Question

111|(abc+bca+cab) ,unde a,b,c sunt numere naturale nenule in baza 10.

Answer 1

abc=a*100+b*10+c
bca=b*100+c*10+a
cab=c*100+a*10+b
111 | (100a+10b+c+100b+10c+a+100c+10a+b)
111 | [ 100 ( a+b+c ) + 10( a+b+c) + (a+b+c ) ]
111 | (a+b+c ) ( 100+10 +1 )
111 | (a+b+c ) * 111
111 | 111 =>
111|(a+b+c )