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So, I have to find the solution to the congruencies

$x \equiv 4 \mod 315$

$x \equiv 9 \mod 715$

I know that I need to divide through by 5 to make the moduli coprime, and I know how to solve for $x$ from there. What I'm having trouble with is figuring out how to divide through by 5. I was given the following hint:

Hint: Let $y = x−4$, show that y must be divisible by $5$.

However, I'm not sure how I should show that $5 ~|~ x−4$.

Any help would be appreciated.

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    $\begingroup$ From $5\mid315$, it follows that, if $315\mid x-4$, then $5\mid x-4$. Do you know the Chinese remainder theorem? $\endgroup$ May 24 at 1:45
  • $\begingroup$ Cf. this question $\endgroup$ May 24 at 1:53
  • $\begingroup$ @J.W.Tanner I do know the Chinese Remainder Theorem, I just wasn't sure how to get $5|x-4$ but clearly I was overthinking things. Thanks for your help. $\endgroup$ May 24 at 2:03
  • $\begingroup$ "However, I'm not sure how I should show that 5|x−4.". Well....$315| x-4$ so $5|x-4$ (and $63|x-4$ but that's not relevant). Likewise $715|x-9$ so $5|x-9$ so $5|x-9+5 = x-4$ (and $143|x-9$ and so $143|x-9+143=x+134$ but that's not relevant either) $\endgroup$
    – fleablood
    May 24 at 3:12
  • $\begingroup$ A good rule is if $a \equiv b \mod n$ and $m|n$ then $a\equiv b \pmod m$ as well. Do you see why?[1] Conversely. If $c\equiv d \pmod m$ then $c \equiv d + km$ for some integer $k: 0 \le k < \frac nm$. [1] The reason why is $a\equiv b \pmod n \iff n| a-b$ but if $m|n$ and $n|a-b$ then $m|a-b$. $\endgroup$
    – fleablood
    May 24 at 3:15
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I'm not sure how I should show that $5 \mid x−4$.

You have $x\equiv 4\bmod 315$. That means $315\mid x-4$.

Since $5\mid315$, it follows from the transitive property of divisibility that $5\mid x-4$.

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We know that $x \cong 4 \mod 315$, and therefore $x-4 \cong 4-4 \mod 315$ (because addition and subtraction preserve congruence).

$y \cong 0 \mod 315$ is equivalent to $315 \mid y$, which is equivalent to $y = 315 \times p$ for some natural $p$. If we break 315 down into prime factors, we get that $y = 3 \times 3 \times 5 \times 7 \times p$, and thus $5 \mid y$.

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