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GRE Weekly Challenge #7

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GRE Weekly Challenge #7 [#permalink] New post 18 Nov 2011, 09:07
GMAT Club invites you to test your GRE knowledge for a chance to win! Each week, we will post a new Challenge Problem for you to attempt. If you submit the correct answer, you will be entered into that week’s drawing for a free Manhattan GRE Strategy guide (e-book). What are you waiting for? Get out your scrap paper and start solving! Click here to view contest & prize details


This week's question:
If x is even, which of the following COULD BE odd?

A.) (x^2+x)(x+2)/4

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

C.) x^2-4x+6

D.) (x^3+x^2)/4

E.) (x-2)(x^2+2x)/4

Please post your answer, along with the explanation, below. Get cracking! :)

Solution
Some clever factoring and a knowledge of Number Properties are necessary to solving this problem (or else some very laborious plugging in of numbers—not a strategy we recommend in the time-limited environment of the GRE!)
A basic familiarity with this problem type should clue us in that we want to factor rather than distribute. For instance, x2 – x should be factored as x(x-1), to see that (provided that x is an integer, which we know that it is because it is even) what we really have on our hands is two consecutive integers multiplied together.
If you multiply two consecutive integers together, you always get an even result, since even times odd (or odd times even) is even.
Let’s factor each answer choice and draw conclusions.
  • A.
    \frac{(x^2+x)(x+2)}{4}
    \frac{x(x+1)(x+2)}{4}
    The numerator is composed of three consecutive integers multiplied together. Since x is even, x + 2 is also even. Hence, EVEN × ODD × EVEN. When we divide by 4, is it possible to get an odd answer?
    If all we knew were (EVEN × ODD × EVEN) / 4, we might conclude that the two evens could cancel out with the 4, leaving behind an odd. However, because the two evens are CONSECUTIVE even integers, at least one of them IS ALREADY a multiple of 4. For instance, if x = 2, then x + 2 = 4. Or if x = 6, then x + 2 = 8. (For any two consecutive evens, exactly one of them is a multiple of 4).
    Therefore, either the x or the x + 2 will completely cancel out the 4, leaving behind an even product. Hence, A will always be even.
  • B.
    x^3–3x^2+2x
    x(x^2–3x+2)
    x(x – 2)(x – 1)
    (x – 2)(x – 1) x
    Once factored and put “in order,” choice B reveals itself to be three consecutive integers multiplied together. Even without knowing that x is even, we would be able to conclude that choice B is always even, because multiplying two or more consecutive integers together always yields an even result (since, when multiplying, just one even makes the entire product even).
  • C.
    x^2–4x+6
    This one doesn’t factor so easily! Try as you might, but you will not be able to come up with two numbers that multiply to 6 and add to –4. However, we can factor the x out of the two terms that contain x:
    x(x – 4) + 6
    We have an even times another even (if x is even, then so is any number generated by adding or subtracting an even from x; thus x – 4 is also even), plus an even.
    (EVEN × EVEN) × EVEN will always be even.
  • D.
    \frac{x^2+x^3}{4}
    \frac{x^2(x+1)}{4}
    We cannot do any more factoring here, although it might be helpful to expand x^2 as (x)(x), as follows:
    \frac{(x)(x)(x+1)}{4}
    Thus, we have (EVEN)(EVEN)(ODD) divided by 4. This looks a little like choice A, where we had a similar situation. However, in choice D, the two evens are not consecutive—rather, they are the SAME number. Therefore, it is NOT guaranteed that one of them is a multiple of 4, as was the case with choice A.
    It is certainly possible that choice D could be even (such as if x = 4), but it is also possible that it could be odd (if x = 2):
    If x = 4, , and the result, 20, is EVEN.
    If x = 2, , and the result, 3, is ODD.
    Because choice D COULD BE odd, D is the correct answer.
  • E.
    \frac{(x-2)(x^2+2x)}{4}
    \frac{(x-2)(x)(x+2)}{4}

    If x is even, then x – 2 is the previous even and x + 2 is the next even—that is, the numerator here is composed of three consecutive even integers multiplied together. For any three consecutive even integers, at least one will be a multiple of 4. Whichever integer is a multiple of 4 will cancel out the 4 on the bottom completely, leaving behind two evens in the numerator. Thus, choice E will always yield an even result.
    The correct answer is D.


The winner of this Week's Challenge is... (drum roll).... dreambeliever. Congratulations! Please send me a pm with your name, email address and choice of Manhattan GRE Guide


Details on the next challenge are available in the Weekly Challenge Master thread... Stay tuned!
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Re: GRE Weekly Challenge #7 [#permalink] New post 18 Nov 2011, 11:38
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Re: GRE Weekly Challenge #7 [#permalink] New post 18 Nov 2011, 12:12
Answer is D althought I do not know bet way to solve.

(x^3+X^2)/4

if X = -2, the solution to this formula would be -1 which is odd
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Re: GRE Weekly Challenge #7 [#permalink] New post 18 Nov 2011, 12:32
Answer D...

if we assume x=2 we get the answer as 3 (odd)
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Re: GRE Weekly Challenge #7 [#permalink] New post 18 Nov 2011, 13:12
Ans D

x.x(x+1)/4
if x= 2 then
2*2*3/4 = 3 which is odd

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Re: GRE Weekly Challenge #7 [#permalink] New post 18 Nov 2011, 13:16
D is the answer: check for x=2: (8+4)/4=3. and for any x = 2(2n-1), where n is an integer.
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Re: GRE Weekly Challenge #7 [#permalink] New post 18 Nov 2011, 18:43
The answer is D.

Plugin 2 in all the answer choices, only choice D returns an odd number
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Re: GRE Weekly Challenge #7 [#permalink] New post 19 Nov 2011, 01:38
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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.
Re: GRE Weekly Challenge #7   [#permalink] 19 Nov 2011, 01:38
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