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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)

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  • $\begingroup$ 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? $\endgroup$ – hardmath Oct 9 '14 at 23:49
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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


Thanks to hardmath for seeing where to get four more sequences: You get extra solutions by attaching some of the 'dead end' strings at the very end. So you can get four additional solutions ending in 68; 685; 6851; 68517. I think these sequences would start respectively with 4; 6; 9; 2.

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  • $\begingroup$ Thanks! But the answer states 9 possibilities. $\endgroup$ – pirsquare Oct 8 '14 at 9:58
  • $\begingroup$ Interesting, but if I understand the problem correctly, I am not seeing where any other possibilities would come from. Of course I could be missing something. $\endgroup$ – paw88789 Oct 8 '14 at 10:00
  • $\begingroup$ Just trying to comprehend this problem. I thought in 46, 4 and 6 are not consecutive numbers as in the case 23 or 34. $\endgroup$ – pirsquare Oct 8 '14 at 10:22
  • $\begingroup$ I understood consecutive digits to be adjacent digits in the sequence, not consecutive in the sense of ordering the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 $\endgroup$ – paw88789 Oct 8 '14 at 10:24
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    $\begingroup$ 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. $\endgroup$ – hardmath Oct 10 '14 at 5:47

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