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CaliCpa
I definitely see your point. Maybe I am just overthinking this. For these types of problems, is it easiest to just take the LCM, and if it is less than what the question is asking for (i.e. 168), then you know its a "maybe" or "indeterminable?" I feel pretty comfortable with the LCM and GCF calculations.


I wouldn't recommend this method. It is much better to try to understand what the question is actually asking and break it down.


CaliCpa
"It asks the question: Is P divisible 168?

This is asking whether P is divisible by 2^3, 7, and 3

(1) Says P is divisible by 2 and 7 (insufficient, we don't know if it is divisible by the remaining factors of 168)
(2) Says P is divisible by 2^2 and 3 (insufficient, we are still missing a 2 and 7)

Together says P is divisible by 2^2, 3, and 7 (because it is possible that the 2 from statement 1 and one of the 2s from statement 2 are in fact the same 2) so it is still insufficient

For example, if Statement two said P was divisible by 24, (which is divisible by 2^3 and 3) then we could say P is divisible by 168.

It will also help you to know that "factors, prime numbers, and divisibility" all mean the same thing
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Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
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CaliCpa
I definitely see your point. Maybe I am just overthinking this. For these types of problems, is it easiest to just take the LCM, and if it is less than what the question is asking for (i.e. 168), then you know its a "maybe" or "indeterminable?" I feel pretty comfortable with the LCM and GCF calculations.

Yes quite so. But consider this, in some cases we would be able to say that number (p in your case) is not divisible by some number.

Let's consider your case:

Is p divisible by 168?

(1) p is divisible by 14=2*7 Not sufficient
(2) p is divisible by 12=2^2*3 Not sufficient

(1)+(2) LCM 84--> p is for sure divisible by 84, but for 168 we don't know. Still not sufficient.


Let's consider another case say:

Is p divisible by 168?

(1) p is divisible by 84
(2) p is not divisible by any integer n>1 in power of 3

The answer would be different.
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