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Find all integers $x$ such that $$12x + 13 ≡ 10(\mod51)$$.

I have a little problem first treated like a equatio $$12x +13-13 ≡ 10-13 \mod(51)$$ $$12x ≡-3 \mod(51)$$ And i thougt with modular like add the right side $-3 + 51$ $$12x ≡ 48 \mod(51)$$ and i divided with 12 on both sides i got this $$x = 4 \mod(51)$$ but in the answer was like this $x = 4 + 17k$ where did $17k$ come from

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  • $\begingroup$ "where did 17k come from". I came from $51=3\cdot 17$, and $ x\equiv 4\bmod 17$. $\endgroup$ – Dietrich Burde Oct 15 '16 at 16:17
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So we have

$$12x=48+51a$$ where $a$ is any integer

$$\iff12(x-4)=17a$$

As $(12,17)=1$

$$17|(x-4)\iff x\equiv4\pmod{17}$$

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  • $\begingroup$ You move 48 to the left side how does you get 17? $\endgroup$ – user3704516 Oct 15 '16 at 16:30

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