Factors Problem

[leumas]

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 4.

[knight247]

Firstly, make it easier on the eyes by multiplying throughout by 12. So we have,
2k+3m=t

(1)k is a multiple of 3 so 2k has to be a multiple of 3. Remember one ground rule,
If two multiples of a certain number are added/subtracted then the resultant number is also a multiple of the same number. So we have Multiple of 3+Multiple of 3=Multiple of 3. Hence, t is a multiple of 3. And, 12 is also a multiple of 3. So, t and 12 do have a common factor greater than 1 i.e. 3. Sufficient

(2)m is a multiple of 4 so 3m is a multiple of 4. And by default, 3m becomes a multiple of 2 as well.
So we have Multiple of 2+Multiple of 2=Multiple of 2. So t is a multiple of 2. 12 and 2 have a common factor greater than 1 i.e. 2. Hence Sufficient

The answer is D

[navami]

Either of them

[Anurag@Gurome]

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 4.

It is given that k/6 + m/4 = t/12.
Multiply both right and left hand side with 12.
2k + 3m = t.

(1) k is a multiple of 3 implies that k = 3n, n being a positive integer.
6n + 3m = t.
t = 3(m + 2n), which means that t is a multiple of 3.
So, t and 12 will always have a common factor of 3 apart from 1; SUFFICIENT.

(2) m is a multiple of 3 implies that m = 3n, n being a positive integer.
t = 9n + 2k.
This clearly does not indicate whether t and 12 have any common factor apart from 1; NOT sufficient.

The correct answer is A.

[Brent@GMATPrepNow]

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 4.

Here's another question we can solve using prime factorization (I just solved this divisibility question using prime factorization: https://www.beatthegmat.com/divisibility-t91797.html)

First, for questions involving multiples, we can say:
If N is a multiple of w, then w is "hiding" within the prime factorization of N
Examples:
24 is a multiple of 3 <--> 24 = 2x2x2x3
70 is a multiple of 5 <--> 70 = 2x5x7
330 is a multiple of 6 <--> 330 = 2x3x5x11

Okay, let's begin by taking the target question "Do t and 12 have a common factor greater than 1?" and rewrite it using prime factorization. We can reword it as "Do the prime factorizations of t and 12 share a number greater that 1?"

Finally, before we examine the statements, let's take the given information, (k/6)+ (m/4) = (t/12), and make it "nicer" 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 we may or may not have primes other than the 3 in the prime factorization of k. All we can be certain of is that there is one 3 within the prime 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
If 3 is a divisor of t, then 3 is in the prime factorization of t
Since the prime factorization of 12 also has a 3 in it, we can see that we can now answer the reworded target question.
As such, statement 1 is sufficient.



Statement 2: m is a multiple of 4.
In other words, 4 is hiding in the prime factorization of m
So, we know that m =(2)(2)(?)(?)(?)...

From here, we'll take our given information, 2k + 3m = t and replace m with (2)(2)(?)(?)(?) to get: 2k + 3(2)(2)(?)(?)(?) = t
At this point, we can factor out a 2 to get: 2[k + 3(2)(?)(?)(?) = t, which means 2 is a divisor of t
If 2 is a divisor of t, then 2 is in the prime factorization of t
Since the prime factorization of 12 also has a 2 in it, we can see that we can now answer the reworded target question.
As such, statement 2 is sufficient, and the answer is D

Cheers,
Brent