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even/odd integer

by mtripathy » Tue Sep 18, 2012 10:41 pm
Source:OG 13th Ed.

If n is 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 4 whenever n is even

OA:E

I found all the answer options can be negated by plugging integers. Is there any trick in the question?
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by everything's eventual » Wed Sep 19, 2012 12:32 am
How did you negate E) ? Take the smallest +ve even integer viz. 2

2 *3 *4 is divisible by 4. Take n as any even +ve integer and for that value n(n+1)(n+2) will always be divisible by 4.

eg : 4*5*6 , 24*25*26 , 34*35*36 just to name a few of many.

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by GMATGuruNY » Wed Sep 19, 2012 1:09 am
mtripathy wrote:Source:OG 13th Ed.

If n is 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 4 whenever n is even

OA:E

I found all the answer options can be negated by plugging integers. Is there any trick in the question?
If n is even, then n(n+1)(n+2) = (even)(odd)(even).
Since each of the two even factors must be divisible by 2, the product here must be divisible by 2*2 = 4.
Thus, answer choice E must be true:
(n)(n+1)(n+2) is divisible by 4 whenever n is even.

The correct answer is E.

Take-aways:
If n is even, then n and n+2 are CONSECUTIVE EVEN INTEGERS.
The product of any two consecutive even integers will always be a multiple of 4.

If n is a positive integer, then n(n+1)(n+2) is the product of 3 consecutive integers.
Of every 3 consecutive integers, EXACTLY ONE will be a multiple of 3.
Thus, n(n+1)(n+2) will always be a multiple of 3.
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by mtripathy » Wed Sep 19, 2012 4:10 am
(A) even only when n is even
Lets say n is 3 so 3x4x5=60 even although n is odd (Negated)
(B) even only when n is odd
Lets say n is 2 so 2x3x4=24 even although n is even (Negated)
(C) odd whenever n is odd
Statement itself false as result is always even
(D) divisible by 3 only when n is odd
Lets say n is 4 so 4x5x6=120 which is divisible by 4 although n is even (Negated)
(E) divisible by 4 whenever n is even
Lets say n is 3 so 3x4x5=60 which is divisible by 4 although n is odd (Negated)

Hence all the options can be negated. Can anyone (experts) please explain?
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by GMATGuruNY » Wed Sep 19, 2012 4:36 am
mtripathy wrote: (E) divisible by 4 whenever n is even
Lets say n is 3 so 3x4x5=60 which is divisible by 4 although n is odd (Negated)
Whenever means IF.
Thus, E can be rephrased as follows:
IF N IS EVEN, then n(n+1)(n+2) is divisible by 4.
Since n=3 is not even, it is not a valid case here.
Only EVEN values can be tested here.
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by Brent@GMATPrepNow » Wed Sep 19, 2012 6:11 am
mtripathy wrote:Source:OG 13th Ed.

If n is 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 4 whenever n is even
There's a nice rule we can use to eliminate A, B, C and D.
It says, If there are n consecutive positive integers, then 1 of those integers is divisible by n.

The extension to this is even nicer: The product of n positive integers is divisible by n, n-1, n-2, . . . 1

Example: The product of any 5 positive integers will be divisible by 5, 4, 3, 2 and 1.

At this point, all we need to do is recognize that n, n+1 and n+2 are 3 consecutive integers.
As such, the product n(n+1)(n+2) must be divisible by 3, 2 and 1.
Since the product must be divisible by 3, we can eliminate (D) because it places a restriction on when the product is divisible by 3.
Since the product must be divisible by 2 (i.e., even), we can eliminate A, B and C because they suggest otherwise.

By the process of elimination, we're left with E, the correct answer.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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