Natural numbers m, n satisfy m + n = 1991, prove: 10 ^ m + 10 ^ n is a multiple of 11? Who knows?
It is common knowledge that 111001100001 are all multiples of 11. Adding n zeros after these numbers is also a multiple of 11
Let m > N, 10 ^ m and 10 ^ n are all 1 followed by m or n zeros, and the sum of them is
1(000...)1(000...)
(m-n-1) (n)
So we only need to prove that m-n-1 = even number,
That is, M-N = odd,
When m + n = 1991, the difference between M and N is always odd,
So 10 ^ m + 10 ^ n is a multiple of 11
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