Problem 333Published on Saturday, 16th April 2011, 01:00 pm; Solved by 615
All positive integers can be partitioned in such a way that each and every term of the partition can be expressed as 2ix3j, where i,j ≥ 0.
Let's consider only those such partitions where none of the terms can divide any of the other terms.
For example, the partition of 17 = 2 + 6 + 9 = (21x30 + 21x31 + 20x32) would not be valid since 2 can divide 6. Neither would the partition 17 = 16 + 1 = (24x30 + 20x30) since 1 can divide 16. The only valid partition of 17 would be 8 + 9 = (23x30 + 20x32).
Many integers have more than one valid partition, the first being 11 having the following two partitions.
11 = 2 + 9 = (21x30 + 20x32)
11 = 8 + 3 = (23x30 + 20x31)
Let's define P(n) as the number of valid partitions of n. For example, P(11) = 2.
Let's consider only the prime integers q which would have a single valid partition such as P(17).
The sum of the primes q <100 such that P(q)=1 equals 233.
Find the sum of the primes q <1000000 such that P(q)=1.