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If $a \equiv b \mod (nm)$, then is it always true that $$a \equiv b \mod n$$ $$a \equiv b \mod m$$

I feel like it's kind of obvious because if $a-b = k_1nm$, then $a-b = k_2n$, ($k_2 = k_1n$) but I can't find any resources on the topic, so maybe I'm wrong

ALEJANDRO PEREZ MORENO
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    You're not only correct but have just provided a proof! You have that $nm|(a-b)$, so $n|(a-b)$, and therefore the modular equivalencies hold. – Steven Stadnicki Sep 11 '19 at 16:54
  • Yes, congruences persist mod factors of the modulus, and your proof is correct (it essentially employs transitivity of "divides", i.e. $, n\mid nm\mid a!-!b,\Rightarrow, n\mid a!-!b),,$ where $,x\mid y,$ means $,x,$ divides $,y.,$ See the linked dupe for more. – Bill Dubuque Sep 11 '19 at 17:03
  • See also here. – Bill Dubuque Sep 11 '19 at 17:16

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Well, if $m,n\geq 2$ are relatively prime, then the mapping $${\Bbb Z}_{mn}\rightarrow{\Bbb Z}_m\times {\Bbb Z}_n:a\mapsto (a\mod m, a\mod n)$$ is a ring isomorphism. In this way, $a\equiv b\mod mn$ is equivalent to $a\equiv b\mod m$ and $a\equiv b\mod n$.

Wuestenfux
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