+0

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-1
119
7
+44

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Nov 9, 2019
edited by htzhang919  Nov 16, 2019

#1
0

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
+44
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deleted

Nov 10, 2019
edited by htzhang919  Nov 16, 2019
#3
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You certainly could! I just thought that was too obvious !!!!.

Nov 10, 2019
#4
+44
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Nov 10, 2019
edited by htzhang919  Nov 16, 2019
#5
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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
#6
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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?

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GA

Nov 16, 2019