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A friend and I are completely stumped on this prompt, and are even having trouble seeing how its statement is true. Any help will be appreciated!

Prove that if $a \equiv b \pmod{3}$, then $2a \equiv 2b \pmod{3}$.

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    If $a-b$ is divisible by 3, then $2a-2b=2\cdot(a-b)$ ... ?2012-10-24

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$a\equiv b\pmod 3 ⇔ 3\mid (a-b)\implies 3\mid n(a-b) ⇔ na\equiv nb\pmod 3$ where $n$ is any integer.

Also, $3\mid n(a-b)\implies 3\mid(a-b)$ if $(n,3)=1$

So, $3\mid (a-b) ⇔ 3\mid n(a-b)$ if $(n,3)=1$

Here $n=2,(2,3)=1,$ so, $3\mid (a-b) ⇔ 3\mid 2(a-b)$

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    @peoplepower, I've generalized the answer.2012-10-24
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Hint $\rm\ \ n\in\Bbb Z\:\Rightarrow\:2n\in\Bbb Z,\ $ i.e. $\rm\ \dfrac{a-b}3\in\Bbb Z \ \Rightarrow\ \dfrac{2a-2b}3\, =\, 2\,\left(\dfrac{a-b}3\right)\in2\,\Bbb Z\subset \Bbb Z$