Re: GRE Weekly Challenge #7
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19 Nov 2011, 02:38
B an C are clearly out as, (even*odd/even=even), and (even + even - even = even).
Now, A, D and E have 4 in the denominator, which means that if the numerator, if expanded, has upto two multiples of 2, all the 2s in the numerator will cancel out with 4 which is 2 x 2.
Alternatively, if it can be proved that the numerator will always have more than 2 multiples of 4, then, only two of the multiples of 2 in the numerator will cancel out, and the remaining multiples of 2 will make the numerator even.
consider E, it can be expanded as, x(x-2)(x+2)/4
[taking x common from x^2 + 2x]
each of these three terms x, x-2, x+2 are even.. therefore, there will be AT LEAST 3 multiples of 2 in the numerator..
Thus, all multiples of 2 will never cancel out, and eventually the number will be even.
Consider A,
It can be expanded as x(x+1)(x+2)/4
Now, of these only 'x' and 'x+2' will be even.
This leaves us with AT least two multiples of 2, to start with, as both are even.
Now, put x=2k, (since it is an even number)
where k can be either odd or even.
This way,
Expression = 2k * (2k+2)/4
Simplifying,
Expression= 2 * 2 * k * (k+1)/4
Now, lets cancel, two 2s in the numerator with the 4 in the denominator.
Expression = k(k+1).
We know k can be odd or even:
If k is odd, k+1 will be even. Thus as even * odd = even. The expression is even.
Also, if k is even, k+1 will be odd, and even * odd = even. The expression is even.
This way, the expression will always be even.
[ Instead of putting x = 2k, one could also note that as x and x+2 are consecutive even numbers, either of them will definitely be a multiple of 4.]
We are left with only option D.
Taking x^2 common out of option D:
Expression = x^2 * (x+1)/4
'x+1' will always be odd, as 'x' is even.
x^2 will have only two multiples of 2, in the case that X has only a single multiple of 2. eg: x=2.
Thus the expression could have only two multiples of 2, which could cancel out completely with 4.
Thus the result will be devoid of any multiple of 2, making it an odd number.