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Divisibility problem

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by gmat009 » Tue Oct 07, 2008 11:05 am
If n is an integer greater than 6, which of the following must be divisible by 3?

A. n (n+1) (n-4)
B. n (n+2) (n-1)
C. n (n+3) (n-5)
D. n (n+4) (n-2)
E. n (n+5) (n-6)

Plz. explain
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by 4meonly » Tue Oct 07, 2008 11:24 am
My approach:
to be sure that the product of 3 numbers is divisible by 3 these 3 numbers must be consequtive. One of them will be a multiply of 3.
A. n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1) - it is a multiply of 3

Draw it on the number line and you'll see

A
Correct?
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by gmat009 » Tue Oct 07, 2008 11:28 am
ans is correct but not clear about explanation.
Can you explain this plz.
"n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1)"
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by PussInBoots » Tue Oct 07, 2008 11:28 am
4meonly is right. It's tricky problem and would take some time to solve. If I get problem like that, I'd come up with some generic number larger than 6, say 10 and do calculations. It'd be faster in some cases
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by PussInBoots » Tue Oct 07, 2008 11:30 am
gmat009 wrote:ans is correct but not clear about explanation.
Can you explain this plz.
"n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1)"
n * (n + 1) * (n - 4) = n(n+1)(n-1-3)=n(n+1)(n-1) - (n+1)*3n
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by gmat009 » Tue Oct 07, 2008 12:08 pm
PussInBoots wrote:
gmat009 wrote:ans is correct but not clear about explanation.
Can you explain this plz.
"n (n+1) (n-4) can be expressed as a product of 3 numbers. It is the same as n (n+1) (n-1)"
n * (n + 1) * (n - 4) = n(n+1)(n-1-3)=n(n+1)(n-1) - (n+1)*3n
Got it..
Thanks....
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by cramya » Tue Oct 07, 2008 2:04 pm
Pick 7 and 8 for n and you can eleiminate everyhting but A.

Like one of the other user mentioned the process of elimination will work here.
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