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**kingwinner** Is it true in general that f(x)≡0 (mod 3) AND f(x)≡0 (mod 5) IMPLIES f(x)≡0 (mod 15)?

Is it true in general that f(x)≡0 (mod m) AND f(x)≡0 (mod n) IMPLIES f(x)≡0 (mod mn)? Why or why not?

$\displaystyle 6= 0\!\!\!\pmod 6\,\,\,and\,\,\,6= 0\!\!\!\pmod 3$ but $\displaystyle 6\neq 0\!\!\!\pmod {6\cdot 3=18}$ , so the proposition isn't true in that generality.

What is true is that $\displaystyle a=0\!\!\!\pmod n\,,\,\,a=0\!\!\!\pmod n\,,\,\,and\,\,\,(n,m)=1\Longrightarrow a=0\!\!\!\pmod{mn}$ ,and in general, without requiring $\displaystyle (n,m)=1$ , we have $\displaystyle a=0\!\!\!\pmod {\frac{mn}{gcd(m,n)}}$

Tonio

Thanks for explaining!