 pm; Solved by 115244; Difficulty rating: 5%

Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, COLIN would obtain a score of 938 × 53 = 49714.

What is the total of all the name scores in the file?

### Problem 23: Non-abundant sumsPublished on Friday, 2nd August 2002, 06:00 pm; Solved by 88655; Difficulty rating: 5%

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

### Problem 24: Lexicographic permutationsPublished on Friday, 16th August 2002, 06:00 pm; Solved by 98534; Difficulty rating: 5%

A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

012   021   102   120   201   210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

### Problem 25: 1000-digit Fibonacci numberPublished on Friday, 30th August 2002, 06:00 pm; Solved by 133401; Difficulty rating: 5%

The Fibonacci sequence is defined by the recurrence relation:

Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.

Hence the first 12 terms will be:

F1 = 1
F2 = 1
F3 = 2
F4 = 3
F5 = 5
F6 = 8
F7 = 13
F8 = 21
F9 = 34
F10 = 55
F11 = 89
F12 = 144

The 12th term, F12, is the first term to contain three digits.

What is the index of the first term in the Fibonacci sequence to contain 1000 digits?

### Problem 26: Reciprocal cyclesPublished on Friday, 13th September 2002, 06:00 pm; Solved by 71445; Difficulty rating: 5%

A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

 1/2 = 0.5 1/3 = 0.(3) 1/4 = 0.25 1/5 = 0.2 1/6 = 0.1(6) 1/7 = 0.(142857) 1/8 = 0.125 1/9 = 0.(1) 1/10 = 0.1

Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.

Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.

### Problem 27: Quadratic primesPublished on Friday, 27th September 2002, 06:00 pm; Solved by 74573; Difficulty rating: 5%

Euler discovered the remarkable quadratic formula:

$n^2 + n + 41$

It turns out that the formula will produce 40 primes for the consecutive integer values $0 \le n \le 39$. However, when $n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41$ is divisible by 41, and certainly when $n = 41, 41^2 + 41 + 41$ is clearly divisible by 41.

The incredible formula $n^2 - 79n + 1601$ was discovered, which produces 80 primes for the consecutive values $0 \le n \le 79$. The product of the coefficients, −79 and 1601, is −126479.

$n^2 + an + b$, where $|a| < 1000$ and $|b| \le 1000$

where $|n|$ is the modulus/absolute value of $n$
e.g. $|11| = 11$ and $|-4| = 4$

Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting with $n = 0$.

### Problem 28: Number spiral diagonalsPublished on Friday, 11th October 2002, 06:00 pm; Solved by 93681; Difficulty rating: 5%

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20  7  8  9 10
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is 101.

What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

### Problem 29: Distinct powersPublished on Friday, 25th October 2002, 06:00 pm; Solved by 89744; Difficulty rating: 5%

Consider all integer combinations of ab for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:

22=4, 23=8, 24=16, 25=32
32=9, 33=27, 34=81, 35=243
42=16, 43=64, 44=256, 45=1024
52=25, 53=125, 54=625, 55=3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

### Problem 30: Digit fifth powersPublished on Friday, 8th November 2002, 06:00 pm; Solved by 93980; Difficulty rating: 5%

Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 14 + 64 + 34 + 44
8208 = 84 + 24 + 04 + 84
9474 = 94 + 44 + 74 + 44

As 1 = 14 is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

### Problem 31: Coin sumsPublished on Friday, 22nd November 2002, 06:00 pm; Solved by 72165; Difficulty rating: 5%

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

It is possible to make £2 in the following way:

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

How many different ways can £2 be made using any number of coins?

### Problem 32: Pandigital productsPublished on Friday, 6th December 2002, 06:00 pm; Solved by 60598; Difficulty rating: 5%

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

### Problem 33: Digit cancelling fractionsPublished on Friday, 20th December 2002, 06:00 pm; Solved by 61291; Difficulty rating: 5%

The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that 49/98 = 4/8, which is correct, is obtained by cancelling the 9s.

We shall consider fractions like, 30/50 = 3/5, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

### Problem 34: Digit factorialsPublished on Friday, 3rd January 2003, 06:00 pm; Solved by 80657; Difficulty rating: 5%

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Note: as 1! = 1 and 2! = 2 are not sums they are not included.

### Problem 35: Circular primesPublished on Friday, 17th January 2003, 06:00 pm; Solved by 72742; Difficulty rating: 5%

The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.

There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.

How many circular primes are there below one million?

### Problem 36: Double-base palindromesPublished on Friday, 31st January 2003, 06:00 pm; Solved by 76574; Difficulty rating: 5%

The decimal number, 585 = 10010010012 (binary), is palindromic in both bases.

Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.

(Please note that the palindromic number, in either base, may not include leading zeros.)

### Problem 37: Truncatable primesPublished on Friday, 14th February 2003, 06:00 pm; Solved by 62953; Difficulty rating: 5%

The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.

### Problem 38: Pandigital multiplesPublished on Friday, 28th February 2003, 06:00 pm; Solved by 53658; Difficulty rating: 5%

Take the number 192 and multiply it by each of 1, 2, and 3:

192 × 1 = 192
192 × 2 = 384
192 × 3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?

### Problem 39: Integer right trianglesPublished on Friday, 14th March 2003, 06:00 pm; Solved by 62448; Difficulty rating: 5%

If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p ≤ 1000, is the number of solutions maximised?

### Problem 40: Champernowne's constantPublished on Friday, 28th March 2003, 06:00 pm; Solved by 68677; Difficulty rating: 5%

An irrational decimal fraction is created by concatenating the positive integers:

0.123456789101112131415161718192021...

It can be seen that the 12th digit of the fractional part is 1.

If dn represents the nth digit of the fractional part, find the value of the following expression.

d1 × d10 × d100 × d1000 × d10000 × d100000 × d1000000

### Problem 41: Pandigital primePublished on Friday, 11th April 2003, 06:00 pm; Solved by 58144; Difficulty rating: 5%

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital and is also prime.

What is the largest n-digit pandigital prime that exists?

### Problem 42: Coded triangle numbersPublished on Friday, 25th April 2003, 06:00 pm; Solved by 64037; Difficulty rating: 5%

The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangle numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a triangle number then we shall call the word a triangle word.

Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?

### Problem 43: Sub-string divisibilityPublished on Friday, 9th May 2003, 06:00 pm; Solved by 50743; Difficulty rating: 5%

The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.

Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:

• d2d3d4=406 is divisible by 2
• d3d4d5=063 is divisible by 3
• d4d5d6=635 is divisible by 5
• d5d6d7=357 is divisible by 7
• d6d7d8=572 is divisible by 11
• d7d8d9=728 is divisible by 13
• d8d9d10=289 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.

### Problem 44: Pentagon numbersPublished on Friday, 23rd May 2003, 06:00 pm; Solved by 49317; Difficulty rating: 5%

Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?

### Problem 45: Triangular, pentagonal, and hexagonalPublished on Friday, 6th June 2003, 06:00 pm; Solved by 61013; Difficulty rating: 5%

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

 Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal Pn=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal Hn=n(2n−1) 1, 6, 15, 28, 45, ...

It can be verified that T285 = P165 = H143 = 40755.

Find the next triangle number that is also pentagonal and hexagonal.

### Problem 46: Goldbach's other conjecturePublished on Friday, 20th June 2003, 06:00 pm; Solved by 52163; Difficulty rating: 5%

It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.

9 = 7 + 2×12
15 = 7 + 2×22
21 = 3 + 2×32
25 = 7 + 2×32
27 = 19 + 2×22
33 = 31 + 2×12

It turns out that the conjecture was false.

What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?

### Problem 47: Distinct primes factorsPublished on Friday, 4th July 2003, 06:00 pm; Solved by 49208; Difficulty rating: 5%

The first two consecutive numbers to have two distinct prime factors are:

14 = 2 × 7
15 = 3 × 5

The first three consecutive numbers to have three distinct prime factors are:

644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19.

Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?

### Problem 48: Self powersPublished on Friday, 18th July 2003, 06:00 pm; Solved by 97632; Difficulty rating: 5%

The series, 11 + 22 + 33 + ... + 1010 = 10405071317.

Find the last ten digits of the series, 11 + 22 + 33 + ... + 10001000.

### Problem 49: Prime permutationsPublished on Friday, 1st August 2003, 06:00 pm; Solved by 49259; Difficulty rating: 5%

The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit increasing sequence.

What 12-digit number do you form by concatenating the three terms in this sequence?

### Problem 50: Consecutive prime sumPublished on Friday, 15th August 2003, 06:00 pm; Solved by 53019; Difficulty rating: 5%

The prime 41, can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

### Problem 51: Prime digit replacementsPublished on Friday, 29th August 2003, 06:00 pm; Solved by 27922; Difficulty rating: 15%

By replacing the 1st digit of the 2-digit number *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.

By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.

Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.

### Problem 52: Permuted multiplesPublished on Friday, 12th September 2003, 06:00 pm; Solved by 56017; Difficulty rating: 5%

It can be seen that the number, 125874, and its double, 251748, contain exactly the same digits, but in a different order.

Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits.

### Problem 53: Combinatoric selectionsPublished on Friday, 26th September 2003, 06:00 pm; Solved by 50478; Difficulty rating: 5%

There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, 5C3 = 10.

In general,

 nCr = n!r!(n−r)! ,where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1.

It is not until n = 23, that a value exceeds one-million: 23C10 = 1144066.

How many, not necessarily distinct, values of  nCr, for 1 ≤ n ≤ 100, are greater than one-million?

### Problem 54: Poker handsPublished on Friday, 10th October 2003, 06:00 pm; Solved by 29941; Difficulty rating: 10%

In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:

• High Card: Highest value card.
• One Pair: Two cards of the same value.
• Two Pairs: Two different pairs.
• Three of a Kind: Three cards of the same value.
• Straight: All cards are consecutive values.
• Flush: All cards of the same suit.
• Full House: Three of a kind and a pair.
• Four of a Kind: Four cards of the same value.
• Straight Flush: All cards are consecutive values of same suit.
• Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.

The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.

If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.

