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I am trying to figure out whether it is true or false that if $10\mid a+b$ and $5\mid b$, then $5\mid a$ in $\mathbb{Z}$.

I am trying to demonstrate it that it is a true statement like so:

$a+b=10q$ and $b=5q'$ therefore $a=5q''$

Substituting, we get that $5q''+5q'=10q$

Therefore, $5(q''+q')=10q$

Simplifying, $q''+q'=2q$

Therefore, $5k=10q$ ($k=q''+q'$) where $2\mid k$ proves that $10\mid 5k$, which means $10\mid5q''+5q'$, making the initial statement true.

Antonio De Angelis
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    Not following. Using your notation, we get $a+5q'=10q$ which implies that $a=5\times(2q-q')$ which is all you need. – lulu Feb 02 '24 at 16:05
  • @lulu Understood, thank you! – Antonio De Angelis Feb 02 '24 at 16:06
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    If $10\mid a+b$, then $5\mid a+b.$. If $5\mid a+b$ and $5\mid b,$ then $5\mid a$ – J. W. Tanner Feb 02 '24 at 16:08
  • I changed my proof, would it be valid this time? You have given me correct proofs but haven't said if mine was correct or not. – Antonio De Angelis Feb 02 '24 at 16:48
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    The structure of your proof seems wrong to me. "...therefore $a=5q''$..." seems to be assuming what you are trying to prove. You then conclude your "proof" with "$10 \mid 5q''+5q'$...". But this last is not what you are trying to prove. You're trying to prove that $5\mid a$. – paw88789 Feb 02 '24 at 17:04
  • You made the initial statement true, but you want to make the final statement $5\mid a$ (or $q''\in\mathbb Z$) true. – peterwhy Feb 02 '24 at 18:50
  • $5\mid a!+!b,,b,\Rightarrow, 5\mid (a!+!b)-b = a,$ by here in the linked dupe. $ \ \ $ – Bill Dubuque Feb 03 '24 at 02:10

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