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If x is a positive integer...

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If x is a positive integer...

by lizakatherine » Fri Feb 12, 2016 8:37 am
Anyone have an effective approach to the following problem? Thanks in advance :)

If x is a positive integer and an odd multiple of 3, and y=x^2 + 3x, then which of the following must be true?

A. y is a multiple of 4.
B. y is a multiple of 8.
C. y is odd.
D. y is a multiple of 18.
E. y is a multiple of 36.
Last edited by lizakatherine on Fri Feb 12, 2016 10:59 am, edited 1 time in total.

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by Marty Murray » Fri Feb 12, 2016 9:04 am
lizakatherine wrote:Anyone have an effective approach to the following problem? Thanks in advance :)

If x is a positive integer and an odd multiple of 3, and y = x² + 3x, then which of the following must be true?

A. y is a multiple of 4.
B. y is a multiple of 8.
C. y is odd.
D. y is a multiple of 18.
E. y is a multiple of 36.
Try plugging in the simplest x that could be a positive integer and an odd multiple of 3, 3 itself.

If x = 3, y = 3² + (3 * 3) = 18

A does not work.

B does not work.

C does not work.

D does work, and may be the answer.

E does not work.

If more than were to have worked when we used x = 3, then we would have had to do some more work to figure out which is the only one that must be true, but only one worked.

The correct answer is C.

Alternate Method

If x is an odd multiple of 3, then the prime factorization of x will include only odd numbers, and at least one 3.

So if y = x² + 3x, y = (some odd number * 3)² + 3(some odd number * 3).

y = (some odd number² * 9) + (some odd number * 9)

y = some odd number[(some odd number * 9) + 9]

y = some odd number(an even number of 9's)

An even number of 9's is always a multiple of 18. So y must be a multiple of 18.

The correct answer is D.
Last edited by Marty Murray on Fri Feb 12, 2016 11:06 pm, edited 1 time in total.
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by Rich.C@EMPOWERgmat.com » Fri Feb 12, 2016 10:07 am
Hi lizakatherine,

EDIT: Now that the entire prompt has been posted, here's how you can solve it....

As it stands, this question can be solved by TESTing VALUES.

We're told that X is a positive integer and an ODD multiple of 3. We're then told that Y = X^2 + 3X. We're asked which of the following answers MUST be true.

Let's TEST X = 3.

IF...
X = 3
Y = (3^2) + 3(3) = 9+9 = 18

18 is NOT a multiple of 4, 8 or 36 and it's NOT odd. There's only one answer left.

Final Answer: D

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Last edited by Rich.C@EMPOWERgmat.com on Fri Feb 12, 2016 4:52 pm, edited 1 time in total.
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by Brent@GMATPrepNow » Fri Feb 12, 2016 10:17 am
lizakatherine wrote:Anyone have an effective approach to the following problem? Thanks in advance :)

If x is a positive integer and an odd multiple of 3, and , then which of the following must be true?

A. y is a multiple of 4.
B. y is a multiple of 8.
C. y is odd.
D. y is a multiple of 18.
E. y is a multiple of 36.
3, and ???,

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by lizakatherine » Fri Feb 12, 2016 11:00 am
Edited the original post. Apologies - on iPad and copy and pasting did not go smoothly..

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by DavidG@VeritasPrep » Fri Feb 12, 2016 11:10 am
lizakatherine wrote:Anyone have an effective approach to the following problem? Thanks in advance :)

If x is a positive integer and an odd multiple of 3, and y=x^2 + 3x, then which of the following must be true?

A. y is a multiple of 4.
B. y is a multiple of 8.
C. y is odd.
D. y is a multiple of 18.
E. y is a multiple of 36.
Say x = 3. y = 3^2 + 3*3 = 18. Only D is true.
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by Matt@VeritasPrep » Sun Feb 14, 2016 9:29 pm
We could also explore this conceptually:

If a number is an odd multiple of 3, we can write it as 3*(something odd) or 3 * (some even integer + 1), or 3*(2k + 1), where k is some integer we don't care about. (It just stands for any integer.)

So our number is of the form 6k + 3.

We know that y = (6k + 3)² + 3(6k + 3), or y = 36k² + 36k + 9 + 18k + 9, or y = 36k² + 54k + 18.

We can see that 18 factors out: y = 18 * (2k² + 3k + 1)

but that 4, 8, and 36 do not, so D is our pick.