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 Nov 9, 2019
edited by htzhang919  Nov 16, 2019
 #1
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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

 Nov 9, 2019
 #2
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 Nov 10, 2019
edited by htzhang919  Nov 16, 2019
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You certainly could! I just thought that was too obvious !!!!.

 Nov 10, 2019
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 Nov 10, 2019
edited by htzhang919  Nov 16, 2019
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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.

 Nov 10, 2019
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 Nov 10, 2019
edited by htzhang919  Nov 16, 2019
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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?

 

 

----

GA

 Nov 16, 2019

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