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

By the way, here's another question we can solve using prime factorization: https://www.beatthegmat.com/divisibility-t91797.html