3. True or not n(n+1) will always be
even, as one out of n or n+1 must be
even
Answers
Step-by-step explanation:
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GMAT Club Forum Index Problem Solving (PS)
If n is a positive integer, then n(n+1)(n+2) is : Problem Solving (PS)
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avdxz
Updated on: Oct 30, 2013


00:00
A
B
C
D
E
DIFFICULTY:
   15% (low)
QUESTION STATS:
based on 1200 sessions
80% (01:22) correct
20% (01:30) wrong
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 4 whenever n is even
Spoiler: OA
Last edited by Bunuel on 30 Oct 2013, 23:52, edited 1 time in total.
RENAMED THE TOPIC.
Kudos
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Most Helpful Expert Reply
Bunuel
EXPERT'S
POST
Feb 24, 2011
Baten80 wrote:224. 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 4 whenever n is even
n(n + 1)(n + 2) is the product of 3 consecutive integers. The product of 3 consecutive integers is ALWAYS divisible by 2 and 3 (generally the product of k consecutive integers is always divisible by k!, check this: defined-functions-108309.html), so n(n + 1)(n + 2) is always even and always divisible by 3: A, B, C and D are out.
Answer: E.
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Kudos
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General Discussion
game over
Jul 10, 2006
(E)
Assume: n is even, then either n or n+2 is a multiple of 4. Hence, n(n+1)(n+2) is divisible by 4.
Therefore: whenever n is even, the term above is divisble by 4.
Kudos
haas_mba07
Jul 10, 2006
(E) divisible by 4 whenever n is even
If n is even => even x odd x even (Prod of two even numbers always divisible by 2x2)
Kudos
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pike
Jul 1, 2011
Manhattan NP covers these well.
Will be of the form
Odd, even, odd = even
Even, odd, even = even
Can quickly rule out all but E
Posted from my mobile device
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ankushjain
Jul 1, 2011
siddhans wrote:How to solve this?
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 4 whenever n is even
n(n+1)(n+2) will always be even as n is a +ve integer so that rules out A, B & C. Atleast one of n, n+1 & n+2 will be even as they are consecutive integers.
even * even is always even e.g 2*4 = 8 or 6*10 = 60 always even
even * odd is always even e.g 2*3 = 6 or 5 * 8 = 40 always even
Either of n, n+1 & n+2 will always be divisible by 3 till the time n is a +ve integer and they are consecutive integers.
Hence that leaves us with E as answer.
Also we can prove it like this way also,
First +ve even integer is 2 and not 0 (0 is neither +ve nor -ve).
so n*n+1*n+2 = 2*3*4 divisible by 4.
or if n=6 then 6*7*8 again divisible by 4.
so is E.
Answer:
n(n + 1)(n + 2) is the product of 3 consecutive integers. The product of 3 consecutive integers is ALWAYS divisible by 2 and 3 (generally the product of k consecutive integers is always divisible by k!, check this: defined-functions-108309.html), so n(n + 1)(n + 2) is always even and always divisible by 3: A, B, C and D are out.