Consider the following five hands dealt to two players:

 Hand Player 1 Player 2 Winner 1 5H 5C 6S 7S KDPair of Fives 2C 3S 8S 8D TDPair of Eights Player 2 2 5D 8C 9S JS ACHighest card Ace 2C 5C 7D 8S QHHighest card Queen Player 1 3 2D 9C AS AH ACThree Aces 3D 6D 7D TD QDFlush with Diamonds Player 2 4 4D 6S 9H QH QCPair of QueensHighest card Nine 3D 6D 7H QD QSPair of QueensHighest card Seven Player 1 5 2H 2D 4C 4D 4SFull HouseWith Three Fours 3C 3D 3S 9S 9DFull Housewith Three Threes Player 1

The file, poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order, and in each hand there is a clear winner.

How many hands does Player 1 win?

### Problem 55: Lychrel numbersPublished on Friday, 24th October 2003, 06:00 pm; Solved by 45920; Difficulty rating: 5%

If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337

That is, 349 took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.

How many Lychrel numbers are there below ten-thousand?

NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.

### Problem 56: Powerful digit sumPublished on Friday, 7th November 2003, 06:00 pm; Solved by 49769; Difficulty rating: 5%

A googol (10100) is a massive number: one followed by one-hundred zeros; 100100 is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.

Considering natural numbers of the form, ab, where a, b < 100, what is the maximum digital sum?

### Problem 57: Square root convergentsPublished on Friday, 21st November 2003, 06:00 pm; Solved by 35004; Difficulty rating: 5%

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

### Problem 58: Spiral primesPublished on Friday, 5th December 2003, 06:00 pm; Solved by 33873; Difficulty rating: 5%

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18  5  4  3 12 29
40 19  6  1  2 11 28
41 20  7  8  9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?

### Problem 59: XOR decryptionPublished on Friday, 19th December 2003, 06:00 pm; Solved by 35405; Difficulty rating: 5%

Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.

A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.

For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both "halves", it is impossible to decrypt the message.

Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.

Your task has been made easy, as the encryption key consists of three lower case characters. Using cipher.txt (right click and 'Save Link/Target As...'), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.

### Problem 60: Prime pair setsPublished on Friday, 2nd January 2004, 06:00 pm; Solved by 21857; Difficulty rating: 20%

The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.

Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.

### Problem 61: Cyclical figurate numbersPublished on Friday, 16th January 2004, 06:00 pm; Solved by 21173; Difficulty rating: 20%

Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

 Triangle P3,n=n(n+1)/2 1, 3, 6, 10, 15, ... Square P4,n=n2 1, 4, 9, 16, 25, ... Pentagonal P5,n=n(3n−1)/2 1, 5, 12, 22, 35, ... Hexagonal P6,n=n(2n−1) 1, 6, 15, 28, 45, ... Heptagonal P7,n=n(5n−3)/2 1, 7, 18, 34, 55, ... Octagonal P8,n=n(3n−2) 1, 8, 21, 40, 65, ...

The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

1. The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
2. Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set.
3. This is the only set of 4-digit numbers with this property.

Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

### Problem 62: Cubic permutationsPublished on Friday, 30th January 2004, 06:00 pm; Solved by 26154; Difficulty rating: 15%

The cube, 41063625 (3453), can be permuted to produce two other cubes: 56623104 (3843) and 66430125 (4053). In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.

Find the smallest cube for which exactly five permutations of its digits are cube.

### Problem 63: Powerful digit countsPublished on Friday, 13th February 2004, 06:00 pm; Solved by 36546; Difficulty rating: 5%

The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is a ninth power.

How many n-digit positive integers exist which are also an nth power?

### Problem 64: Odd period square rootsPublished on Friday, 27th February 2004, 06:00 pm; Solved by 18375; Difficulty rating: 20%

All square roots are periodic when written as continued fractions and can be written in the form:

 √N = a0 + 1 a1 + 1 a2 + 1 a3 + ...

For example, let us consider √23:

 √23 = 4 + √23 — 4 = 4 + 1 = 4 + 1 1√23—4 1 + √23 – 37

If we continue we would get the following expansion:

 √23 = 4 + 1 1 + 1 3 + 1 1 + 1 8 + ...

The process can be summarised as follows:

 a0 = 4, 1√23—4 = √23+47 = 1 + √23—37 a1 = 1, 7√23—3 = 7(√23+3)14 = 3 + √23—32 a2 = 3, 2√23—3 = 2(√23+3)14 = 1 + √23—47 a3 = 1, 7√23—4 = 7(√23+4)7 = 8 + √23—4 a4 = 8, 1√23—4 = √23+47 = 1 + √23—37 a5 = 1, 7√23—3 = 7(√23+3)14 = 3 + √23—32 a6 = 3, 2√23—3 = 2(√23+3)14 = 1 + √23—47 a7 = 1, 7√23—4 = 7(√23+4)7 = 8 + √23—4

It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:

√2=[1;(2)], period=1
√3=[1;(1,2)], period=2
√5=[2;(4)], period=1
√6=[2;(2,4)], period=2
√7=[2;(1,1,1,4)], period=4
√8=[2;(1,4)], period=2
√10=[3;(6)], period=1
√11=[3;(3,6)], period=2
√12= [3;(2,6)], period=2
√13=[3;(1,1,1,1,6)], period=5

Exactly four continued fractions, for N ≤ 13, have an odd period.

How many continued fractions for N ≤ 10000 have an odd period?

### Problem 65: Convergents of ePublished on Friday, 12th March 2004, 06:00 pm; Solved by 25270; Difficulty rating: 15%

The square root of 2 can be written as an infinite continued fraction.

 √2 = 1 + 1 2 + 1 2 + 1 2 + 1 2 + ...

The infinite continued fraction can be written, √2 = [1;(2)], (2) indicates that 2 repeats ad infinitum. In a similar way, √23 = [4;(1,3,1,8)].

It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for √2.

 1 + 1 = 3/2 2
 1 + 1 = 7/5 2 + 1 2
 1 + 1 = 17/12 2 + 1 2 + 1 2
 1 + 1 = 41/29 2 + 1 2 + 1 2 + 1 2

Hence the sequence of the first ten convergents for √2 are:

1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...

What is most surprising is that the important mathematical constant,
e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...].

The first ten terms in the sequence of convergents for e are:

2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...

The sum of digits in the numerator of the 10th convergent is 1+4+5+7=17.

Find the sum of digits in the numerator of the 100th convergent of the continued fraction for e.

### Problem 66: Diophantine equationPublished on Friday, 26th March 2004, 06:00 pm; Solved by 16329; Difficulty rating: 25%

Consider quadratic Diophantine equations of the form:

x2 – Dy2 = 1

For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.

It can be assumed that there are no solutions in positive integers when D is square.

By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:

32 – 2×22 = 1
22 – 3×12 = 1
92 – 5×42 = 1
52 – 6×22 = 1
82 – 7×32 = 1

Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.

Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.

### Problem 67: Maximum path sum IIPublished on Friday, 9th April 2004, 06:00 pm; Solved by 83395; Difficulty rating: 5%

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.

NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 299 altogether! If you could check one trillion (1012) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)

### Problem 68: Magic 5-gon ringPublished on Friday, 23rd April 2004, 06:00 pm; Solved by 17282; Difficulty rating: 25%

Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.

Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example), each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.

It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.

 Total Solution Set 9 4,2,3; 5,3,1; 6,1,2 9 4,3,2; 6,2,1; 5,1,3 10 2,3,5; 4,5,1; 6,1,3 10 2,5,3; 6,3,1; 4,1,5 11 1,4,6; 3,6,2; 5,2,4 11 1,6,4; 5,4,2; 3,2,6 12 1,5,6; 2,6,4; 3,4,5 12 1,6,5; 3,5,4; 2,4,6

By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.

Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string for a "magic" 5-gon ring?

### Problem 69: Totient maximumPublished on Friday, 7th May 2004, 06:00 pm; Solved by 28937; Difficulty rating: 10%

Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.

 n Relatively Prime φ(n) n/φ(n) 2 1 1 2 3 1,2 2 1.5 4 1,3 2 2 5 1,2,3,4 4 1.25 6 1,5 2 3 7 1,2,3,4,5,6 6 1.1666... 8 1,3,5,7 4 2 9 1,2,4,5,7,8 6 1.5 10 1,3,7,9 4 2.5

It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.

Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.

### Problem 70: Totient permutationPublished on Friday, 21st May 2004, 06:00 pm; Solved by 18646; Difficulty rating: 20%

Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to n which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
The number 1 is considered to be relatively prime to every positive number, so φ(1)=1.

Interestingly, φ(87109)=79180, and it can be seen that 87109 is a permutation of 79180.

Find the value of n, 1 < n < 107, for which φ(n) is a permutation of n and the ratio n/φ(n) produces a minimum.

### Problem 71: Ordered fractionsPublished on Friday, 4th June 2004, 06:00 pm; Solved by 24482; Difficulty rating: 10%

Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that 2/5 is the fraction immediately to the left of 3/7.

By listing the set of reduced proper fractions for d ≤ 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.

### Problem 72: Counting fractionsPublished on Friday, 18th June 2004, 06:00 pm; Solved by 18653; Difficulty rating: 20%

Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that there are 21 elements in this set.

How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?

### Problem 73: Counting fractions in a rangePublished on Friday, 2nd July 2004, 06:00 pm; Solved by 21192; Difficulty rating: 15%

Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that there are 3 fractions between 1/3 and 1/2.

How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?

### Problem 74: Digit factorial chainsPublished on Friday, 16th July 2004, 06:00 pm; Solved by 22450; Difficulty rating: 15%

The number 145 is well known for the property that the sum of the factorial of its digits is equal to 145:

1! + 4! + 5! = 1 + 24 + 120 = 145

Perhaps less well known is 169, in that it produces the longest chain of numbers that link back to 169; it turns out that there are only three such loops that exist:

169 → 363601 → 1454 → 169
871 → 45361 → 871
872 → 45362 → 872

It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

69 → 363600 → 1454 → 169 → 363601 (→ 1454)
78 → 45360 → 871 → 45361 (→ 871)
540 → 145 (→ 145)

Starting with 69 produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.

How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?

