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

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Re: n is a positive integer [#permalink] New post 01 Sep 2011, 18:53
The product of two even integer is always divisible by 4. So, the definite answer is E.
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Re: If n is positive integer, then n(n+)(n+2) is [#permalink] New post 27 Aug 2013, 01:56
I am assuming the expression is n(n+1)(n+2)

a) This not true. If n is even, (n+1) will be odd and if n is odd (n+1) will be even. It is sufficient for any one of the terms to be even to make the entire expression even.

b) For the same reasons as "a" this is alos not true.

c) This is also not true. It is sufficient for (n+1) alone to be a multiple of 3 for the entire expression to be divisible by 3. eg : n=2

e) Whenever n is even, (n+2) will also be even. Hence the expression will have two even integers and hence is always divisible by 4.
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Re: PS: Odd/Even and Divisibility [#permalink] New post 30 Oct 2013, 12:08
avdxz 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 4 whenever n is even


E..whenever ..the equation is a product of three consecutive numbers and when the first is even the last will also be an even nmber...thus we will have two two's in the final product and thus the number will be divisible by four...
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Re: 224. Arithmetic operation [#permalink] New post 22 Apr 2014, 05:07
Bunuel wrote:
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.


can we consider 0 as positive integer. If yes then whole number will become 0 and we can not divide 0 by 4.

Please clarify
Thanks.
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Re: 224. Arithmetic operation [#permalink] New post 22 Apr 2014, 06:16
Expert's post
PathFinder007 wrote:
Bunuel wrote:
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.


can we consider 0 as positive integer. If yes then whole number will become 0 and we can not divide 0 by 4.

Please clarify
Thanks.


Three remarks here:

1. Zero is neither positive, nor negative integer.

2. Division by zero is not allowed (number/0 is undefined) but we can divide 0 by any non-zero number (0/number=0).

3. Zero is divisible by EVERY integer except zero itself, since 0/integer=0=integer (or, which is the same, zero is a multiple of every integer except zero itself). Thus even if we were told that n is integer (so if n could be a negative integer, zero or a positive integer) n(n + 1)(n + 2) would still be divisible by 4 for any even n.

Hope it's clear.
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Re: 224. Arithmetic operation   [#permalink] 22 Apr 2014, 06:16
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