If a and b are positive integers, what is the remainder when 4^{2a+1+b} is divided by 10?This is
a classic "C trap" question: "C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together. When you see such question you should be extremely cautious when choosing C for an answer.
Back to the question: 4 in positive integer power can have only 2 last digits: 4, when the power is odd or 6 when the power is even. Hence, to get the remainder of 4^x/10 we should know whether the power is odd or even: if it's odd the remainder will be 4 and if it's even the remainder will be 6.
(1) a = 1 -->
4^{2a+1+b}=4^{3+b} depending on b the power can be even or odd. Not sufficient.
(2) b = 2 -->
4^{2a+1+b}=4^{2a+3}=4^{even+odd}=4^{odd} --> the remainder upon division of
4^{odd} by 10 is 4. Sufficient.
Answer: B.
enigma123 wrote:
2a (even) + 1 (odd) + b (even) = ODD. So the exponent to 4 is ODD. So I understand that if we put 3, 5 etc I get the remainder 4, but why can't I put exponent as 1 as 1 is ODD too. Can you please help?
The power of 4 is
2a+3 and since
a is a positive integer then the lowest value of
2a+3 is 5, for
a=1. Next, even if the power were 1 then 4^1=4 and the remainder upon division of 4 by 10 would still be 4.
Hope it's clear.
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