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??
DS
This topic has expert replies
-
- Legendary Member
- Posts: 1799
- Joined: Wed Dec 24, 2008 3:03 am
- Thanked: 36 times
- Followed by:2 members
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.
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.
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.
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.
-
- Legendary Member
- Posts: 1799
- Joined: Wed Dec 24, 2008 3:03 am
- Thanked: 36 times
- Followed by:2 members