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If $x \equiv 0 \pmod{a}$ and $x \equiv 0 \pmod{b}$ then $x \equiv 0 \pmod{ab}$?

Use of CRT, but that doesnt seem to work here?

Jimit
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    What about $x=b=20$, $a=10$ – wet Oct 26 '16 at 02:25
  • It works, $x \equiv 0$,but how to prove it? – Jimit Oct 26 '16 at 02:27
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    It doesn't work, read @wet's counterexample again. Or consider that if it worked it would imply that if $x$ is a multiple of $a$ then $x$ must be a multiple of $a^2$, which is obviously false. You are missing an assumption. – dxiv Oct 26 '16 at 02:32
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    @dxiv The given statement is true only if (a, b) = 1? Should I put this as an answer? – Parcly Taxel Oct 26 '16 at 02:41
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    @ParclyTaxel I can't guess what's missing. The conclusion holds if $(a,b)=1$, but it also holds if you replace $\pmod{ab}$ with $\pmod{\text{lcm}(a,b)}$ for example. – dxiv Oct 26 '16 at 02:46
  • @ParclyTaxel yes, make an answer – Will Jagy Oct 26 '16 at 02:56
  • Note that this statement holds if $(a,b)$ are prime numbers...(prime factors of $x$) – Soham Oct 27 '16 at 15:24

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Note that $x\equiv0\bmod a$ and $x\equiv0\bmod b$ can be written more concisely as $a\mid x$ and $b\mid x$. Given these two conditions, only $\operatorname{lcm}(a,b)\mid x$ is always true; $ab\mid x$ is true if $\gcd(a,b)=1$.

For example, if $a=6$, $b=15$ ($\operatorname{lcm}(a,b)=30$) and $x=120$, $a\mid x$ and $b\mid x$ are true, but $ab\mid x$ is not true; only $30\mid x$ is true.

Parcly Taxel
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