tapan22 wrote:
ps_dahiya wrote:
Either there is a typo or I am a jerk
Following is the given information:
1. m and p are integers
2. 2<m<p
3. m is not a factor of p.
If r is the remainder when p is divided by m, is r>1? I think r will always be greater than 1. Because
1. p is atleast 1 greater than m because of conditions 1 and 2 above.
2. r can not be zero because of condition 3 above.
3. a remainder is never a real number.
So we don't need to read the statements to answer the question?????
consider 2<3<4
4/3--> remainder = 1 which is not greater than 1.
r will not always be greater.
I would go with A here.
St1: GCD is 2 so that means they are both even and will differ by at least 2. So Suff.
St2.: Stands for 10 and 3 but fails for 5 and 6 so Insuff.
Hence A
Just to verify my thought process...
So C1. states that the GCD will also be even but with a multiple that is both odd and not factors of each other. (the oddness takes care of the C1 because every even number is divisible by 2 and since they are multiples that are not factors of each other eliminates the possibility that they are factors.)
So for example
6 (2*3) and 10 (2*5) 10/6 R=4
10 (2*5) and 12 (2*2*3) 12/10 R=2
So R>1 because you are dividing two different odd multiples that are not factors of each other (3,5,7,11,13), hence the difference will always be at least 2.
C2. gives 5,6 ---> 6/5 R=1
or 15/6 15/6 ---> R=3
INSUFFICIENT
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