Stmt I

This tells us that m and p are even since they both share a common factor 2

even/even we always get a even remiander if there is a remainder. We know the remiander is not 0 and it cant be 1 so it has to be greater than 1

SUFF

StmtII

x=5 y=6 REMAINDER > 1 NO
x=10 y=15 REMAINDER > 1 YES

INSUFF

A

Given :
m, p integers &
2 < m < p &
p/m = NQ + r ; N - integer, Q - quotient, r - remainder

Is r > 1 ?

Stmt 1)

GCD (m,p) = 2
Plugging numbers -
m p
GCD( 4, 6) = 2
GCD( 8, 10) = 2
So, m, p have to be consecutive even integers for GCD to be 2 & m not a factor of p. Hence, r = 2.
Stmt 1) sufficient.

Stmt 2)
LCM(m, p ) = 30
Plugging numbers -

LCM( 5, 6) = 30
LCM( 3, 10) = 30
So, r = 1

I guess this is what even Sumit arrived at. I see that the OA is A). So we need to find which combination is not satisfying [ LCM(m,p) as 30 and r = 1] .

Experts, opinions please.

Thanks cramya

10, 15... this was the combination !!

But, can you please tell me how you arrived at the combination 10,15 ? Is there an easier way to determine this ?

I could get only (3,10) & (5,6).

Look at the possible combinations of factors of 30

30 = 2*3*5

2*5 = 10
3*5 = 15

General Tip:

For number properties questions try factorizing whats given in to their prime factors. This way its easier to see the connections.It will more often help than not.

Good luck, Syr.

Regards,
CR

Good trick to note here is Even divides Even leaves remainder as even: any mathematical concept to proove this, Ian please?