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KudoshinichixKaitoKid

$a:5$ (dư 1)

$\Rightarrow5n+1\left(n\in\mathbb{N}\right)$

$b:5$ (dư 2)

$\Rightarrow5m+2\left(m\in\mathbb{N}\right)$

$a.b=\left(5n+1\right)\left(5m+2\right)$

$=5n\left(5m+2\right)+\left(5m+2\right)$

$=25nm+10n+5m+2$

$=5\left(5nm+2n+m\right)+2$

Vì $5\vdots5$

$\Rightarrow5\left(5nm+2n+m\right)\vdots5$

$\Rightarrow5\left(5nm+2n+m\right)+2$ chia 5 dư 2 (đpcm).