### Problem 75: Singular integer right trianglesPublished on Friday, 30th July 2004, 06:00 pm; Solved by 14913; Difficulty rating: 25%

It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.

12 cm: (3,4,5)
24 cm: (6,8,10)
30 cm: (5,12,13)
36 cm: (9,12,15)
40 cm: (8,15,17)
48 cm: (12,16,20)

In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.

120 cm: (30,40,50), (20,48,52), (24,45,51)

Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?

### Problem 76: Counting summationsPublished on Friday, 13th August 2004, 06:00 pm; Solved by 24081; Difficulty rating: 10%

It is possible to write five as a sum in exactly six different ways:

4 + 1
3 + 2
3 + 1 + 1
2 + 2 + 1
2 + 1 + 1 + 1
1 + 1 + 1 + 1 + 1

How many different ways can one hundred be written as a sum of at least two positive integers?

### Problem 77: Prime summationsPublished on Friday, 27th August 2004, 06:00 pm; Solved by 15864; Difficulty rating: 25%

It is possible to write ten as the sum of primes in exactly five different ways:

7 + 3
5 + 5
5 + 3 + 2
3 + 3 + 2 + 2
2 + 2 + 2 + 2 + 2

What is the first value which can be written as the sum of primes in over five thousand different ways?

### Problem 78: Coin partitionsPublished on Friday, 10th September 2004, 06:00 pm; Solved by 13971; Difficulty rating: 30%

Let p(n) represent the number of different ways in which n coins can be separated into piles. For example, five coins can be separated into piles in exactly seven different ways, so p(5)=7.

 OOOOO OOOO   O OOO   OO OOO   O   O OO   OO   O OO   O   O   O O   O   O   O   O

Find the least value of n for which p(n) is divisible by one million.

### Problem 79: Passcode derivationPublished on Friday, 17th September 2004, 06:00 pm; Solved by 36052; Difficulty rating: 5%

A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.

The text file, keylog.txt, contains fifty successful login attempts.

Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible secret passcode of unknown length.

### Problem 80: Square root digital expansionPublished on Friday, 8th October 2004, 06:00 pm; Solved by 16679; Difficulty rating: 20%

It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.

The square root of two is 1.41421356237309504880..., and the digital sum of the first one hundred decimal digits is 475.

For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.

### Problem 81: Path sum: two waysPublished on Friday, 22nd October 2004, 06:00 pm; Solved by 29375; Difficulty rating: 10%

In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down, is indicated in bold red and is equal to 2427.

$$\begin{pmatrix} \color{red}{131} & 673 & 234 & 103 & 18\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\ 630 & 803 & \color{red}{746} & \color{red}{422} & 111\\ 537 & 699 & 497 & \color{red}{121} & 956\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$

Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by only moving right and down.

### Problem 82: Path sum: three waysPublished on Friday, 5th November 2004, 06:00 pm; Solved by 17810; Difficulty rating: 20%

NOTE: This problem is a more challenging version of Problem 81.

The minimal path sum in the 5 by 5 matrix below, by starting in any cell in the left column and finishing in any cell in the right column, and only moving up, down, and right, is indicated in red and bold; the sum is equal to 994.

$$\begin{pmatrix} 131 & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\ 630 & 803 & 746 & 422 & 111\\ 537 & 699 & 497 & 121 & 956\\ 805 & 732 & 524 & 37 & 331 \end{pmatrix}$$

Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing a 80 by 80 matrix, from the left column to the right column.

### Problem 83: Path sum: four waysPublished on Friday, 19th November 2004, 06:00 pm; Solved by 15232; Difficulty rating: 25%

NOTE: This problem is a significantly more challenging version of Problem 81.

In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right, by moving left, right, up, and down, is indicated in bold red and is equal to 2297.

$$\begin{pmatrix} \color{red}{131} & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\ \color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & \color{red}{150}\\ 630 & 803 & 746 & \color{red}{422} & \color{red}{111}\\ 537 & 699 & 497 & \color{red}{121} & 956\\ 805 & 732 & 524 & \color{red}{37} & \color{red}{331} \end{pmatrix}$$

Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing a 80 by 80 matrix, from the top left to the bottom right by moving left, right, up, and down.

### Problem 84: Monopoly oddsPublished on Friday, 3rd December 2004, 06:00 pm; Solved by 10390; Difficulty rating: 35%

In the game, Monopoly, the standard board is set up in the following way:

 GO A1 CC1 A2 T1 R1 B1 CH1 B2 B3 JAIL H2 C1 T2 U1 H1 C2 CH3 C3 R4 R2 G3 D1 CC3 CC2 G2 D2 G1 D3 G2J F3 U2 F2 F1 R3 E3 E2 CH2 E1 FP

A player starts on the GO square and adds the scores on two 6-sided dice to determine the number of squares they advance in a clockwise direction. Without any further rules we would expect to visit each square with equal probability: 2.5%. However, landing on G2J (Go To Jail), CC (community chest), and CH (chance) changes this distribution.

In addition to G2J, and one card from each of CC and CH, that orders the player to go directly to jail, if a player rolls three consecutive doubles, they do not advance the result of their 3rd roll. Instead they proceed directly to jail.

At the beginning of the game, the CC and CH cards are shuffled. When a player lands on CC or CH they take a card from the top of the respective pile and, after following the instructions, it is returned to the bottom of the pile. There are sixteen cards in each pile, but for the purpose of this problem we are only concerned with cards that order a movement; any instruction not concerned with movement will be ignored and the player will remain on the CC/CH square.

• Community Chest (2/16 cards):
2. Go to JAIL
• Chance (10/16 cards):
2. Go to JAIL
3. Go to C1
4. Go to E3
5. Go to H2
6. Go to R1
7. Go to next R (railway company)
8. Go to next R
9. Go to next U (utility company)
10. Go back 3 squares.

The heart of this problem concerns the likelihood of visiting a particular square. That is, the probability of finishing at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability of finishing on it is zero, the CH squares will have the lowest probabilities, as 5/8 request a movement to another square, and it is the final square that the player finishes at on each roll that we are interested in. We shall make no distinction between "Just Visiting" and being sent to JAIL, and we shall also ignore the rule about requiring a double to "get out of jail", assuming that they pay to get out on their next turn.

By starting at GO and numbering the squares sequentially from 00 to 39 we can concatenate these two-digit numbers to produce strings that correspond with sets of squares.

Statistically it can be shown that the three most popular squares, in order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO (3.09%) = Square 00. So these three most popular squares can be listed with the six-digit modal string: 102400.

If, instead of using two 6-sided dice, two 4-sided dice are used, find the six-digit modal string.

### Problem 85: Counting rectanglesPublished on Friday, 17th December 2004, 06:00 pm; Solved by 20878; Difficulty rating: 15%

By counting carefully it can be seen that a rectangular grid measuring 3 by 2 contains eighteen rectangles:

Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the nearest solution.

### Problem 86: Cuboid routePublished on Friday, 7th January 2005, 06:00 pm; Solved by 10496; Difficulty rating: 35%

A spider, S, sits in one corner of a cuboid room, measuring 6 by 5 by 3, and a fly, F, sits in the opposite corner. By travelling on the surfaces of the room the shortest "straight line" distance from S to F is 10 and the path is shown on the diagram.

However, there are up to three "shortest" path candidates for any given cuboid and the shortest route doesn't always have integer length.

It can be shown that there are exactly 2060 distinct cuboids, ignoring rotations, with integer dimensions, up to a maximum size of M by M by M, for which the shortest route has integer length when M = 100. This is the least value of M for which the number of solutions first exceeds two thousand; the number of solutions when M = 99 is 1975.

Find the least value of M such that the number of solutions first exceeds one million.

### Problem 87: Prime power triplesPublished on Friday, 21st January 2005, 06:00 pm; Solved by 17491; Difficulty rating: 20%

The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way:

28 = 22 + 23 + 24
33 = 32 + 23 + 24
49 = 52 + 23 + 24
47 = 22 + 33 + 24

How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?

### Problem 88: Product-sum numbersPublished on Friday, 4th February 2005, 06:00 pm; Solved by 8140; Difficulty rating: 40%

A natural number, N, that can be written as the sum and product of a given set of at least two natural numbers, {a1, a2, ... , ak} is called a product-sum number: N = a1 + a2 + ... + ak = a1 × a2 × ... × ak.

For example, 6 = 1 + 2 + 3 = 1 × 2 × 3.

For a given set of size, k, we shall call the smallest N with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, k = 2, 3, 4, 5, and 6 are as follows.

k=2: 4 = 2 × 2 = 2 + 2
k=3: 6 = 1 × 2 × 3 = 1 + 2 + 3
k=4: 8 = 1 × 1 × 2 × 4 = 1 + 1 + 2 + 4
k=5: 8 = 1 × 1 × 2 × 2 × 2 = 1 + 1 + 2 + 2 + 2
k=6: 12 = 1 × 1 × 1 × 1 × 2 × 6 = 1 + 1 + 1 + 1 + 2 + 6

Hence for 2≤k≤6, the sum of all the minimal product-sum numbers is 4+6+8+12 = 30; note that 8 is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for 2≤k≤12 is {4, 6, 8, 12, 15, 16}, the sum is 61.

What is the sum of all the minimal product-sum numbers for 2≤k≤12000?

### Problem 89: Roman numeralsPublished on Friday, 18th February 2005, 06:00 pm; Solved by 18182; Difficulty rating: 20%

For a number written in Roman numerals to be considered valid there are basic rules which must be followed. Even though the rules allow some numbers to be expressed in more than one way there is always a "best" way of writing a particular number.

For example, it would appear that there are at least six ways of writing the number sixteen:

IIIIIIIIIIIIIIII
VIIIIIIIIIII
VVIIIIII
XIIIIII
VVVI
XVI

However, according to the rules only XIIIIII and XVI are valid, and the last example is considered to be the most efficient, as it uses the least number of numerals.

The 11K text file, roman.txt (right click and 'Save Link/Target As...'), contains one thousand numbers written in valid, but not necessarily minimal, Roman numerals; see About... Roman Numerals for the definitive rules for this problem.

