What will be the remainder when ($$99^{99}$$​ + 99) is divisible by 100?

Correct option is B

Given:

Find the remainder when $$99^{99}$$​ + 99 is divided by 100.

Concept Used:

If $$(a – 1)^n$$​is divided by a then, remainder is $$(-1)^n$$​, where n is odd

Solution:

Using the binomial theorem for $$99^{99}:$$​​

​$$99^{99} = (100 - 1)^{99}$$​​

When expanded, all terms except the last term $$(-1)^{99}$$​will be divisible by 100.

So, the term left:

$$(-1)^{99}$$​+ 99 = -1 + 99 = 98

Thus, The remainder is 98