What is the remainder when p - q is divided by 100, if p=299+499+699+899+.....

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

Options:

A . 50

B . 10

C . 20

D . 30

Answer: Option A
:
A
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

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