Find the number of characters saved by writing each of these in their minimal form.

Note: You can assume that all the Roman numerals in the file contain no more than four consecutive identical units.

### Problem 90: Cube digit pairsPublished on Friday, 4th March 2005, 06:00 pm; Solved by 9153; Difficulty rating: 40%

Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of 2-digit numbers.

For example, the square number 64 could be formed:

In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: 01, 04, 09, 16, 25, 36, 49, 64, and 81.

For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube.

However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7} allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09.

In determining a distinct arrangement we are interested in the digits on each cube, not the order.

{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}

But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit numbers.

How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?

### Problem 91: Right triangles with integer coordinatesPublished on Friday, 18th March 2005, 06:00 pm; Solved by 12847; Difficulty rating: 25%

The points P (x1, y1) and Q (x2, y2) are plotted at integer co-ordinates and are joined to the origin, O(0,0), to form ΔOPQ.

There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between 0 and 2 inclusive; that is,
0 ≤ x1, y1, x2, y2 ≤ 2.

Given that 0 ≤ x1, y1, x2, y2 ≤ 50, how many right triangles can be formed?

### Problem 92: Square digit chainsPublished on Friday, 1st April 2005, 06:00 pm; Solved by 35795; Difficulty rating: 5%

A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.

For example,

44 → 32 → 13 → 10 → 11
85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89

Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.

How many starting numbers below ten million will arrive at 89?

### Problem 93: Arithmetic expressionsPublished on Friday, 15th April 2005, 06:00 pm; Solved by 9631; Difficulty rating: 35%

By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four arithmetic operations (+, −, *, /) and brackets/parentheses, it is possible to form different positive integer targets.

For example,

8 = (4 * (1 + 3)) / 2
14 = 4 * (3 + 1 / 2)
19 = 4 * (2 + 3) − 1
36 = 3 * 4 * (2 + 1)

Note that concatenations of the digits, like 12 + 34, are not allowed.

Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different target numbers of which 36 is the maximum, and each of the numbers 1 to 28 can be obtained before encountering the first non-expressible number.

Find the set of four distinct digits, a < b < c < d, for which the longest set of consecutive positive integers, 1 to n, can be obtained, giving your answer as a string: abcd.

### Problem 94: Almost equilateral trianglesPublished on Friday, 29th April 2005, 06:00 pm; Solved by 10052; Difficulty rating: 35%

It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the almost equilateral triangle 5-5-6 has an area of 12 square units.

We shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.

Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed one billion (1,000,000,000).

### Problem 95: Amicable chainsPublished on Friday, 13th May 2005, 06:00 pm; Solved by 11741; Difficulty rating: 30%

The proper divisors of a number are all the divisors excluding the number itself. For example, the proper divisors of 28 are 1, 2, 4, 7, and 14. As the sum of these divisors is equal to 28, we call it a perfect number.

Interestingly the sum of the proper divisors of 220 is 284 and the sum of the proper divisors of 284 is 220, forming a chain of two numbers. For this reason, 220 and 284 are called an amicable pair.

Perhaps less well known are longer chains. For example, starting with 12496, we form a chain of five numbers:

12496 → 14288 → 15472 → 14536 → 14264 (→ 12496 → ...)

Since this chain returns to its starting point, it is called an amicable chain.

Find the smallest member of the longest amicable chain with no element exceeding one million.

### Problem 96: Su DokuPublished on Friday, 27th May 2005, 06:00 pm; Solved by 14123; Difficulty rating: 25%

Su Doku (Japanese meaning number place) is the name given to a popular puzzle concept. Its origin is unclear, but credit must be attributed to Leonhard Euler who invented a similar, and much more difficult, puzzle idea called Latin Squares. The objective of Su Doku puzzles, however, is to replace the blanks (or zeros) in a 9 by 9 grid in such that each row, column, and 3 by 3 box contains each of the digits 1 to 9. Below is an example of a typical starting puzzle grid and its solution grid.

 0 0 39 0 00 0 1 0 2 03 0 58 0 6 6 0 00 0 14 0 0 0 0 87 0 00 0 6 1 0 20 0 07 0 8 9 0 00 0 82 0 0 0 0 28 0 00 0 5 6 0 92 0 30 1 0 5 0 00 0 93 0 0

 4 8 39 6 72 5 1 9 2 13 4 58 7 6 6 5 78 2 14 9 3 5 4 87 2 91 3 6 1 3 25 6 47 9 8 9 7 61 3 82 4 5 3 7 28 1 46 9 5 6 8 92 5 34 1 7 5 1 47 6 93 8 2

A well constructed Su Doku puzzle has a unique solution and can be solved by logic, although it may be necessary to employ "guess and test" methods in order to eliminate options (there is much contested opinion over this). The complexity of the search determines the difficulty of the puzzle; the example above is considered easy because it can be solved by straight forward direct deduction.

The 6K text file, sudoku.txt (right click and 'Save Link/Target As...'), contains fifty different Su Doku puzzles ranging in difficulty, but all with unique solutions (the first puzzle in the file is the example above).

By solving all fifty puzzles find the sum of the 3-digit numbers found in the top left corner of each solution grid; for example, 483 is the 3-digit number found in the top left corner of the solution grid above.

### Problem 97: Large non-Mersenne primePublished on Friday, 10th June 2005, 06:00 pm; Solved by 37935; Difficulty rating: 5%

The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form 26972593−1; it contains exactly 2,098,960 digits. Subsequently other Mersenne primes, of the form 2p−1, have been found which contain more digits.

However, in 2004 there was found a massive non-Mersenne prime which contains 2,357,207 digits: 28433×27830457+1.

Find the last ten digits of this prime number.

### Problem 98: Anagramic squaresPublished on Friday, 17th June 2005, 06:00 pm; Solved by 9280; Difficulty rating: 35%

By replacing each of the letters in the word CARE with 1, 2, 9, and 6 respectively, we form a square number: 1296 = 362. What is remarkable is that, by using the same digital substitutions, the anagram, RACE, also forms a square number: 9216 = 962. We shall call CARE (and RACE) a square anagram word pair and specify further that leading zeroes are not permitted, neither may a different letter have the same digital value as another letter.

Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, find all the square anagram word pairs (a palindromic word is NOT considered to be an anagram of itself).

What is the largest square number formed by any member of such a pair?

NOTE: All anagrams formed must be contained in the given text file.

### Problem 99: Largest exponentialPublished on Friday, 1st July 2005, 06:00 pm; Solved by 25901; Difficulty rating: 10%

Comparing two numbers written in index form like 211 and 37 is not difficult, as any calculator would confirm that 211 = 2048 < 37 = 2187.

However, confirming that 632382518061 > 519432525806 would be much more difficult, as both numbers contain over three million digits.

Using base_exp.txt (right click and 'Save Link/Target As...'), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.

NOTE: The first two lines in the file represent the numbers in the example given above.

### Problem 100: Arranged probabilityPublished on Friday, 15th July 2005, 06:00 pm; Solved by 13393; Difficulty rating: 30%

If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, P(BB) = (15/21)×(14/20) = 1/2.

The next such arrangement, for which there is exactly 50% chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs.

By finding the first arrangement to contain over 1012 = 1,000,000,000,000 discs in total, determine the number of blue discs that the box would contain.

### Problem 101: Optimum polynomialPublished on Friday, 29th July 2005, 06:00 pm; Solved by 9258; Difficulty rating: 35%

If we are presented with the first k terms of a sequence it is impossible to say with certainty the value of the next term, as there are infinitely many polynomial functions that can model the sequence.

As an example, let us consider the sequence of cube numbers. This is defined by the generating function,
un = n3: 1, 8, 27, 64, 125, 216, ...

Suppose we were only given the first two terms of this sequence. Working on the principle that "simple is best" we should assume a linear relationship and predict the next term to be 15 (common difference 7). Even if we were presented with the first three terms, by the same principle of simplicity, a quadratic relationship should be assumed.

We shall define OP(k, n) to be the nth term of the optimum polynomial generating function for the first k terms of a sequence. It should be clear that OP(k, n) will accurately generate the terms of the sequence for nk, and potentially the first incorrect term (FIT) will be OP(k, k+1); in which case we shall call it a bad OP (BOP).

As a basis, if we were only given the first term of sequence, it would be most sensible to assume constancy; that is, for n ≥ 2, OP(1, n) = u1.

Hence we obtain the following OPs for the cubic sequence:

 OP(1, n) = 1 1, 1, 1, 1, ... OP(2, n) = 7n−6 1, 8, 15, ... OP(3, n) = 6n2−11n+6 1, 8, 27, 58, ... OP(4, n) = n3 1, 8, 27, 64, 125, ...

Clearly no BOPs exist for k ≥ 4.

By considering the sum of FITs generated by the BOPs (indicated in red above), we obtain 1 + 15 + 58 = 74.

Consider the following tenth degree polynomial generating function:

un = 1 − n + n2n3 + n4n5 + n6n7 + n8n9 + n10

Find the sum of FITs for the BOPs.

### Problem 102: Triangle containmentPublished on Friday, 12th August 2005, 06:00 pm; Solved by 18404; Difficulty rating: 15%

Three distinct points are plotted at random on a Cartesian plane, for which -1000 ≤ x, y ≤ 1000, such that a triangle is formed.

Consider the following two triangles:

A(-340,495), B(-153,-910), C(835,-947)

X(-175,41), Y(-421,-714), Z(574,-645)

It can be verified that triangle ABC contains the origin, whereas triangle XYZ does not.

Using triangles.txt (right click and 'Save Link/Target As...'), a 27K text file containing the co-ordinates of one thousand "random" triangles, find the number of triangles for which the interior contains the origin.

NOTE: The first two examples in the file represent the triangles in the example given above.

### Problem 103: Special subset sums: optimumPublished on Friday, 26th August 2005, 06:00 pm; Solved by 6506; Difficulty rating: 45%

Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:

1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) > S(C).

