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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. NEED QUICKLY! Thanks!!!

User Tennile
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5 votes

Answer:

Explanation:

very interesting question. The temptation is to say that n should be 11 and that likely is divisible by 11 but it may not be the smallest.

100 + 100^2 = 100 + 10000 = 10100

The pattern of the series goes 101010101 ... 00...

100 / 11 = The remainder is 1/11

10100 / 11 = the remainder is 2/11

1010100 /11 the remainder is 3/11

The pattern suggests that the remainder will be 0 then n = 11

There might be other ways of doing this, but I don't know them.

User Kuro
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