Question

How many of the first 2018 numbers in the sequence 101, 1001, 10001, 100001, ... Are divisible by 101?

Answer 1

Answer:

ans is1

Step-by-step explanation:

because in rest of the numbers in series we will get remainder as 1

Answer 2

Answer:505

Step-by-step explanation: The number 10^{n} +1 is divisible by 101 only if 10^{n}= -1 (mod 101)

we note that (10, 10^{2}, 10^{3}, 10^{4}) = 10, -1, -10,1) (mod 101) so the powers of 10 are 4-periodic mod 101.

It follows that $10^n\equiv -1\pmod{101}$ if and only if $n\equiv 2\pmod 4$.

In the given list, $10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1$, the desired exponents are $2,6,10,\dots,2018$, and there are$10^{}$