If S(A) is minimised for a given n, we shall call it an optimum special sum set. The first five optimum special sum sets are given below.

n = 1: {1}
n = 2: {1, 2}
n = 3: {2, 3, 4}
n = 4: {3, 5, 6, 7}
n = 5: {6, 9, 11, 12, 13}

It seems that for a given optimum set, A = {a1, a2, ... , an}, the next optimum set is of the form B = {b, a1+b, a2+b, ... ,an+b}, where b is the "middle" element on the previous row.

By applying this "rule" we would expect the optimum set for n = 6 to be A = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the optimum set, as we have merely applied an algorithm to provide a near optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25}, with S(A) = 115 and corresponding set string: 111819202225.

Given that A is an optimum special sum set for n = 7, find its set string.

NOTE: This problem is related to Problem 105 and Problem 106.

### Problem 104: Pandigital Fibonacci endsPublished on Friday, 9th September 2005, 06:00 pm; Solved by 13697; Difficulty rating: 25%

The Fibonacci sequence is defined by the recurrence relation:

Fn = Fn−1 + Fn−2, where F1 = 1 and F2 = 1.

It turns out that F541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9, but not necessarily in order). And F2749, which contains 575 digits, is the first Fibonacci number for which the first nine digits are 1-9 pandigital.

Given that Fk is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.

### Problem 105: Special subset sums: testingPublished on Friday, 23rd September 2005, 06:00 pm; Solved by 6516; Difficulty rating: 45%

Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:

1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) > S(C).

For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159, 161, 139, 158} satisfies both rules for all possible subset pair combinations and S(A) = 1286.

Using sets.txt (right click and "Save Link/Target As..."), a 4K text file with one-hundred sets containing seven to twelve elements (the two examples given above are the first two sets in the file), identify all the special sum sets, A1, A2, ..., Ak, and find the value of S(A1) + S(A2) + ... + S(Ak).

NOTE: This problem is related to Problem 103 and Problem 106.

### Problem 106: Special subset sums: meta-testingPublished on Friday, 7th October 2005, 06:00 pm; Solved by 5199; Difficulty rating: 50%

Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:

1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
2. If B contains more elements than C then S(B) > S(C).

For this problem we shall assume that a given set contains n strictly increasing elements and it already satisfies the second rule.

Surprisingly, out of the 25 possible subset pairs that can be obtained from a set for which n = 4, only 1 of these pairs need to be tested for equality (first rule). Similarly, when n = 7, only 70 out of the 966 subset pairs need to be tested.

For n = 12, how many of the 261625 subset pairs that can be obtained need to be tested for equality?

NOTE: This problem is related to Problem 103 and Problem 105.

### Problem 107: Minimal networkPublished on Friday, 21st October 2005, 06:00 pm; Solved by 9184; Difficulty rating: 35%

The following undirected network consists of seven vertices and twelve edges with a total weight of 243.

The same network can be represented by the matrix below.

 A B C D E F G A - 16 12 21 - - - B 16 - - 17 20 - - C 12 - - 28 - 31 - D 21 17 28 - 18 19 23 E - 20 - 18 - - 11 F - - 31 19 - - 27 G - - - 23 11 27 -

However, it is possible to optimise the network by removing some edges and still ensure that all points on the network remain connected. The network which achieves the maximum saving is shown below. It has a weight of 93, representing a saving of 243 − 93 = 150 from the original network.

Using network.txt (right click and 'Save Link/Target As...'), a 6K text file containing a network with forty vertices, and given in matrix form, find the maximum saving which can be achieved by removing redundant edges whilst ensuring that the network remains connected.

### Problem 108: Diophantine reciprocals IPublished on Friday, 4th November 2005, 06:00 pm; Solved by 10695; Difficulty rating: 30%

In the following equation x, y, and n are positive integers.

 1x + 1y = 1n

For n = 4 there are exactly three distinct solutions:

 15 + 120 = 14 16 + 112 = 14 18 + 18 = 14

What is the least value of n for which the number of distinct solutions exceeds one-thousand?

NOTE: This problem is an easier version of Problem 110; it is strongly advised that you solve this one first.

### Problem 109: DartsPublished on Friday, 18th November 2005, 06:00 pm; Solved by 6686; Difficulty rating: 45%

In the game of darts a player throws three darts at a target board which is split into twenty equal sized sections numbered one to twenty.

The score of a dart is determined by the number of the region that the dart lands in. A dart landing outside the red/green outer ring scores zero. The black and cream regions inside this ring represent single scores. However, the red/green outer ring and middle ring score double and treble scores respectively.

At the centre of the board are two concentric circles called the bull region, or bulls-eye. The outer bull is worth 25 points and the inner bull is a double, worth 50 points.

There are many variations of rules but in the most popular game the players will begin with a score 301 or 501 and the first player to reduce their running total to zero is a winner. However, it is normal to play a "doubles out" system, which means that the player must land a double (including the double bulls-eye at the centre of the board) on their final dart to win; any other dart that would reduce their running total to one or lower means the score for that set of three darts is "bust".

When a player is able to finish on their current score it is called a "checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s and double bull).

There are exactly eleven distinct ways to checkout on a score of 6:

 D3 D1 D2 S2 D2 D2 D1 S4 D1 S1 S1 D2 S1 T1 D1 S1 S3 D1 D1 D1 D1 D1 S2 D1 S2 S2 D1

Note that D1 D2 is considered different to D2 D1 as they finish on different doubles. However, the combination S1 T1 D1 is considered the same as T1 S1 D1.

In addition we shall not include misses in considering combinations; for example, D3 is the same as 0 D3 and 0 0 D3.

Incredibly there are 42336 distinct ways of checking out in total.

How many distinct ways can a player checkout with a score less than 100?

### Problem 110: Diophantine reciprocals IIPublished on Friday, 2nd December 2005, 06:00 pm; Solved by 6718; Difficulty rating: 40%

In the following equation x, y, and n are positive integers.

 1x + 1y = 1n

It can be verified that when n = 1260 there are 113 distinct solutions and this is the least value of n for which the total number of distinct solutions exceeds one hundred.

What is the least value of n for which the number of distinct solutions exceeds four million?

NOTE: This problem is a much more difficult version of Problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.

### Problem 111: Primes with runsPublished on Friday, 16th December 2005, 06:00 pm; Solved by 5915; Difficulty rating: 45%

Considering 4-digit primes containing repeated digits it is clear that they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by 22, and so on. But there are nine 4-digit primes containing three ones:

1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111

We shall say that M(n, d) represents the maximum number of repeated digits for an n-digit prime where d is the repeated digit, N(n, d) represents the number of such primes, and S(n, d) represents the sum of these primes.

So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit prime where one is the repeated digit, there are N(4, 1) = 9 such primes, and the sum of these primes is S(4, 1) = 22275. It turns out that for d = 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are N(4, 0) = 13 such cases.

In the same way we obtain the following results for 4-digit primes.

 Digit, d M(4, d) N(4, d) S(4, d) 0 2 13 67061 1 3 9 22275 2 3 1 2221 3 3 12 46214 4 3 2 8888 5 3 1 5557 6 3 1 6661 7 3 9 57863 8 3 1 8887 9 3 7 48073

For d = 0 to 9, the sum of all S(4, d) is 273700.

Find the sum of all S(10, d).

### Problem 112: Bouncy numbersPublished on Friday, 30th December 2005, 06:00 pm; Solved by 20838; Difficulty rating: 15%

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand (525) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches 50% is 538.

Surprisingly, bouncy numbers become more and more common and by the time we reach 21780 the proportion of bouncy numbers is equal to 90%.

Find the least number for which the proportion of bouncy numbers is exactly 99%.

### Problem 113: Non-bouncy numbersPublished on Friday, 10th February 2006, 06:00 pm; Solved by 9296; Difficulty rating: 30%

Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, 134468.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, 66420.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, 155349.

As n increases, the proportion of bouncy numbers below n increases such that there are only 12951 numbers below one-million that are not bouncy and only 277032 non-bouncy numbers below 1010.

How many numbers below a googol (10100) are not bouncy?

### Problem 114: Counting block combinations IPublished on Friday, 17th February 2006, 06:00 pm; Solved by 9036; Difficulty rating: 35%

A row measuring seven units in length has red blocks with a minimum length of three units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square. There are exactly seventeen ways of doing this.

How many ways can a row measuring fifty units in length be filled?

NOTE: Although the example above does not lend itself to the possibility, in general it is permitted to mix block sizes. For example, on a row measuring eight units in length you could use red (3), black (1), and red (4).

### Problem 115: Counting block combinations IIPublished on Friday, 24th February 2006, 06:00 pm; Solved by 8261; Difficulty rating: 35%

NOTE: This is a more difficult version of Problem 114.

A row measuring n units in length has red blocks with a minimum length of m units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.

Let the fill-count function, F(m, n), represent the number of ways that a row can be filled.

For example, F(3, 29) = 673135 and F(3, 30) = 1089155.

That is, for m = 3, it can be seen that n = 30 is the smallest value for which the fill-count function first exceeds one million.

In the same way, for m = 10, it can be verified that F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value for which the fill-count function first exceeds one million.

For m = 50, find the least value of n for which the fill-count function first exceeds one million.

### Problem 116: Red, green or blue tilesPublished on Friday, 3rd March 2006, 06:00 pm; Solved by 10107; Difficulty rating: 30%

A row of five black square tiles is to have a number of its tiles replaced with coloured oblong tiles chosen from red (length two), green (length three), or blue (length four).

If red tiles are chosen there are exactly seven ways this can be done.

If green tiles are chosen there are three ways.

And if blue tiles are chosen there are two ways.

Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of replacing the black tiles in a row measuring five units in length.

How many different ways can the black tiles in a row measuring fifty units in length be replaced if colours cannot be mixed and at least one coloured tile must be used?

NOTE: This is related to Problem 117.

### Problem 117: Red, green, and blue tilesPublished on Friday, 10th March 2006, 06:00 pm; Solved by 9267; Difficulty rating: 35%

Using a combination of black square tiles and oblong tiles chosen from: red tiles measuring two units, green tiles measuring three units, and blue tiles measuring four units, it is possible to tile a row measuring five units in length in exactly fifteen different ways.

How many ways can a row measuring fifty units in length be tiled?

NOTE: This is related to Problem 116.

