If m is an integer, is m/102 an integer?

Question

(1) 99 is a factor of 165m.
(2) 34 is a factor of 7m.

Bunuel · Accepted Answer

VERITAS PREP OFFICIAL SOLUTION

Correct Answer: C

Explanation: The prime factorization of 102 is 2 x 3 x 17. Therefore, if m has those three primes as factor, it has 102 as a factor.

(1) The prime factorization of 99 is 3 x 3 x 11. These three primes MUST be factors of 165m. However, the factorization of 165m is 3 x 5 x 11 x m. Since 99 has two 3s as factors, one of the 3s must be a factor of m. Although we can prove that m has 3 as a factor, we cannot prove whether it has a 17 or a 2 as a factor, and the statement is insufficient.

(2) The prime factorization of 34 is 2 x 17. These two primes MUST be factors of 7m, which means they must both be factors of m (since they're obviously not factors of 7.) Knowing that 2 and 17 are factors of m isn't enough to answer this question, however, as we still cannot determine whether m has 3 as a factor.

Together, however, we know that 2, 3, and 17 are factors of m. This means, as stated above, that 102 is a factor of m. Because both statements together are sufficient, the correct answer is C.

sterling19 · Answer

The prime factorization of 102 is  2 * 51 , so m must have these factors for m/102 to yield an integer.

(1) 99 is a factor of 165m.      99 = 3 * 3 * 11 ; we still do not know if m contains factors of 2 and 51
(2) 34 is a factor of 7m      34 = 17 * 2 ; we still do not know if m contains a factor of 51

Both statements together show that m has factors of 2, 3, 11, and 17. We still do not know if m contains a factor of 51 (ex. m = 3366).
The correct answer is  E .

gmat6nplus1 · Answer

Riconsidering, the answer should be C

1) m is a multiple of 3. It might also be divisible by 102

2) m is a multple of 17(2). we don't know whether m is divisible by 3.

1+2) LCM = 17(2)3 so m is definitely divisible by 102.

Answer C.

jaxndaxt145 · Answer

I chose C.

102 = (2*3*17)

1) 165M - (5*3*11)M ; 33 = 3 * 11; 99 = 3*11*(3) M must have a factor of 3

2) 7M - (7)M ;34 = 17*2 none of these are 7 so M must have factors of 17 & 2

3, 17 & 2 must be in M only together do we have both sets of info.

EMPOWERgmatRichC · Answer

Hi sterling19,

Your approach to this question is solid: prime-factoring the data (and question) is a great concept-based approach. I think that missed something though.....51 can be prime-factored down into (3)(17).

Knowing this, would your answer change?

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

arkle · Answer

As soon as you see this question, your first step should be to do prime factorization of 102.

102 = 2x3x17

For m/102 to be integer --> 1) m should be divisible by 2,3,and 17 2) or m should contain 2,3, and 17 in its prime factorization.

After doing this much work, you should look at the answer choices. Not before that!

A) 99 is a factor of 165m.

Again do the prime factorization of 99 and 165.

99 = 3x3x11
165 = 3x5x11

This option tells you that when 165m is divided by 99, the result is an integer. Now we can see we have one extra '3', which is of our interest, but we still don't have 2 and 17. Remember our sole purpose is to prove that m has 2,3, and 17 as its factors. There is also one extra 5, but that is useless info.

This option is maybe i.e. insufficient.

B) 34 is a factor of 7m

34 = 2x17

This option tells you that when 7m is divided by 34, the result is an integer. This option tells us that there is 2 and 17 in m, but 3 is missing. Again insufficient. 7 is also available but that is useless info.

Using both the options together, we get 2,3, and 17 in m. So, (C) is our answer.

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If m is an integer, is m/102 an integer?

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If m is an integer, is m/102 an integer?

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