Question

CMR : abc + bca + cba \(⋮\) 3

Answer 1

Ta có : $\overline{abc}+\overline{bca}+\overline{cba}$

$=100a+10b+c+100b+10c+a+100c+10b+a=111(a+b+c)$

Mà $111\vdots 3=>\overline{abc}+\overline{bca}+\overline{cba}\vdots 3$

Answer 2

Ta có: nếu:

\(abc+bca+cba⋮3\)

\(\Leftrightarrow100a+10b+c+100b+10c+a+100c+10b+a⋮3\)

\(\left(100a+a+a\right)+\left(100b+10b+b\right)+\left(100c+c+10c\right)⋮3\)

\(=102a+111b+111c\)

\(102⋮3\Leftrightarrow102a⋮3\)

\(111⋮3\Leftrightarrow111b⋮3\)

\(111⋮3\Leftrightarrow111c⋮3\)

\(\Leftrightarrow102a+111b+111c⋮3\Leftrightarrow abc+bca+cba⋮3\left(đpcm\right)\)