### Problem 118: Pandigital prime setsPublished on Friday, 24th March 2006, 06:00 pm; Solved by 5646; Difficulty rating: 45%

Using all of the digits 1 through 9 and concatenating them freely to form decimal integers, different sets can be formed. Interestingly with the set {2,5,47,89,631}, all of the elements belonging to it are prime.

How many distinct sets containing each of the digits one through nine exactly once contain only prime elements?

### Problem 119: Digit power sumPublished on Friday, 7th April 2006, 06:00 pm; Solved by 10314; Difficulty rating: 30%

The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 83 = 512. Another example of a number with this property is 614656 = 284.

We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.

You are given that a2 = 512 and a10 = 614656.

Find a30.

### Problem 120: Square remaindersPublished on Friday, 21st April 2006, 06:00 pm; Solved by 11686; Difficulty rating: 25%

Let r be the remainder when (a−1)n + (a+1)n is divided by a2.

For example, if a = 7 and n = 3, then r = 42: 63 + 83 = 728 ≡ 42 mod 49. And as n varies, so too will r, but for a = 7 it turns out that rmax = 42.

For 3 ≤ a ≤ 1000, find rmax.

### Problem 121: Disc game prize fundPublished on Friday, 19th May 2006, 06:00 pm; Solved by 8097; Difficulty rating: 35%

A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted. After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.

The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game.

If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker should allocate for winning in this game would be £10 before they would expect to incur a loss. Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9.

Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played.

### Problem 122: Efficient exponentiationPublished on Friday, 2nd June 2006, 06:00 pm; Solved by 6412; Difficulty rating: 40%

The most naive way of computing n15 requires fourteen multiplications:

n × n × ... × n = n15

But using a "binary" method you can compute it in six multiplications:

n × n = n2
n2 × n2 = n4
n4 × n4 = n8
n8 × n4 = n12
n12 × n2 = n14
n14 × n = n15

However it is yet possible to compute it in only five multiplications:

n × n = n2
n2 × n = n3
n3 × n3 = n6
n6 × n6 = n12
n12 × n3 = n15

We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.

For 1 ≤ k ≤ 200, find m(k).

### Problem 123: Prime square remaindersPublished on Friday, 16th June 2006, 06:00 pm; Solved by 9715; Difficulty rating: 30%

Let pn be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder when (pn−1)n + (pn+1)n is divided by pn2.

For example, when n = 3, p3 = 5, and 43 + 63 = 280 ≡ 5 mod 25.

The least value of n for which the remainder first exceeds 109 is 7037.

Find the least value of n for which the remainder first exceeds 1010.

### Problem 124: Ordered radicalsPublished on Friday, 14th July 2006, 06:00 pm; Solved by 11714; Difficulty rating: 25%

The radical of n, rad(n), is the product of the distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.

If we calculate rad(n) for 1n ≤ 10, then sort them on rad(n), and sorting on n if the radical values are equal, we get:

 Unsorted Sorted n rad(n) n rad(n) k 1 1 1 1 1 2 2 2 2 2 3 3 4 2 3 4 2 8 2 4 5 5 3 3 5 6 6 9 3 6 7 7 5 5 7 8 2 6 6 8 9 3 7 7 9 10 10 10 10 10

Let E(k) be the kth element in the sorted n column; for example, E(4) = 8 and E(6) = 9.

If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000).

### Problem 125: Palindromic sumsPublished on Friday, 4th August 2006, 06:00 pm; Solved by 11453; Difficulty rating: 25%

The palindromic number 595 is interesting because it can be written as the sum of consecutive squares: 62 + 72 + 82 + 92 + 102 + 112 + 122.

There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is 4164. Note that 1 = 02 + 12 has not been included as this problem is concerned with the squares of positive integers.

Find the sum of all the numbers less than 108 that are both palindromic and can be written as the sum of consecutive squares.

### Problem 126: Cuboid layersPublished on Friday, 18th August 2006, 06:00 pm; Solved by 3783; Difficulty rating: 55%

The minimum number of cubes to cover every visible face on a cuboid measuring 3 x 2 x 1 is twenty-two.

If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.

However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.

We shall define C(n) to represent the number of cuboids that contain n cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) = 8.

It turns out that 154 is the least value of n for which C(n) = 10.

Find the least value of n for which C(n) = 1000.

### Problem 127: abc-hitsPublished on Friday, 1st September 2006, 06:00 pm; Solved by 5077; Difficulty rating: 50%

The radical of n, rad(n), is the product of distinct prime factors of n. For example, 504 = 23 × 32 × 7, so rad(504) = 2 × 3 × 7 = 42.

We shall define the triplet of positive integers (a, b, c) to be an abc-hit if:

1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1
2. a < b
3. a + b = c

For example, (5, 27, 32) is an abc-hit, because:

1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1
2. 5 < 27
3. 5 + 27 = 32
4. rad(4320) = 30 < 32

It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for c < 1000, with ∑c = 12523.

Find ∑c for c < 120000.

### Problem 128: Hexagonal tile differencesPublished on Friday, 29th September 2006, 06:00 pm; Solved by 4046; Difficulty rating: 55%

A hexagonal tile with number 1 is surrounded by a ring of six hexagonal tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an anti-clockwise direction.

New rings are added in the same fashion, with the next rings being numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows the first three rings.

By finding the difference between tile n and each of its six neighbours we shall define PD(n) to be the number of those differences which are prime.

For example, working clockwise around tile 8 the differences are 12, 29, 11, 6, 1, and 13. So PD(8) = 3.

In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and 10, hence PD(17) = 2.

It can be shown that the maximum value of PD(n) is 3.

If all of the tiles for which PD(n) = 3 are listed in ascending order to form a sequence, the 10th tile would be 271.

Find the 2000th tile in this sequence.

### Problem 129: Repunit divisibilityPublished on Friday, 27th October 2006, 06:00 pm; Solved by 5192; Difficulty rating: 45%

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.

Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.

The least value of n for which A(n) first exceeds ten is 17.

Find the least value of n for which A(n) first exceeds one-million.

### Problem 130: Composites with prime repunit propertyPublished on Friday, 27th October 2006, 06:00 pm; Solved by 4887; Difficulty rating: 45%

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.

Given that n is a positive integer and GCD(n, 10) = 1, it can be shown that there always exists a value, k, for which R(k) is divisible by n, and let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = 5.

You are given that for all primes, p > 5, that p − 1 is divisible by A(p). For example, when p = 41, A(41) = 5, and 40 is divisible by 5.

However, there are rare composite values for which this is also true; the first five examples being 91, 259, 451, 481, and 703.

Find the sum of the first twenty-five composite values of n for which
GCD(n, 10) = 1 and n − 1 is divisible by A(n).

### Problem 131: Prime cube partnershipPublished on Friday, 10th November 2006, 06:00 pm; Solved by 6118; Difficulty rating: 40%

There are some prime values, p, for which there exists a positive integer, n, such that the expression n3 + n2p is a perfect cube.

For example, when p = 19, 83 + 82×19 = 123.

What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred.

How many primes below one million have this remarkable property?

### Problem 132: Large repunit factorsPublished on Friday, 1st December 2006, 06:00 pm; Solved by 5256; Difficulty rating: 45%

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k.

For example, R(10) = 1111111111 = 11×41×271×9091, and the sum of these prime factors is 9414.

Find the sum of the first forty prime factors of R(109).

### Problem 133: Repunit nonfactorsPublished on Friday, 1st December 2006, 06:00 pm; Solved by 4540; Difficulty rating: 50%

A number consisting entirely of ones is called a repunit. We shall define R(k) to be a repunit of length k; for example, R(6) = 111111.

Let us consider repunits of the form R(10n).

Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is divisible by 17. Yet there is no value of n for which R(10n) will divide by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four primes below one-hundred that can be a factor of R(10n).

Find the sum of all the primes below one-hundred thousand that will never be a factor of R(10n).

### Problem 134: Prime pair connectionPublished on Friday, 15th December 2006, 06:00 pm; Solved by 5781; Difficulty rating: 45%

Consider the consecutive primes p1 = 19 and p2 = 23. It can be verified that 1219 is the smallest number such that the last digits are formed by p1 whilst also being divisible by p2.

In fact, with the exception of p1 = 3 and p2 = 5, for every pair of consecutive primes, p2 > p1, there exist values of n for which the last digits are formed by p1 and n is divisible by p2. Let S be the smallest of these values of n.

Find ∑ S for every pair of consecutive primes with 5 ≤ p1 ≤ 1000000.

### Problem 135: Same differencesPublished on Friday, 29th December 2006, 06:00 pm; Solved by 5250; Difficulty rating: 45%

Given the positive integers, x, y, and z, are consecutive terms of an arithmetic progression, the least value of the positive integer, n, for which the equation, x2y2z2 = n, has exactly two solutions is n = 27:

342 − 272 − 202 = 122 − 92 − 62 = 27

It turns out that n = 1155 is the least value which has exactly ten solutions.

How many values of n less than one million have exactly ten distinct solutions?

### Problem 136: Singleton differencePublished on Friday, 29th December 2006, 06:00 pm; Solved by 4605; Difficulty rating: 50%

The positive integers, x, y, and z, are consecutive terms of an arithmetic progression. Given that n is a positive integer, the equation, x2y2z2 = n, has exactly one solution when n = 20:

132 − 102 − 72 = 20

In fact there are twenty-five values of n below one hundred for which the equation has a unique solution.

How many values of n less than fifty million have exactly one solution?

### Problem 137: Fibonacci golden nuggetsPublished on Friday, 12th January 2007, 06:00 pm; Solved by 4458; Difficulty rating: 50%

Consider the infinite polynomial series AF(x) = xF1 + x2F2 + x3F3 + ..., where Fk is the kth term in the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... ; that is, Fk = Fk−1 + Fk−2, F1 = 1 and F2 = 1.

For this problem we shall be interested in values of x for which AF(x) is a positive integer.

 Surprisingly AF(1/2) = (1/2).1 + (1/2)2.1 + (1/2)3.2 + (1/2)4.3 + (1/2)5.5 + ... = 1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ... = 2

The corresponding values of x for the first five natural numbers are shown below.

 x AF(x) √2−1 1 1/2 2 (√13−2)/3 3 (√89−5)/8 4 (√34−3)/5 5

We shall call AF(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 10th golden nugget is 74049690.

