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I'm not sure how to prove $n\in \mathbb{Z}^+$ is divisible by $d \in \mathbb{N}$ such that $n \equiv o \pmod d$

I get that I have to prove the cases of (1) if $d\mid n$ then $n\equiv 0 \pmod d$, and (2) if $n\equiv 0 \pmod d$, then $d\mid n$. But I don't know how to prove them. So, what's the proof for case 1? I think I can figure out 2 from there.

Arturo Magidin
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ZeroPhase
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2 Answers2

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One says that $n \pmod{d} = r$ when $ n = q d + r $ for some integer $q$ and $r \in \{0,\dots, d-1\}$. Thus, $r=0$ means $ n = q d $, which is just another way of saying that $d$ divides $n$.

quid
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  • But OP did not once use the notation “$n\pmod d=r$”. Instead, OP used the notation “$n\equiv0\pmod d$” . – Lubin Jul 27 '22 at 16:17
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First I would like to state the definition of $a≡b($mod $m)$

As first used by Gauss, the above notation implies that $$m|a-b$$

Now as you mentioned, the question says $n≡0($mod $d)$

$\implies d|n-0$ $\implies d|n$

which I think provides the answer to part $(ii)$ of your question

Now, you can easily solve part $(i)$ as well using the definition of congruence.

Gloona
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