Question

If ${22}^3+{23}^3+{24}^3+.........+{88}^3$is divided by 110 then the remainder will be

• A. $55$
• B. $1$
• C. $0$
• D. $44$

Answer 1

The correct option is A $55$

(a^n + b^n ) is divisible by a+b , given that n is odd.
we need to form pairs of numbers whose sum is 110.
(22^3 + 88^3 ) + (23^3 + 87^3 ) ...... + 55^3
so all pairs are divisibly by 110 , which leave 55^3/110 , and the remainder for this is 55.
so IMO A.