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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +

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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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New post Updated on: 01 Jul 2017, 01:49
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C
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Question Stats:

66% (02:03) correct 34% (01:59) wrong based on 146 sessions

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If n is an integer greater than 10, then the expression (n^2 - 2n)(n + 1)(n - 1) MUST be divisible by which of the following?

I. 4
II. 6
III. 18

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

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Originally posted by mkrishnabdrr on 01 Jul 2017, 01:30.
Last edited by Bunuel on 01 Jul 2017, 01:49, edited 1 time in total.
Edited the question.
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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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New post 01 Jul 2017, 01:47
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The expression (\(n^2\)-2n)(n+1)(n-1) can be expressed as n(n-2)(n-1)(n+1),
which is the product of 4 consecutive integers. It has also been given that the number n>10

We know that the product of 4 consecutive numbers is always divisible by \(24(2^3*3)\)
Rule : The product of any n consecutive integers will be always divisible by n!

Hence, the expression must be divisible by \(4(2^2)\) and \(6(2*3)\) but not \(18(2*3^2)\)
The answer is Option C(i and ii only)
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Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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New post 20 Jul 2017, 08:11
pushpitkc wrote:
The expression (\(n^2\)-2n)(n+1)(n-1) can be expressed as n(n-2)(n-1)(n+1),
which is the product of 4 consecutive integers. It has also been given that the number n>10

We know that the product of 4 consecutive numbers is always divisible by \(24(2^3*3)\)
Rule : The product of any n consecutive integers will be always divisible by n!

Hence, the expression must be divisible by \(4(2^2)\) and \(6(2*3)\) but not \(18(2*3^2)\)
The answer is Option C(i and ii only)


pushpitkc can you please share a link related to shared rule please
TIA
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If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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New post 21 Jul 2017, 08:30
ehsan090 wrote:
pushpitkc wrote:
The expression (\(n^2\)-2n)(n+1)(n-1) can be expressed as n(n-2)(n-1)(n+1),
which is the product of 4 consecutive integers. It has also been given that the number n>10

We know that the product of 4 consecutive numbers is always divisible by \(24(2^3*3)\)
Rule : The product of any n consecutive integers will be always divisible by n!

Hence, the expression must be divisible by \(4(2^2)\) and \(6(2*3)\) but not \(18(2*3^2)\)
The answer is Option C(i and ii only)


pushpitkc can you please share a link related to shared rule please
TIA

ehsan090 , the rule is under Consecutive Integers, here:

https://gmatclub.com/forum/math-number-theory-88376.html
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Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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New post 24 Jul 2017, 03:13
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I took two values, i.e. 11 and 12 and applied them to the given formula. Only 4 and 6 satisfied both. While 18 satisfied only when n=11.
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Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +  [#permalink]

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Re: If n is an integer greater than 10, then the expression (n^2 - 2n)(n +   [#permalink] 29 Mar 2019, 11:08
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