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=10nk (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.
-
n−9 is a multiple of 99. The least n>10 is 108.
-
n+10 is a multiple of 99. The least n>10 is 89.
-
n+10 is a multiple of 9 while n−9 is a multiple of 11. The least n>10 is 53.
-
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