Question

If D = ABC+BCA+CAB where A, B and C are decimal digits, then D is duvisible by??

Answer 1

Answer:

Step-by-step explanation:

ABC=100A+10B+C

BCA=100B+10C+A

CAB=100C+10A+B

Adding these 3 numbers we get

ABC+BCA+CAB

=100(A+B+C)+10(A+B+C)+(A+B+C)

=(100+10+1)(A+B+C)

=111(A+B+C)

this implies

D=111(A+B+C)

D is a multiple of A+B+C

then D is divisible by A+B+C

Also D is divisble by 111

Answer 2

Answer:

37 but not 29