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Clock sequence

Problem 506 Published on Sunday, 8th March 2015, 04:00 am; Solved by 677;
Difficulty rating: 30%

Consider the infinite repeating sequence of digits:
1234321234321234321...

Amazingly, you can break this sequence of digits into a sequence of integers such that the sum of the digits in the n'th value is n.

The sequence goes as follows:
1, 2, 3, 4, 32, 123, 43, 2123, 432, 1234, 32123, ...

Let vn be the n'th value in this sequence. For example, v2 = 2, v5 = 32 and v11 = 32123.

Let S(n) be v1 + v2 + ... + vn. For example, S(11) = 36120, and S(1000) mod 123454321 = 18232686.

Find S(1014) mod 123454321.