Let n be a positive integer. Prove that if the number 99...9(n 9's) is divisible by n, then the number 11...1(n 1's) is also divisible by n
Let 99...9 = a.
Let 11...1 = b.
Then a/b=9, so b divides a and the factors of a and b, other than 9 = 3^2, are identical.
Assuming n is not a divisor of b, then it must be of the form 3k or 9k.
Note that by basic divisibility rules, if the SOD of n is divisible by 3 or 9, then n is.
But if so, then by the terms of the problem, there are 3k or 9k 1's in b, guaranteeing that b is divisible accordingly, a contradiction.
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Posted by broll
on 2021-06-07 22:45:03 |
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