Is 3 a factor of the positive integer m?

Hi All,

This type of question can be solved by using 'prime factorization.' Before getting into the specifics of this question, here's how prime factorization "works"....

For a number to be a FACTOR of another number, the larger number has to have the same "prime pieces" that the smaller number has.

For example...
10 is a factor of 30 because....
10 = (2)(5) and 30 = (2)(3)(5)

Notice how 30 has a "2" and a "5"? That's why 10 is a factor of 30

10 is NOT a factor of 35 though, because....
10 = (2)(5) and 35 = (5)(7)

Notice how 35 does NOT have the "2" that we need? That's why 10 is NOT a factor of 35.

This same principle is at play in this question. We just have to account for the variable M and what it MUST (or MAY) contain.

We're asked if 3 is a factor of the positive integer M. So this question is really asking "when we prime factor M, is there a "3" in there?" This is a YES/NO question.

Fact 1: 12 is a factor of 25M

Using the above "steps", we know....
12 = (2)(2)(3)
25M = (5)(5)(M)

Since we're told that 12 IS a factor of 25M, the "2", the other "2" and the "3" have to be somewhere. They MUST be in the M. This means that the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT.

Fact 2: 12 is a factor of 15M

12 = (2)(2)(3)
15M = (3)(5)(M)

Here, since 12 is a factor of 15M, the "2" and the other "2" MUST be in the M, but there's already a "3" there. If there's ANOTHER "3" in the M, then the answer to the question is YES. But there might not be one in there; in that case, the answer to the question is NO.
Fact 2 is INSUFFICIENT.

Final Answer:

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

(1) 12 is a factor of 25m ; as 12 and 25 doesnt have any factor in common (except 1) so m surely has 12 in it. SUFFICIENT.
(2) 12 is a factor of 15m[/quote] as 15 and 12 share 3 as acommon factor so m may ot my not have 3 in it . INSUFFICIENT.

answer a

Here From statement 1 => m=12*P for some integer P
so m/3 must be an integer so => sufficient
from statement 2 => m=4*P so m/3 may or may not be an integer .
hence A

Target question: Is 3 a factor of the positive integer m?

This is a good candidate for rephrasing the target question.

-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:

If N is a factor by k, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)
-----BACK TO THE QUESTION!---------------------

The above concept allows us to REPHRASE the target question as...
REPHRASED target question: Is there a 3 hiding in the prime factorization of m?

Statement 1: 12 is a factor of 25m
12 = (2)(2)(3)
In other words, statement 1 is telling us that there are two 2's and one 3 hiding in the prime factorization of 25m
25m = (5)(5)(m)
Since there are no 3's hiding in (5)(5), it must be the case that there's a 3 hiding in the prime factorization of m
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: 12 is a factor of 15m
12 = (2)(2)(3)
In other words, statement 2 is telling us that there are two 2's and one 3 hiding in the prime factorization of 15m
15m = (3)(5)(m)
As we can see, we already have a 3 hiding in the prime factorization of 15m, so we can't say for sure whether there's a 3 hiding in the prime factorization of m.
To more certain, consider these two counter-examples:
Case a: m = 4. This means 15m = (15)(4) = 60, and 12 IS a factor of 60. In this case, 3 is NOT a factor of m
Case b: m = 12. This means 15m = (15)(12) = 180, and 12 IS a factor of 180. In this case, 3 IS a factor of m
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

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