0
$\begingroup$

I was wondering if it could be proved without using the Chinese remainder theorem, arithmetically.

Thank you

$\endgroup$
2
  • 1
    $\begingroup$ If $2\mid a-b$ and $3\mid a-b$, then $6\mid a-b$ $\endgroup$
    – J. W. Tanner
    Nov 9, 2020 at 11:15
  • 2
    $\begingroup$ Technically you don't need the full power of the Chinese remainder theorem. You can make an argument from a simple lemma involving prime divisors and/or prime factorizations. $\endgroup$
    – Derek Luna
    Nov 9, 2020 at 11:22

2 Answers 2

Reset to default
4
$\begingroup$

$2,3$ are relatively prime so $2|(m-n)$, $3|(m-n)$ implies $6|(m-n)$ meaning $m\equiv n \mod 6$.

$\endgroup$
0
$\begingroup$

If $c$ is a common multiple of $a$ and $b$, then $c$ is a multiple of their least common multiple $\operatorname{lcm}(a,b)$.

Apply this to $c=m-n$ and $a=2$ and $b=3$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged .