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If k, m, and t are positive integers and

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If k, m, and t are positive integers and

by mitzwillrockgmat » Mon May 24, 2010 10:35 am
If k, m, and t are positive integers and K/6 + M/4 = T/12 , do T and 12 have a common factor greater than 1 ?

(1) k is a multiple of 3.
(2) m is a multiple of 3.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Pleaassssse help! :)
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by asamaverick » Mon May 24, 2010 11:00 am
K/6 + M/4 = T/12

Multiplying each term by 12 gives us:
(2K + 3M) = T

We have to identify if T has any one of these as its factors: 2,3,4,6, 12 (these are the factors of 12 > 1)

(1) K is a multiple of 3. Let K = 3x (x is an integer)
This means T = 2K + 3M = 6x + 3 m = 3 (2x + m) will always be a multiple of 3. Hence 3 will always be a factor of 12. Hence yes, they have a common factor greater than 1.

(2) M is a multiple of 3.
Let M = 3y (y is an integer)
T = 2K + 3(3y) = 2K + 9y.
Consider K = 1, y = 1. This gives T = 11, no common factor > 1.
Consider K = 1, y = 2. This gives T = 20, has 2 as a common factor with 12.
Hence (2) alone is insufficient.

Therefore, IMO the answer should be A. What's the OA?
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by Patrick_GMATFix » Mon May 24, 2010 11:03 am
The equation in the prompt can be rewritten as 2k/12 + 3m/12 = t/12. So we can deduce that 2k+3m=t

Question is: do t and 12 have common factors greater than 1? REPHRASE: Is t a multiple of 2, 3, 4, 6 or 12?

(1) Since k is a multiple of 3, 2k must also be a multiple of 3. t = 2k+3m, so we can deduce that t = (multiple of 3) + (multiple of 3). Therefore, t is a multiple of 3. This definitively answers the rephrase. (1) is sufficient.

(2) Because m is a multiple of 3, 3m must be a multiple of 9. t = 2k+3m, so we can deduce that t = 2k + (multiple of 9). If t = 11, it will have NO common factors with 12. Otherwise if t=20, then it will have several common factors with 12. We don't have enough data to answer.

Correct answer is A

TAKE-AWAY: Algebraically, the only way to prove that an expression is a multiple of x is to show that the expression can be written as the product of (x) and an integer. For instance, statement (1) allows us to replace k with 3*int and to rewrite t=2k+3m as t=2*3*int+3m ---> t = 3(2*int+m). Notice that t must be a multiple of 3 since it equals 3 times the integer (2*int+m).

You can find similar questions if you have access to the Solutions Engine by searching for questions of topic="Number Properties" and difficulty="600-700 AND 700+"

-Best of luck,
Patrick
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by albatross86 » Mon May 24, 2010 11:04 am
Here's my guess:

First job is reducing the question stem to something workable:

k/6 + m/4 = t/12
Multiply throughout by 12
=> t = 2k + 3m

We need to find out if t shares a factor with 12, greater than 1.

Statement 1:
k is a multiple of 3. So let's say k is 3x where x is k/3 which would be an integer.

=> t = 2*3x + 3m
=> t = 3*(2x + 3m)

Therefore 3 is a factor of t. 3 is also a factor of 12. SUFFICIENT.


Statement 2:
m is a multiple of 3.
You can see clearly that this doesn't really help us do much with the question stem to determine anything about t's factors. INSUFFICIENT.

Pick A.
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by mitzwillrockgmat » Tue May 25, 2010 11:26 am
albatross86 wrote:Here's my guess:

First job is reducing the question stem to something workable:

k/6 + m/4 = t/12
Multiply throughout by 12
=> t = 2k + 3m

We need to find out if t shares a factor with 12, greater than 1.

Statement 1:
k is a multiple of 3. So let's say k is 3x where x is k/3 which would be an integer.

=> t = 2*3x + 3m
=> t = 3*(2x + 3m)

Therefore 3 is a factor of t. 3 is also a factor of 12. SUFFICIENT.


Statement 2:
m is a multiple of 3.
You can see clearly that this doesn't really help us do much with the question stem to determine anything about t's factors. INSUFFICIENT.

Pick A.
thanks everyone! really good explainations! :)
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