If x is even, which of the following could be odd ?

Question

A.  \frac{(x^2 + x)(x + 2)}{4}

B.  x^3 - 3x^2 + 2x

C.  x^2 - 4x + 6

D.  \frac{x^3 + x^2}{4}

E.  \frac{(x - 2)(x^2 + 2x)}{4}

shameekv1989 · Accepted Answer

a) A can be rewritten in the form  \frac{x(x+1)(x+2)}{4}  i.e. 3 consecutive integers -> Since x is even then the (x+2) will be divisible by 4 (whenever it is possible; won't be when x is -2 or 0 but the expression won't be odd in such cases) and the expression will be even.

b)  x^3−3x^2+2x  =  x*(x^2-3x+2)  =  x*(x-1)*(x-2)  -> Again multiplication of 3 consecutive integers -> Always even

c)  x^2-4x+6  -> Since x is even this term will always be even as even*even = even and even + even = even

d)  \frac{x^3+x^2}{4}  =  \frac{x^2*(x+1)}{4}  ->  x^2  will always be even and will always be divisible by 4 since x is a multiple of 2. x+1 will always be odd since x is even. Hence this could be ODD ---> Answer

e)  \frac{(x-2)(x^2+2x)}{4}  =  \frac{(x-2)x(x+2)}{4}  -> again since x is even x+2 will be divisible by 4 as in case a)

sujoykrdatta · Answer

We will work with options here (since the question clearly asks - which of the "following" could be odd)

Let x = 2k, where k is an integer (since x is even, it must be a multiple of 2)

Option A:  (x^2 + x)(x + 2)/4  =  (4k^2 + 2k)(2k + 2)/4  =  2k(2k + 1) * 2(k + 1)/4  =  k(2k + 1) * (k + 1) 
Here, the first term (2k + 1) is odd; however, k(k + 1) is always even
Thus, Option A is even

Option B:  x^3 - 3x^2 + 2x  = x(x - 1)(x - 2) 
If x is even, the value is even 
Thus, Option B is even

Option C:  x^2 - 4x + 6  = (x - 2)(x - 2) + 2
If x is even, the value is (even) * (even) + 2, which is even
Thus, Option C is even

Option D:  (x^3 + x^2)/4  =  (8k^3 + 4k^2)/4  =  (2k^3 + k^2) 
If k is odd, the value is odd
Thus, Option D COULD be odd

Option E:  (x^2 + 2x)(x - 2)/4  =  (4k^2 + 4k)(2k - 2)/4  =  4k(2k + 1) * 2(k - 1)/4  =  2k(2k + 1) * (k - 1) 
This is clearly even
Thus, Option E is even

Answer D

achloes · Answer

Given that x is even, the quickest approach is plug in 2 for x.

Only D can produce an odd value.

JoeAa · Answer

are there any way I needn't try one by one?

Bunuel · Answer

Given that the question asks to find an option that could be odd, we need to evaluate each option until you reach an option that could produce an odd result. Unless, of course, you right away stumble to the correct option.

Ayushi0002 · Answer

Even X Odd = Even ; So, it can't produce odd, unless I found out 2^2X3/4 -- in that case only it produces odd. I think if one is unable to get that -- they could mark the wrong one. Is there any other efficient way as it took me 3+ mins to do it.

lawyeroffduty · Answer

can we just take x as 2, and basis that we can solve it? because the word could is mentioned in the question, but whereas if must is present, can we do with the same way or not?

If x is even, which of the following could be odd ?

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GMAT Problem Solving (PS) Questions

If x is even, which of the following could be odd ?

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