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I am solving a problem and at some point in the solution I arrived at $$s-t=kn$$ for some $k\in\mathbb{Z}$. Note that $s, t, n$ are also whole numbers.

Can someone show me how do we conclude from here that $s$ and $t$ have the same remainder when divided by $n$?

I mean we can write $s=t+kn$ and $t=s-kn$, but doesn't this mean that the remainder when we devide $s$ by $n$ is $k$? And the same for $t$: the remainder is $s$. I am not sure what's exactly going on.

Thanks!

SAQ
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  • Modulo $n$ you have $s-t\equiv 0$, so $s\equiv t$. – Dietrich Burde Jan 27 '24 at 21:10
  • @DietrichBurde, I am not really familiar with that concept. May you explain it in another way? – SAQ Jan 27 '24 at 21:11
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    $a\equiv b\bmod n$ just means that $n$ divides $b-a$. So $n$ divides $s-t$. If $s=q_1n+r_1$ and $t=q_2n+r_2$, then $n$ divides $s-t=(q_1-q_2)n+(r_1-r_2)$, so $n$ divides $r_1-r_2$, which are smaller than $n$, so $r_1=r_2$. – Dietrich Burde Jan 27 '24 at 21:14

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