Problem:
Let a10​=10, and for each integer n>10 let an​=100an−1​+n. Find the least n>10 such that an​ is a multiple of 99.
Solution:
Because 100≡1(mod99),an​≡an−1​+n(mod99). Thus an​≡∑k=10n​k (mod99), which is equivalent to 2(n+10)(n−9)​. Because n+10 and n−9 cannot both be multiples of 3,an​ is a multiple of 99 if and only if one of the following holds.
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n−9 is a multiple of 99. The least n>10 is 108.
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n+10 is a multiple of 99. The least n>10 is 89.
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n+10 is a multiple of 9 while n−9 is a multiple of 11. The least n>10 is 53.
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n+10 is a multiple of 11 while n−9 is a multiple of 9. The least n>10 is 45.
The requested minimum is 45​.
The problems on this page are the property of the MAA's American Mathematics Competitions