Question

If ${22}^3+{23}^3+{24}^3+\cdots +{87}^3+{88}^3$ is divided by 110, then the remainder will be:

• A. 55
• B. 1
• C. 44

Answer 1

The correct option is A 55

$(22{)}^3+(23{)}^3+(24{)}^3+\cdots +(87{)}^3+(88{)}^3=({22}^3+{88}^3)+({23}^3+{87}^3)+({24}^3+{86}^3)+({25}^3+{85}^3)+\cdots +({54}^3+{56}^3)+{55}^3$
Now since we know that $a^n+b^n$ is divisible by (a+b) when n. is an odd number. Therefore all the terms, except $(55{)}^3$, is divisible by 110.
Now the remainder when $(55{)}^3$ is divided by 110 is 55.
Hence the required remainder of the whole expression is 55.