Question

va rog A =abc +bca +cab este divizibil cu 111

Answer 1

$A=\overline{abc}+\overline{bca}+\overline{cab}\\A=100\cdot a+10\cdot b+c+100\cdot b+10\cdot c+a+100\cdot c+10\cdot a+b\\A=(100\cdot a+10\cdot a+a) +(100\cdot b+10\cdot b+b)+(100\cdot c+10\cdot c+c)\\A=111\cdot a+111\cdot b+111\cdot c\\A=111(a+b+c)\\Deci~ A~ \vdots~ 111$

Answer 2

A=(100a+10b+c) +(100b+10c+a)+(100c+10a+b)
A=111a+111b+111c
A=111(a+b+c)
A este divizibil cu 111