How many of the first 2018 numbers in the sequence 101,1001,10001,100001,… are divisible by 101?
Answer choices
Show solution
Solution
The k-th term is 10k+1+1, which 101 divides iff 10k+1≡−1(mod101). Notice 102=100≡−1(mod101). So 10m≡−1 exactly when m≡2(mod4), meaning k+1≡2, that is k≡1(mod4). Among k=1,2,…,2018, the values 1,5,…,2017 number 505. Thus, C is the correct answer.