Question

3) 4) 5) if a, b and c are digits, then abc + bca + cab is divisible by:

Answer 1

hey mate
here's the solution...

Answer 2

Answer:

The answer is 111

Step-by-step explanation:

If a ,b ,c are digits then the numbers formed can be:

abc = 100a + 10b+ c

bca = 100b + 10c + a

cab = 100c + 10a + b

On adding these three numbers ;

abc + bca + cab = ( 100a + 10b+ c ) + ( 100b + 10c + a ) +  ( 100c + 10a + b)

abc+  bca + cab = ( 100a + 10a+ a ) + ( 100b + 10b + b ) +  ( 100c + 10c + c)

abc+  bca + cab = 111 a + 111 b+ 111 c

abc+  bca + cab = 111 ( a + b + c)

Hence it is always divisible by 111.