I am studying for an exam and came upon the problem of solving the linear congruence $10 x = 15 \pmod{45}$. I am stuck as the $10$ and $45$ are not coprime but have the $5$ as a common factor.
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$ak = bk \pmod n \implies a\equiv b \pmod {\frac n{\gcd(k,n)}}$. So $10x \equiv 15\pmod {45}\implies 2x \equiv 3\pmod 9$ and not $2$ is coprime to $9$ and we can continue. And when in doubt. Just convert to arithmetic. $10x = 15 +45k$ so $x = \frac {15 + 45k}{10} = \frac {3 +9k}{2}$ so $k=2m+1$ is odd so $x = \frac {3 +9(2m+1)}2 = \frac {12+18m}2 = 6+9m$ so $x \equiv 6 \pmod 9$ – fleablood Jan 14 '21 at 21:30
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Clue:
$$45|10x-15\iff 9|2x-3$$
J. W. Tanner
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Following your hint, am I right in saying that x = 6 (mod 9). Thanks – Joann Sammut Jan 14 '21 at 21:04
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and, @JoannSammut, you can easily check that $10(9k+6)=90k+60\equiv15\bmod45$ – J. W. Tanner Jan 14 '21 at 21:09
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A useful theorem to tuck under your belt: $ak \equiv bk \pmod n \implies a\equiv b \pmod {\frac n{\gcd(k,n)}}$. the hint J.W.Tanner gave you should be a hint as to how to prove the theorem. – fleablood Jan 14 '21 at 21:31