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m mod 9 = 6

m = 9 + 6 = 15

n mod 9 = 0

n = 9

m x n = 15 x 9 = 135

The largest integer that mn is divisible by is:

9 x 6 = 45

You certainly could! I just thought that was too obvious !!!!.

Here are the divisors of 135 = (1, 3, 5, 9, 15, 27, 45, 135) >>Total = 8

I don't really know what they mean!!. Try one of the divisors from 9 and up and see what you get.

Here’s a copy of the original question.

Given that \(m\) and \(n\) are positive integers such that \(m\equiv 6\pmod 9\) and \(n\equiv 0\pmod 9\), what is the largest integer that \(mn\) is necessarily divisible by?

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GA