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I have the following question :

Find $m\in \mathbb{N}$ such that $12m=10\pmod {80}$.

The answer is that such an $m$ does not exist.

How could one conclude that without a calculator?

Any help will be appreciated.

3 Answers 3

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Note $$12m \equiv 10 \pmod{80} \implies 12m \equiv 10 \pmod{4}$$ Now $$12m \equiv 0 \pmod 4 $$ However $$10 \equiv 2 \not \equiv 0 \pmod 4$$ So no such $m \in \mathbb{N}$ exists.

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    Why is this correct? $12m \equiv 10 \pmod{80} \implies 12m \equiv 10 \pmod{4}$ – 2017-01-28
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    Because $12m \equiv 10 \pmod {80}$ implies that $12m-10=80k$ for some integer $k$. $12m-10=80k=4(20k)$, so $12m \equiv 10 \pmod {4}$. – 2017-01-28
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    @JaVaPG Sorry, forgot to ping you. See my earlier comment. – 2017-01-28
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    Sorry took me some time to understand now I get it, Thank you! – 2017-01-28
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The condition $12m\equiv 10\pmod{80}$ is equivalent to $6m\equiv5\pmod{40}$, which cannot be satisfied because, for any $m,k\in\mathbb{Z}$, the integers $6m-5$ and $40k$ are respectively odd and even.

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Suppose such an $m\in\mathbb{N}$ does exist.

We now have $12m\equiv10 \pmod{80}$, so write $12m-10$ as a multiple of $80$; thus $12m-10=80$ for some $k\in\mathbb{Z}$. Now $12m-10=4\ell$ for $\ell=40k$, so that $$12m\equiv10\pmod{4}.\tag{1}$$

But then $$\begin{align} 0&\equiv12m\pmod{4} \\ &\stackrel{(1)}{\equiv}10 \pmod{4} \\ &\equiv2 \pmod{4}, \end{align}$$ a contradiction.

Thus no such $m$ exists.