Any thoughts on how to solve this problem? [closed]

How many numbers do there exist having 2013 digits, in which every two-digit number composed of two consecutive digits is a multiple of either 17 or 23? (Taken from Singapore and Asian Schools Math Olympiad)

closed as off-topic by Did, Hamou, user147263, Davide Giraudo, KasperOct 8 '14 at 11:56

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• Have you made a list of all two-digit numbers that are multiples of 17 or 23? Which pairs of these can form consecutive pairs, overlapping as required by the problem? – hardmath Oct 9 '14 at 23:49

Allowable pairs of consecutive digits are $17, 23, 34, 46, 51, 68, 69, 85, 92, 00$. Since the second number in each pair must be the first number of a new pair, the digits, $1, 5, 7, 8$ quickly lead to dead ends.
This leaves us with a cycle $\overline{23469}$. Starting where you wish in this cycle gives $5$ possibilities. You probably would not count the sequence of $2013$ zeros, as a $2013$-digit number
• Some additional solutions arise because when we near the end of the required 2013 digits, we can depart from the repetition of $\overline{23469}$. For example, instead of $69$ we could put $68$, and either end there or add further $85$ and end there, or add further $51$ and end there, or add further $17$. These four extra possibilities plus the five already identified make 9 solutions. – hardmath Oct 10 '14 at 5:47