Data Sufficiency

Hi BlueDragon2010,

This DS question can be dealt with using Number Properties and/or TESTing Values. Doing a bit of algebra at the beginning will make the question easier to solve though:

We're told that K, M and T are POSITIVE INTEGERS; we're also told that K/6 + M/4 = T/12. We're asked if T and 12 have a common factor greater than 1? This is YES/NO question.

From the given equation, we can use common denominators and simplify:

2K/12 + 3M/12 = T/12

2K + 3M = T

This equation will make the work a bit easier.

Fact 1: K is a multiple of 3

This Number Property tells us that 2K will be a multiple of 3 (since K is a multiple of 3). 3M is ALSO a multiple of 3 (since M is an integer). When you add a "multiple of 3" to another "multiple of 3", you'll end up with an EVEN NUMBER.

This tells us that T MUST be EVEN. Since T is even, T and 12 WILL have a common factor greater than 1 (the number 2, at the very least) and the answer to the question is YES.
Fact 1 is SUFFICIENT

Fact 2: M is a multiple of 3

This Number Property tells us that 3M will be a multiple of 3. 2K will be a multiple of 2. When you add a "multiple of 3" to a "multiple of 2", you MIGHT get an even, but you might get an odd....

3 + 2 = 5 --> 5 and 12 do NOT have a common factor greater than 1 (so the answer is NO)
6 + 6 = 12 --> 12 and 12 DO have a common factor greater than 1 (the number 12, so the answer is YES).
Fact 2 is INSUFFICIENT.

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

BlueDragon2010 wrote: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

Target question: Do t and 12 have a common factor (divisor) greater than 1?


A little background information
For questions involving factors (aka "divisors"), we can say:
If k is a divisor of N, then k is "hiding" within the prime factorization of N
Examples:
3 is a divisor of 24 <--> 24 = (2)(2)(2)(3)
5 is a divisor of 70 <--> 70 = (2)(5)(7)
8 is a divisor of 56 <--> 56 = (2)(2)(2)(7)

Similarly, for questions involving multiples, we can say:
If N is a multiple of k, then k is "hiding" within the prime factorization of N
Examples:
24 is a multiple of 3 <--> 24 = (2)(2)(2)(3)
70 is a multiple of 5 <--> (2)(5)(7)
330 is a multiple of 6 <--> 330 = (2)(3)(5)(11)

Okay, now let's solve the question....

Given: (k/6)+ (m/4) = (t/12)
Simplify this by multiplying both sides by 12 to get 2k + 3m = t

Statement 1: k is a multiple of 3.
In other words, 3 is hiding in the prime factorization of k
So, we know that k =(3)(?)(?)(?)...

Aside: notice that the prime factorization of k may or may not have any primes other than the 3. All we can be certain of is that there is one 3 within the prime factorization (thus the question marks in the factorization)

From here, we'll take our given information, 2k + 3m = t, and replace k with (3)(?)(?)(?) to get: (2)(3)(?)(?)(?) + 3m = t
At this point, we can factor out a 3 to get: 3[(2)(?)(?)(?) + m] = t, which means 3 is a divisor of t.
Since 3 is also a divisor of 12, we can see that t and 12 have a common factor (divisor) greater than 1.
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: m is a multiple of 3.
In other words, 3 is hiding in the prime factorization of m
So, we know that m =(3)(?)(?)(?)...
From here, we'll take our given information, 2k + 3m = t, and replace m with (3)(?)(?)(?) to get: 2k + 3(3)(?)(?)(?) = t
At this point, we cannot factor out any number from the expression.
So, we cannot determine whether t has any divisors (factors) in common with 12
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Aside:
To demonstrate that statement 2 is NOT SUFFICIENT, consider these two contradictory sets of values for k, m and t.
case a: k=3, m=3, t=15, in which case t and 12 have a common factor (divisor) greater than 1.
case b: k=1, m=3, t=11, in which case t and 12 do not have a common factor (divisor) greater than 1.

Cheers,
Brent