Question

8
1. If
abc is

a three digit number then, abc+cab+ bca is always divisible by​

Answer 1

Answer:

abc = 100a + 10b + c.

bca = 100b + 10c + a.

cab = 100c + 10a + b.

⇒ 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 = 111a + 111b + 111c.

⇒ abc + bca + cab = 111( a + b + c).

⇒ abc + bca + cab is divisible by 111 and (a + b + c) .

Also abc + bca + cab will be divisible by factors of 111

Option (a) ⇒ abc + bca + cab is divisible by (a + b + c) because abc + bca + cab = 111(a + b + c).

Option (b) ⇒ 3 is divisible by 111 because sum of digits of 111 is 3 and 3 is divisible by 3( Divisibility test of 3).

Option (c) ⇒ 37 is the factor of 111 gives 0 as remainder and ( 37 x 4 = 111) therefore 111 is divisible by 37

Option (d) ⇒ 9 is not factor of 111 because not gives 0 as remainder. Therefore 111 is not divisible by 9 .