Problem:
How many of the first 2018 numbers in the sequence 101,1001,10001,100001,… are divisible by 101?
Answer Choices:
A. 253
B. 504
C. 505
D. 506
E. 1009
Solution:
The numbers in the given sequence are of the form 10n+1 for n=2,3,…,2019. If n is even, say n=2k for some positive integer k, then 10n+1=100k+1≡(−1)k+1(mod101). Thus 10n+1 is divisible by 101 if and only if k is odd, which means n=2,6,10,…,2018. There are 41(2018−2)+1=(C)505 such values. On the other hand, if n is odd, say n=2k+1 for some positive integer k, then
10n+1=10⋅10n−1+1=10⋅100k+1≡10⋅(−1)k+1(mod101),
which is congruent to 9 or 11, and 10n+1 is not divisible by 101 in this case.
The problems on this page are the property of the MAA's American Mathematics Competitions