Find the 15th golden nugget.

### Problem 138: Special isosceles trianglesPublished on Saturday, 20th January 2007, 11:00 am; Solved by 4764; Difficulty rating: 45%

Consider the isosceles triangle with base length, b = 16, and legs, L = 17.

By using the Pythagorean theorem it can be seen that the height of the triangle, h = √(172 − 82) = 15, which is one less than the base length.

With b = 272 and L = 305, we get h = 273, which is one more than the base length, and this is the second smallest isosceles triangle with the property that h = b ± 1.

Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1 and b, L are positive integers.

### Problem 139: Pythagorean tilesPublished on Saturday, 27th January 2007, 09:00 am; Solved by 4592; Difficulty rating: 50%

Let (a, b, c) represent the three sides of a right angle triangle with integral length sides. It is possible to place four such triangles together to form a square with length c.

For example, (3, 4, 5) triangles can be placed together to form a 5 by 5 square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5 square can be tiled with twenty-five 1 by 1 squares.

However, if (5, 12, 13) triangles were used then the hole would measure 7 by 7 and these could not be used to tile the 13 by 13 square.

Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to take place?

### Problem 140: Modified Fibonacci golden nuggetsPublished on Saturday, 3rd February 2007, 07:00 am; Solved by 3468; Difficulty rating: 55%

Consider the infinite polynomial series AG(x) = xG1 + x2G2 + x3G3 + ..., where Gk is the kth term of the second order recurrence relation Gk = Gk−1 + Gk−2, G1 = 1 and G2 = 4; that is, 1, 4, 5, 9, 14, 23, ... .

For this problem we shall be concerned with values of x for which AG(x) is a positive integer.

The corresponding values of x for the first five natural numbers are shown below.

 x AG(x) (√5−1)/4 1 2/5 2 (√22−2)/6 3 (√137−5)/14 4 1/2 5

We shall call AG(x) a golden nugget if x is rational, because they become increasingly rarer; for example, the 20th golden nugget is 211345365.

Find the sum of the first thirty golden nuggets.

### Problem 141: Investigating progressive numbers, n, which are also squarePublished on Saturday, 17th February 2007, 05:00 am; Solved by 3128; Difficulty rating: 60%

A positive integer, n, is divided by d and the quotient and remainder are q and r respectively. In addition d, q, and r are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.

For example, 58 divided by 6 has quotient 9 and remainder 4. It can also be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common ratio 3/2).
We will call such numbers, n, progressive.

Some progressive numbers, such as 9 and 10404 = 1022, happen to also be perfect squares.
The sum of all progressive perfect squares below one hundred thousand is 124657.

Find the sum of all progressive perfect squares below one trillion (1012).

### Problem 142: Perfect Square CollectionPublished on Saturday, 24th February 2007, 01:00 am; Solved by 4815; Difficulty rating: 45%

Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x − y, x + z, x − z, y + z, y − z are all perfect squares.

### Problem 143: Investigating the Torricelli point of a trianglePublished on Friday, 2nd March 2007, 09:00 pm; Solved by 2145; Difficulty rating: 65%

Let ABC be a triangle with all interior angles being less than 120 degrees. Let X be any point inside the triangle and let XA = p, XC = q, and XB = r.

Fermat challenged Torricelli to find the position of X such that p + q + r was minimised.

Torricelli was able to prove that if equilateral triangles AOB, BNC and AMC are constructed on each side of triangle ABC, the circumscribed circles of AOB, BNC, and AMC will intersect at a single point, T, inside the triangle. Moreover he proved that T, called the Torricelli/Fermat point, minimises p + q + r. Even more remarkable, it can be shown that when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO also intersect at T.

If the sum is minimised and a, b, c, p, q and r are all positive integers we shall call triangle ABC a Torricelli triangle. For example, a = 399, b = 455, c = 511 is an example of a Torricelli triangle, with p + q + r = 784.

Find the sum of all distinct values of p + q + r ≤ 120000 for Torricelli triangles.

### Problem 144: Investigating multiple reflections of a laser beamPublished on Friday, 9th March 2007, 05:00 pm; Solved by 5087; Difficulty rating: 50%

In laser physics, a "white cell" is a mirror system that acts as a delay line for the laser beam. The beam enters the cell, bounces around on the mirrors, and eventually works its way back out.

The specific white cell we will be considering is an ellipse with the equation 4x2 + y2 = 100

The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing, allowing the light to enter and exit through the hole.

The light beam in this problem starts at the point (0.0,10.1) just outside the white cell, and the beam first impacts the mirror at (1.4,-9.6).

Each time the laser beam hits the surface of the ellipse, it follows the usual law of reflection "angle of incidence equals angle of reflection." That is, both the incident and reflected beams make the same angle with the normal line at the point of incidence.

In the figure on the left, the red line shows the first two points of contact between the laser beam and the wall of the white cell; the blue line shows the line tangent to the ellipse at the point of incidence of the first bounce.

The slope m of the tangent line at any point (x,y) of the given ellipse is: m = −4x/y

The normal line is perpendicular to this tangent line at the point of incidence.

The animation on the right shows the first 10 reflections of the beam.

How many times does the beam hit the internal surface of the white cell before exiting?

### Problem 145: How many reversible numbers are there below one-billion?Published on Friday, 16th March 2007, 01:00 pm; Solved by 13985; Difficulty rating: 20%

Some positive integers n have the property that the sum [ n + reverse(n) ] consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, and 904 are reversible. Leading zeroes are not allowed in either n or reverse(n).

There are 120 reversible numbers below one-thousand.

How many reversible numbers are there below one-billion (109)?

### Problem 146: Investigating a Prime Pattern Published on Saturday, 24th March 2007, 09:00 am; Solved by 4102; Difficulty rating: 50%

The smallest positive integer n for which the numbers n2+1, n2+3, n2+7, n2+9, n2+13, and n2+27 are consecutive primes is 10. The sum of all such integers n below one-million is 1242490.

What is the sum of all such integers n below 150 million?

### Problem 147: Rectangles in cross-hatched gridsPublished on Saturday, 31st March 2007, 06:00 am; Solved by 2411; Difficulty rating: 65%

In a 3x2 cross-hatched grid, a total of 37 different rectangles could be situated within that grid as indicated in the sketch.

There are 5 grids smaller than 3x2, vertical and horizontal dimensions being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is cross-hatched, the following number of different rectangles could be situated within those smaller grids:

1x1: 1
2x1: 4
3x1: 8
1x2: 4
2x2: 18

Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles could be situated within 3x2 and smaller grids.

How many different rectangles could be situated within 47x43 and smaller grids?

### Problem 148: Exploring Pascal's trianglePublished on Saturday, 7th April 2007, 02:00 am; Solved by 4209; Difficulty rating: 50%

We can easily verify that none of the entries in the first seven rows of Pascal's triangle are divisible by 7:

 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

However, if we check the first one hundred rows, we will find that only 2361 of the 5050 entries are not divisible by 7.

Find the number of entries which are not divisible by 7 in the first one billion (109) rows of Pascal's triangle.

### Problem 149: Searching for a maximum-sum subsequencePublished on Friday, 13th April 2007, 10:00 pm; Solved by 3991; Difficulty rating: 50%

Looking at the table below, it is easy to verify that the maximum possible sum of adjacent numbers in any direction (horizontal, vertical, diagonal or anti-diagonal) is 16 (= 8 + 7 + 1).

 −2 5 3 2 9 −6 5 1 3 2 7 3 −1 8 −4 8

Now, let us repeat the search, but on a much larger scale:

First, generate four million pseudo-random numbers using a specific form of what is known as a "Lagged Fibonacci Generator":

For 1 ≤ k ≤ 55, sk = [100003 − 200003k + 300007k3] (modulo 1000000) − 500000.
For 56 ≤ k ≤ 4000000, sk = [sk−24 + sk−55 + 1000000] (modulo 1000000) − 500000.

Thus, s10 = −393027 and s100 = 86613.

The terms of s are then arranged in a 2000×2000 table, using the first 2000 numbers to fill the first row (sequentially), the next 2000 numbers to fill the second row, and so on.

Finally, find the greatest sum of (any number of) adjacent entries in any direction (horizontal, vertical, diagonal or anti-diagonal).

### Problem 150: Searching a triangular array for a sub-triangle having minimum-sumPublished on Friday, 13th April 2007, 10:00 pm; Solved by 3243; Difficulty rating: 55%

In a triangular array of positive and negative integers, we wish to find a sub-triangle such that the sum of the numbers it contains is the smallest possible.

In the example below, it can be easily verified that the marked triangle satisfies this condition having a sum of −42.

We wish to make such a triangular array with one thousand rows, so we generate 500500 pseudo-random numbers sk in the range ±219, using a type of random number generator (known as a Linear Congruential Generator) as follows:

t := 0
for k = 1 up to k = 500500:
t := (615949*t + 797807) modulo 220
sk := t−219

Thus: s1 = 273519, s2 = −153582, s3 = 450905 etc

Our triangular array is then formed using the pseudo-random numbers thus:

s1
s2  s3
s4  s5  s6
s7  s8  s9  s10
...

Sub-triangles can start at any element of the array and extend down as far as we like (taking-in the two elements directly below it from the next row, the three elements directly below from the row after that, and so on).
The "sum of a sub-triangle" is defined as the sum of all the elements it contains.
Find the smallest possible sub-triangle sum.

### Problem 151: Paper sheets of standard sizes: an expected-value problemPublished on Friday, 20th April 2007, 06:00 pm; Solved by 4506; Difficulty rating: 50%

A printing shop runs 16 batches (jobs) every week and each batch requires a sheet of special colour-proofing paper of size A5.

Every Monday morning, the foreman opens a new envelope, containing a large sheet of the special paper with size A1.

He proceeds to cut it in half, thus getting two sheets of size A2. Then he cuts one of them in half to get two sheets of size A3 and so on until he obtains the A5-size sheet needed for the first batch of the week.

