Let abc be a three-digit number. Then abc – cba is not divisible by

Question

Philoid · Accepted Answer

General form of abc is 100a + 10b + c and general form of cba is 100c + 10b + a .

Now abc – cba = (100a + 10b + c) – (100c + 10b + a)

⇒ abc – cba = 100a – a + 10b – 10b + c – 100c.

⇒ abc – cba = 99a -99c = 99(a-c)

⇒ abc – cba is divisible by 99 because 99 is factor of abc – cba.

Hence all numbers which are factors of 99 will also be divisible by abc – cba.

Option (a) ⇒ 9 is the factor of 99 ( 11 x 9 = 99). Therefore abc-cba is divisible by 9.

Option (b) ⇒ 11 is the factor of 99 ( 11 x 9 = 99). Therefore abc-cba is divisible by 11

Option (c) ⇒ 18 is not the factor of 99 because does not give 0 remainder. Therefore abc - cba is not divisible by 18.

Option (d) ⇒ 33 is the factor of 99 (33 x 3 = 99). Therefore abc-cba is divisible by 33.