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Source: — Data Sufficiency |
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by cramya » Sat Apr 11, 2009 6:04 pm
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
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stuck at stmt 2

by syr » Sat Apr 11, 2009 6:04 pm
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.
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by syr » Sat Apr 11, 2009 6:08 pm
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).
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by cramya » Sat Apr 11, 2009 6:43 pm
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
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by maihuna » Sat May 09, 2009 11:35 am
Good trick to note here is Even divides Even leaves remainder as even: any mathematical concept to proove this, Ian please?
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by Ian Stewart » Sun May 10, 2009 9:41 am
maihuna wrote:Good trick to note here is Even divides Even leaves remainder as even: any mathematical concept to proove this, Ian please?
Yes, that's certainly true. From the definition of quotients and remainders, when you divide n by d, we have

n = qd + r

where r is the remainder, and q is the quotient. You could rewrite the above:

r = n - qd

and if n is even, and d is even, then r must be even (since on the right side, we have even - even = even).

It's often easier to understand these types of abstract equations by plugging in a few numbers to see how they work.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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