gprep ds remainder problem

[raghavsarathy]

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

[tohellandback]

IMO A

1)The greatest common factor of m and p is 2 .
both m and p are even. if m is not a factor of p, remainder will always be atleast 2.
ex, 4,6
4,10
8,20
20,22
20,98

Sufficient

2)The least common multiple of m and p is 30.

m=5,p=6, r=1
m=6,p=15,r=3
not sufficient

[cubicle_bound_misfit]

what the question is asking ?

1. m and p both greater than 2

now as p is not a factor of m then

p%m can have values like 1, 2, .... m-1

so only case where p%m will have a value of 1 is when m and p are consecutive integers , i.e 4,5 ... 6,7.... 9,10 etc and their GCF is 1. and LCM is = m*p

stmt 1 says

GCF is 2 therefore remainder is always >1. SUFF

stmt 2 says LCM is 30 , it can have multiple possibilities ie m = 5 , p =6 remainder is 1
but m =3 and p =10 remainder is >1. Hence Insufficient.

Hope that helps.

[raghavsarathy]

Thanks a lot guys.

OA - A

I made a mistake in choosing D. For statement 2 I took a couple of number pairs and always go a remainder of 1. I had not taken 6 and 15 !