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by sunilrawat » Thu Oct 13, 2011 7:50 am
N is a positive integer and is even. Then (N^2 - 4) (N - 2) + 2 (N - 2)^2 + 12 (N + 2) + 24 is divisible by:
2
4
8
16
24

I guess the minimum value N can take is 2. And for 2, the expression IS divisible by 24.
Where am I wrong?
OA C
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by baladon99 » Thu Oct 13, 2011 8:02 am
Test for N=4 . The answer is not divisible by 24.
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by Anurag@Gurome » Thu Oct 13, 2011 8:06 am
sunilrawat wrote:N is a positive integer and is even. Then (N^2 - 4) (N - 2) + 2 (N - 2)^2 + 12 (N + 2) + 24 is divisible by...
# (N² - 4)(N - 2) + 2(N - 2)² + 12(N + 2) + 24
= (N - 2)²(N + 2) + 2(N - 2)² + 12(N + 2) + 24
= (N - 2)²[(N + 2) + 2] + 12[(N + 2) + 2]
= [(N - 2)² + 12](N + 4)
= (N² - 2N + 16)(N + 4)

As N is a positive even integer, we can replace N with 2k, where k is any positive integer.

Hence, the expression can written as,
# (N² - 2N + 16)(N + 4)
= (4k² - 4k + 16)(2k + 4)
= 8*(k² - k + 4)(k + 2)

Hence, the expression is always divisible by 8.

The correct answer is C.
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