Data Sufficiency

Ahhh, I can't believe I did that. Thanks for correcting me.

Ron, as my post stated above, do you know of a similar algebraic approach that we can apply to S2. As I mentioned, the first time around I got the question wrong and chose D because I failed to consider the right numbers for LCM 30.

Thanks,

Hello experts,

I was going through Ron's solution to this question. I realized that we have to pick numbers whose LCM is 30. Could you please let me know how to pick such numbers? Its easy to find the LCM when the numbers are given but to find those 2 numbers when the LCM is given is a bit confusing.

Regards,
Amit

It is quite simple.

The question asks us if the remainder is greater than one or not?

lets assume the remainder is 0 or 1.

Scenario 1 (remainder is zero):
This is not possible due to the fact that p is not divisible by m.

Scenario 2 (remainder is one):
This is not also possible because consecutive numbers can have a remainder of one. and since both numbers are even (GCF is 2), the remainder is greater than one.

As a result, Statement 1 is sufficient for answering the question.

Lets take Statement 2:
LCM is 30. Lets state all of possible numbers that m and p can have:
1 2 3 5 6 10 15 30

Well, 1 and 2 will be omitted (the question tells us m and p are greater than 2) and the others will remain:

if p=10 and m=6, the remainder will be greater than 1
and
if p=6 and m=5, the remainder will be one.

So statement 2 is insufficient for answering the problem and the answer will be A.

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