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

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.

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.

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

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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