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If n is a positive integer, then n(n+1)(n+2) is

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If n is a positive integer, then n(n+1)(n+2) is  [#permalink]

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New post 11 Aug 2018, 09:58
If n is a positive integer, then n(n+1)(n+2) is

A. even only when n is even
B. even only when n is odd
C. odd whenever n is odd
D. divisible by 3 only when n is odd
E. divisible by 12 whenever n is even


I understand why A through D can be eliminated because the product above will always be even since there is an even number in the product itself. n(n+1)(n+2) will also always be divisible by 3 because in any given consecutive series you will always have atleast 1 multiple of 3 included.

However for answer E, N can even AND odd and be divisible by 12:

n = 2, then 2*3*4 is divisible by 12,

BUT N can also be odd and divisible by 12:
n = 3, then 3*4*5 is divisible by 12.

So why doesn't E include that n can also be odd?
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Re: If n is a positive integer, then n(n+1)(n+2) is  [#permalink]

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New post 11 Aug 2018, 10:15
kungfuquant wrote:
If n is a positive integer, then n(n+1)(n+2) is

A. even only when n is even
B. even only when n is odd
C. odd whenever n is odd
D. divisible by 3 only when n is odd
E. divisible by 12 whenever n is even


I understand why A through D can be eliminated because the product above will always be even since there is an even number in the product itself. n(n+1)(n+2) will also always be divisible by 3 because in any given consecutive series you will always have atleast 1 multiple of 3 included.

However for answer E, N can even AND odd and be divisible by 12:

n = 2, then 2*3*4 is divisible by 12,

BUT N can also be odd and divisible by 12:
n = 3, then 3*4*5 is divisible by 12.

So why doesn't E include that n can also be odd?


Option E states WHENEVER and not ONLY WHEN
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Re: If n is a positive integer, then n(n+1)(n+2) is   [#permalink] 11 Aug 2018, 10:15

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