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Find all solutions for $x$ in the congruence equation $30x\equiv40(\text{mod } 50), 0\leq x\leq 50$

30 does not have a modular inverse since $\text{gcd}(30,50)=10>1$. However, we can employ a "trick": $$\frac{30x}{10}\equiv\frac{40}{10}\left(\text{mod } \frac{50}{10}\right)\implies3x\equiv4(\text{mod } 5)\implies x\equiv3(\text{mod } 5)$$

And, now all numbers $0\leq x\leq 50$ that are in the same congruence class as $x\equiv3(\text{mod } 5)$ work: $$x\in{3,8,...,48}$$

This is useful in non-trivial/competition problems as well:

Let $a_n=6^{n}+8^{n}$. Determine the remainder upon dividing $a_ {83}$ by $49$. (AIME 1983/6) (See Solution 3, where it is used to work mod 7, and then get the answer back in terms of mod 49)

Is there a standard name for this trick? Is it a property? When/where can it be used/not used? I am looking for this online, but I am only able to find references to modular inverses.

Bill Dubuque
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Starlight
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  • This is certainly a theorem (or lemma) in elementary number theory textbooks that treat congruences; I've seen it in Niven/Zuckerman/Montgomery for example. – Greg Martin Jan 11 '22 at 05:07
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    The trick is that $10(ax) \equiv 10(b) \pmod{10c} \iff$ [by definition] $10c ~| ~[10(b) - 10(ax)] \iff c ~| ~[b - ax] \iff b \equiv ax \pmod{c}.$ This is nothing more than the definition that $r \equiv s \pmod{t} \iff t ~| ~(s - r).$ – user2661923 Jan 11 '22 at 05:13
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    There is no "standard" name for this congruence rule, but some call it the Congruence Cancellation Rule - which is consistent with general ring theoretic terminology - see the linked dupe. – Bill Dubuque Jan 11 '22 at 08:27
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    Btw, the contest problem is better done as $,14n\bmod 49 = 7(2n\bmod 7),$ by the mod Distributive Law. See the many "Linked" posts there for many further examples. – Bill Dubuque Jan 11 '22 at 08:59

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