The number N = 100+100^2+100^3 + ... + 100^n. Find the least possible value of n such that the number N is divisible by 11.

It's a big alphabet, why confuse us with N and n?

Anyway, this one is a bit easier than it looks. We're going to look at it mod 11, meaning we're only interested in the remainders when we divide by 11. The nice thing about multiplication and addition is if we're going to take the remainder at the end, we can also take the remainders at the beginning and along the way and get the same answer, with smaller numbers.

That all sounds very complicated. It's easier explained.

100 = 9*11 + 1 so has remainder 1 when divided by 11. So if we multiply by 100 that's not going to change the remainder when we divided by 11; it's like multiplying by 1.

So 100, 100^2, 100^3, etc all have remainder 1 when divided by 11. So we need to add up 11 of them to get a number that's divisible by 11.

Answer: 11