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.
If x is a positive integer...
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Last edited by lizakatherine on Fri Feb 12, 2016 10:59 am, edited 1 time in total.
 Marty Murray
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Try plugging in the simplest x that could be a positive integer and an odd multiple of 3, 3 itself.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 = xÂ² + 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.
If x = 3, y = 3Â² + (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 = xÂ² + 3x, y = (some odd number * 3)Â² + 3(some odd number * 3).
y = (some odd numberÂ² * 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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 Rich.C@EMPOWERgmat.com
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
Last edited by Rich.C@EMPOWERgmat.com on Fri Feb 12, 2016 4:52 pm, edited 1 time in total.
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 Brent@GMATPrepNow
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3, and ???,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.
Cheers,
Brent
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Say x = 3. y = 3^2 + 3*3 = 18. Only D is true.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.
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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)Â² + 3(6k + 3), or y = 36kÂ² + 36k + 9 + 18k + 9, or y = 36kÂ² + 54k + 18.
We can see that 18 factors out: y = 18 * (2kÂ² + 3k + 1)
but that 4, 8, and 36 do not, so D is our pick.
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)Â² + 3(6k + 3), or y = 36kÂ² + 36k + 9 + 18k + 9, or y = 36kÂ² + 54k + 18.
We can see that 18 factors out: y = 18 * (2kÂ² + 3k + 1)
but that 4, 8, and 36 do not, so D is our pick.