Prove $\forall n \in\mathbb{N}, (n|105 \wedge n|70) \implies 5|n$

Question

I know the definition of divides into is $$a|b \equiv \exists a\in\mathbb{Z}, b = ac$$ however I'm not sure how to manipulate this to prove $$\forall n \in\mathbb{N}, (n|105 \wedge n|70) \implies 5|n$$

Any help to lead me in the right way would be appreciated! I'm new to proofs so please explain any steps I could take!

Yes. Yes it would. $n=7$ is also. Further $a\mid b$ if and only if $\exists \color{red}{c}\in\Bbb Z$ such that $b=ac$. — JMoravitz, Jun 06 '16 at 16:31
@AndréNicolas that would also be false using $n=7$ or $n=35$ as counterexamples. — JMoravitz, Jun 06 '16 at 16:36

score 1 · Answer 1 · answered Jun 06 '16 at 16:38

1

$n=1$ and $n=7$ are counterexamples.

Since $\gcd(105,70)=35$, $1,5,7$, and $35$ can be candidates of $n$, but $1$ and $7$ are not multiple of $5$.

answered Jun 06 '16 at 16:38

choco_addicted

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Prove $\forall n \in\mathbb{N}, (n|105 \wedge n|70) \implies 5|n$

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