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by ketkoag » Fri Mar 27, 2009 7:18 am
Q. The integers m and p are such that 2<m<p and m is not a factor of p. If r is the remainder when p is divided by m, is r > 1 ?

1. the greatest common factor of m and p is 2

2. the least common multiple of m and p is 30

Ans: d
How??
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Source: — Data Sufficiency |
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by goelmohit2002 » Fri Mar 27, 2009 8:07 am
Why not "A" alone is sufficient ?

It says both no.s are +ve and both have minimum difference of 2. Since GCD = 2.

So any two numbers with minimum diff = 2 will yield minimum remainder = 2 IMO.

e.g. 10, 12
6, 10

Is there any set of number where remainder can be 1. It cannot be zero...since it is already told that m is not a factor of p.

Please tell if i am missing something here.
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by 2ndShot » Fri Mar 27, 2009 3:47 pm
From the question we can conclude that the remainder must be at least 1.

Statement (1) tells us that (p-m)/2 > 1, so the remainder of p/m must be greater than 1. SUFFICIENT

Statement (2) tells us the least common multiple of m and p is 30. So let's break it down into prime factors: 30 = 2*3*5
This leaves two possibilities for m and p: m=3 and p=10 or m=5 and p=6
Hence, the remainder of p/m is 1. SUFFICIENT

So the answer is D.
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by goelmohit2002 » Fri Mar 27, 2009 7:25 pm
IMO 2nd statement alone is not sufficient.

For e.g. if m = 6, p =10, then LCM = 30 But R = 4.
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