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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everything's eventual
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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.
Regards.
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.
Regards.
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If n is even, then n(n+1)(n+2) = (even)(odd)(even).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?
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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(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?
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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Whenever means IF.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)
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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There's a nice rule we can use to eliminate A, B, C and D.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
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
