LCD and GCM

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LCD and GCM

by NehaPathak » Sat Sep 24, 2011 5:22 am
If x and y are positive integer then xy is a multiple of 8?

A)Least common divisor is 10.

B)Greatest common multiple is 100.




Please help me i really got confused......

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by cans » Sat Sep 24, 2011 5:30 am
A) x=y=10,not
x=2,y=5, not
sufficient

B) x=y=100, yes. thus sufficient
IMO D
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by shankar.ashwin » Sat Sep 24, 2011 5:34 am
Just remember;

Product of 2 Nos = [LCM of the 2 Nos] * [HCF of the 2 Nos]

(1) & (2) gives you both individually but you need both.

So, C IMO
NehaPathak wrote:If x and y are positive integer then xy is a multiple of 8?

A)Least common divisor is 10.

B)Greatest common multiple is 100.




Please help me i really got confused......

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by Brent@GMATPrepNow » Sat Sep 24, 2011 6:38 am
NehaPathak wrote:If x and y are positive integer then xy is a multiple of 8?

A)Least common divisor is 10.
B)Greatest common multiple is 100.

Please help me i really got confused......
A few problems with the wording here. Least common divisor? Greatest common multiple?

I'm assuming that the question should read:

If x and y are positive integers, is xy a multiple of 8?
1)The greatest common divisor of x and y is 10
2)The least common multiple of x and y is 100


Let's find contradictory values for x and y to show that each statement alone is not sufficient.

Statement 1:
Consider 2 different cases that satisfy the condition that the greatest common divisor of x and y is 10
case a: x=10 and y=10, in which case xy is not a multiple of 8
case b: x=10 and y=100, in which case xy is a multiple of 8
Statement 1 is not sufficient

Statement 2:
Consider 2 different cases that satisfy the condition that the least common multiple of x and y is 100
case a: x=4 and y=25, in which case xy is not a multiple of 8
case b: x=10 and y=100, in which case xy is a multiple of 8
Statement 2 is not sufficient

Statements 1 & 2:
We have a nice rule that says "[GCD of x and y][LCM of x and y] = xy"
So, from statements 1 and 2, we know that "[10][100] = xy"
So, xy = 1000 and 1000 is a multiple of 8
Since we can now answer the target question with certainty, we can see that the answer is C.

Cheers,
Brent
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by NehaPathak » Sun Sep 25, 2011 6:03 am
Thank you so much Brent. It really helps alot.