Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.3 - Page 177: 5


Yes, $6m(2m+10)$ is divisible by $4$. To see this, note that $6m(2m+10)=12m(m+5)=4(3m)(m+5)$. Clearly, $4(3m)(m+5)$ is divisible by $4$, so we get that $4\times(3m)(m+5)=6m(2m+10)$, so $4$ divides $6m(2m+10)$ by definition.

Work Step by Step

Recall that $a|b$, where $a$ and $b$ are integers such that $a\ne0$, if and only if $b=ka$ for some integer $k$. Here, $k=3m(m+5)$, which is an integer because $m$, $3$, and $5$ are integers.
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