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LCM

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by knight247 » Mon Jul 18, 2011 11:04 pm
If a and b are consecutive positive integers, is LCM(ab, b+1) divisible by 6?
(1) b is divisible by 3
(2) a<b

OA is D. I'm somehow not able to prove that statement (2) is enough. Detailed explanations would be appreciated.
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Source: — Data Sufficiency |
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by amit2k9 » Mon Jul 18, 2011 11:33 pm
a check for a,b = 2,3 ; 4,3 ; 5,6; 7,6. yes for all. sufficient.

b check for a,b = 1,2; 3,4; 5,4 for 1st two it is yes and for 5,4 it is no.

hence sufficient that a<b.

thus D it is.
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by Frankenstein » Mon Jul 18, 2011 11:34 pm
Hi,
From(2): a < b
As a and b are consecutive positive integers, a = b-1
So, ab = b(b-1). One of b, b-1 is always even. So, b(b-1) is even
Now, b-1,b,b+1 are consecutive numbers. So, one of them should definitely be divisible by 3.
So, either b(b-1) or b+1 is multiple of 3.
So, LCM of these numbers will be a multiple of 2*3 = 6
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by Fiver » Tue Jul 19, 2011 2:39 am
knight247 wrote:If a and b are consecutive positive integers, is LCM(ab, b+1) divisible by 6?
(1) b is divisible by 3
(2) a<b

OA is D. I'm somehow not able to prove that statement (2) is enough. Detailed explanations would be appreciated.
For a number to be divisible by 6, it needs to be divisible by atleast both 2 & 3.
If the 3 numbers in question are in the order a,b,b+1, then obviously the LCM (ab, b+1) is divisible by 6 because atleast 1 is even & atleast 1 is a multiple of 3.

If b is odd & divisible by 3, then both a & b+1 are even and the LCM is divisible by 6.
If b is even & divisible by 3, then itself is divisible by 6.

Given that both a & b are +'ve & consecutive, the only situation when LCM(ab, b+1) is not divisible by 6 is when a & b+1 are the same value & b is an even no. (not divisible by 3) & also preceeding an odd number not divisible by 3.

e.g. 4 & 5, 1 & 2

This situation is ruled out by virtue of statement 2.

Hence D appears correct.
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by srikanthb.69 » Tue Jul 19, 2011 8:03 am
LCM of ab,b+1 is a*b*(b+1) and given that a & b are consecutive +ive integers.
we can say that one of a,b is even.
if a is even then b+1 is also even and all are consecutive numbers so for every three consecutive numbers one of the number is divisible by 3. if a is 3..then LCM is divisible by 6 bec a is even and is divisible by 3 as well.
If b is divisible by 3.. then also the LCM is divisible by 6 bec a is even.

HENCE the question itself contains the answer. SO this might not be a GMAT question.
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by HeintzC2 » Tue Sep 20, 2011 1:23 pm
This is only true given that a<b.
Take a=5, b=4. a=5,b=4,b+1=5, LCM(ab,b+1) = LCM(20,5) = 20. Which is not divisible by 6.
srikanthb.69 wrote:LCM of ab,b+1 is a*b*(b+1) and given that a & b are consecutive +ive integers.
we can say that one of a,b is even.
if a is even then b+1 is also even and all are consecutive numbers so for every three consecutive numbers one of the number is divisible by 3. if a is 3..then LCM is divisible by 6 bec a is even and is divisible by 3 as well.
If b is divisible by 3.. then also the LCM is divisible by 6 bec a is even.

HENCE the question itself contains the answer. SO this might not be a GMAT question.
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