Question

What is the remainder when p - q is divided by 100, if $p=2^{99}+4^{99}+6^{99}+8^{99}+.........+{100}^{99}$ and $q=1^{99}+3^{99}+5^{99}+7^{99}+..........+{99}^{99}?$

• A. 50
• B. 10
• C. 20
• D. 30

Answer 1

The correct option is A

50

99=4k+3
Thus, if you take the individual differences (upto 10)
For $(2^{99}+4^{99}+6^{99}+8^{99}+{10}^{99})-(1^{99}+3^{99}+5^{99}+7^{99}+9^{99})\Rightarrow$
Last digit of this expression = (8+4+6+2+0) -(1+7+5+3+9) = -5 = 5 (10-5)
However, since there are 10 such groups; (since ${10}^{99}$ or multiples of ${10}^{99}$ are exactly divisible by 100)
$=5\times 10=50$ (which will always end in zero)

Thus the remainder when divided by 100 is the last two digits = 50 itself