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Please, help me on this question. I decided, but I need to know if my answer is correct.

I thought of calculating m for the values ​​of the divisors of 24, that is, making m belonging to D (24).

So I found

$m=2 \Rightarrow \{2k+1;k \in \mathbb{Z}\}$

$m=3 \Rightarrow \{3k+2;k \in \mathbb{Z}\}$

$m=4 \Rightarrow \{4k+1;k \in \mathbb{Z}\}$

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    $\begingroup$ I'm not sure I understand the question. Is the question that $17\equiv 24\mod m$, and you need to find out $m$? $\endgroup$ – Guy Apr 29 '17 at 14:55
  • $\begingroup$ $17 \in \bar{a}$ $\endgroup$ – Felipe Maia Apr 29 '17 at 15:02
  • $\begingroup$ @FelipeMaia What is $\bar{a}$? $\endgroup$ – kccu Apr 29 '17 at 15:04
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    $\begingroup$ @kccu Set consisting of all integers that are congruent to the integer $a\; mod\; m$ $\endgroup$ – Felipe Maia Apr 29 '17 at 18:33
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Definition of residue. The number $r$ in the congruence $a\equiv r\pmod m$ is called the residue of $a\pmod m$. In the case at hand $r=17$ and $ a=24 $.

This means that for some integer $k$ the following equality holds $24=17+km $. You should then have $km=7$, where $ k $ and $ m $ are positive integers. This implies that $ m=7 $, because $7 $ is a prime number, that is, it has no divisors, except $ 1 $ and $7 $.

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Hint: you have $24\equiv 17\pmod{m}$ if and only if $m\mid(24-17)$.

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  • $\begingroup$ So who is $ m$? $\endgroup$ – Felipe Maia Apr 29 '17 at 19:34
  • $\begingroup$ I understood that $24 = m \cdot k + r$. For this reason, the question "who is $m$?" $\endgroup$ – Felipe Maia Apr 29 '17 at 19:39
  • $\begingroup$ @FelipeMaia What are the divisors of $24-17=7$? The formula “$a\mid b$” is read “$a$ is a divisor of $b$”. $\endgroup$ – egreg Apr 29 '17 at 20:07
  • $\begingroup$ You did not understand my speech. I understand the concept of divisor, otherwise it would not have said "I thought of calculating m for the values ​​of the divisors of $24$"... Anyway, what I meant is that in this situation $24=m \cdot k + r$, how to get $m$?! $\endgroup$ – Felipe Maia Apr 29 '17 at 23:38
  • $\begingroup$ @FelipeMaia You can't, in general. The problem at hand is not to compute $r$, but you have the information that $24$ and $17$ are congruent modulo $m$. $\endgroup$ – egreg Apr 30 '17 at 8:31

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