All the unused sheets are placed back in the envelope.

At the beginning of each subsequent batch, he takes from the envelope one sheet of paper at random. If it is of size A5, he uses it. If it is larger, he repeats the 'cut-in-half' procedure until he has what he needs and any remaining sheets are always placed back in the envelope.

Excluding the first and last batch of the week, find the expected number of times (during each week) that the foreman finds a single sheet of paper in the envelope.

Give your answer rounded to six decimal places using the format x.xxxxxx .

### Problem 152: Writing 1/2 as a sum of inverse squaresPublished on Friday, 27th April 2007, 02:00 pm; Solved by 2092; Difficulty rating: 65%

There are several ways to write the number 1/2 as a sum of inverse squares using distinct integers.

For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used:

In fact, only using integers between 2 and 45 inclusive, there are exactly three ways to do it, the remaining two being: {2,3,4,6,7,9,10,20,28,35,36,45} and {2,3,4,6,7,9,12,15,28,30,35,36,45}.

How many ways are there to write the number 1/2 as a sum of inverse squares using distinct integers between 2 and 80 inclusive?

### Problem 153: Investigating Gaussian IntegersPublished on Saturday, 5th May 2007, 10:00 am; Solved by 2064; Difficulty rating: 65%

As we all know the equation x2=-1 has no solutions for real x.
If we however introduce the imaginary number i this equation has two solutions: x=i and x=-i.
If we go a step further the equation (x-3)2=-4 has two complex solutions: x=3+2i and x=3-2i.
x=3+2i and x=3-2i are called each others' complex conjugate.
Numbers of the form a+bi are called complex numbers.
In general a+bi and abi are each other's complex conjugate.

A Gaussian Integer is a complex number a+bi such that both a and b are integers.
The regular integers are also Gaussian integers (with b=0).
To distinguish them from Gaussian integers with b ≠ 0 we call such integers "rational integers."
A Gaussian integer is called a divisor of a rational integer n if the result is also a Gaussian integer.
If for example we divide 5 by 1+2i we can simplify in the following manner:
Multiply numerator and denominator by the complex conjugate of 1+2i: 1−2i.
The result is .
So 1+2i is a divisor of 5.
Note that 1+i is not a divisor of 5 because .
Note also that if the Gaussian Integer (a+bi) is a divisor of a rational integer n, then its complex conjugate (abi) is also a divisor of n.

In fact, 5 has six divisors such that the real part is positive: {1, 1 + 2i, 1 − 2i, 2 + i, 2 − i, 5}.
The following is a table of all of the divisors for the first five positive rational integers:

 n Gaussian integer divisors with positive real part Sum s(n) of these divisors 1 1 1 2 1, 1+i, 1-i, 2 5 3 1, 3 4 4 1, 1+i, 1-i, 2, 2+2i, 2-2i,4 13 5 1, 1+2i, 1-2i, 2+i, 2-i, 5 12

For divisors with positive real parts, then, we have: .

For 1 ≤ n ≤ 105, ∑ s(n)=17924657155.

What is ∑ s(n) for 1 ≤ n ≤ 108?

### Problem 154: Exploring Pascal's pyramidPublished on Saturday, 12th May 2007, 06:00 am; Solved by 2093; Difficulty rating: 65%

A triangular pyramid is constructed using spherical balls so that each ball rests on exactly three balls of the next lower level.

Then, we calculate the number of paths leading from the apex to each position:

A path starts at the apex and progresses downwards to any of the three spheres directly below the current position.

Consequently, the number of paths to reach a certain position is the sum of the numbers immediately above it (depending on the position, there are up to three numbers above it).

The result is Pascal's pyramid and the numbers at each level n are the coefficients of the trinomial expansion (x + y + z)n.

How many coefficients in the expansion of (x + y + z)200000 are multiples of 1012?

### Problem 155: Counting Capacitor CircuitsPublished on Saturday, 19th May 2007, 02:00 am; Solved by 2883; Difficulty rating: 60%

An electric circuit uses exclusively identical capacitors of the same value C.
The capacitors can be connected in series or in parallel to form sub-units, which can then be connected in series or in parallel with other capacitors or other sub-units to form larger sub-units, and so on up to a final circuit.

Using this simple procedure and up to n identical capacitors, we can make circuits having a range of different total capacitances. For example, using up to n=3 capacitors of 60 F each, we can obtain the following 7 distinct total capacitance values:

If we denote by D(n) the number of distinct total capacitance values we can obtain when using up to n equal-valued capacitors and the simple procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ...

Find D(18).

Reminder : When connecting capacitors C1, C2 etc in parallel, the total capacitance is CT = C1 + C2 +...,
whereas when connecting them in series, the overall capacitance is given by:

### Problem 156: Counting DigitsPublished on Friday, 25th May 2007, 10:00 pm; Solved by 1886; Difficulty rating: 70%

Starting from zero the natural numbers are written down in base 10 like this:
0 1 2 3 4 5 6 7 8 9 10 11 12....

Consider the digit d=1. After we write down each number n, we will update the number of ones that have occurred and call this number f(n,1). The first values for f(n,1), then, are as follows:

 n f(n,1) 0 0 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 10 2 11 4 12 5

Note that f(n,1) never equals 3.
So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The next solution is n=199981.

In the same manner the function f(n,d) gives the total number of digits d that have been written down after the number n has been written.
In fact, for every digit d ≠ 0, 0 is the first solution of the equation f(n,d)=n.

Let s(d) be the sum of all the solutions for which f(n,d)=n.
You are given that s(1)=22786974071.

Find ∑ s(d) for 1 ≤ d ≤ 9.

Note: if, for some n, f(n,d)=n for more than one value of d this value of n is counted again for every value of d for which f(n,d)=n.

### Problem 157: Solving the diophantine equation 1/a+1/b= p/10nPublished on Friday, 1st June 2007, 06:00 pm; Solved by 2140; Difficulty rating: 65%

Consider the diophantine equation 1/a+1/b= p/10n with a, b, p, n positive integers and ab.
For n=1 this equation has 20 solutions that are listed below:

 1/1+1/1=20/10 1/1+1/2=15/10 1/1+1/5=12/10 1/1+1/10=11/10 1/2+1/2=10/10 1/2+1/5=7/10 1/2+1/10=6/10 1/3+1/6=5/10 1/3+1/15=4/10 1/4+1/4=5/10 1/4+1/20=3/10 1/5+1/5=4/10 1/5+1/10=3/10 1/6+1/30=2/10 1/10+1/10=2/10 1/11+1/110=1/10 1/12+1/60=1/10 1/14+1/35=1/10 1/15+1/30=1/10 1/20+1/20=1/10

How many solutions has this equation for 1 ≤ n ≤ 9?

### Problem 158: Exploring strings for which only one character comes lexicographically after its neighbour to the leftPublished on Friday, 15th June 2007, 02:00 pm; Solved by 2949; Difficulty rating: 55%

Taking three different letters from the 26 letters of the alphabet, character strings of length three can be formed.
Examples are 'abc', 'hat' and 'zyx'.
When we study these three examples we see that for 'abc' two characters come lexicographically after its neighbour to the left.
For 'hat' there is exactly one character that comes lexicographically after its neighbour to the left. For 'zyx' there are zero characters that come lexicographically after its neighbour to the left.
In all there are 10400 strings of length 3 for which exactly one character comes lexicographically after its neighbour to the left.

We now consider strings of n ≤ 26 different characters from the alphabet.
For every n, p(n) is the number of strings of length n for which exactly one character comes lexicographically after its neighbour to the left.

What is the maximum value of p(n)?

### Problem 159: Digital root sums of factorisationsPublished on Saturday, 30th June 2007, 02:00 pm; Solved by 2659; Difficulty rating: 60%

A composite number can be factored many different ways. For instance, not including multiplication by one, 24 can be factored in 7 distinct ways:

24 = 2x2x2x3
24 = 2x3x4
24 = 2x2x6
24 = 4x6
24 = 3x8
24 = 2x12
24 = 24

Recall that the digital root of a number, in base 10, is found by adding together the digits of that number, and repeating that process until a number is arrived at that is less than 10. Thus the digital root of 467 is 8.

We shall call a Digital Root Sum (DRS) the sum of the digital roots of the individual factors of our number.
The chart below demonstrates all of the DRS values for 24.

FactorisationDigital Root Sum
2x2x2x3
9
2x3x4
9
2x2x6
10
4x6
10
3x8
11
2x12
5
24
6

The maximum Digital Root Sum of 24 is 11.
The function mdrs(n) gives the maximum Digital Root Sum of n. So mdrs(24)=11.
Find ∑mdrs(n) for 1 < n < 1,000,000.

### Problem 160: Factorial trailing digitsPublished on Friday, 7th September 2007, 02:00 pm; Solved by 2800; Difficulty rating: 60%

For any N, let f(N) be the last five digits before the trailing zeroes in N!.
For example,

9! = 362880 so f(9)=36288
10! = 3628800 so f(10)=36288
20! = 2432902008176640000 so f(20)=17664

Find f(1,000,000,000,000)

### Problem 161: TriominoesPublished on Friday, 21st September 2007, 06:00 pm; Solved by 1647; Difficulty rating: 70%

A triomino is a shape consisting of three squares joined via the edges. There are two basic forms:

If all possible orientations are taken into account there are six:

Any n by m grid for which nxm is divisible by 3 can be tiled with triominoes.
If we consider tilings that can be obtained by reflection or rotation from another tiling as different there are 41 ways a 2 by 9 grid can be tiled with triominoes:

In how many ways can a 9 by 12 grid be tiled in this way by triominoes?

### Problem 162: Hexadecimal numbersPublished on Friday, 5th October 2007, 10:00 pm; Solved by 4496; Difficulty rating: 45%

In the hexadecimal number system numbers are represented using 16 different digits:

0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F

The hexadecimal number AF when written in the decimal number system equals 10x16+15=175.

In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1 and A are all present.
Like numbers written in base ten we write hexadecimal numbers without leading zeroes.

How many hexadecimal numbers containing at most sixteen hexadecimal digits exist with all of the digits 0,1, and A present at least once?