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If n is an integer greater than 50, then the expression (n^2

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by Mike@Magoosh » Mon May 12, 2014 2:50 pm
If n is an integer greater than 50, then the expression (n^2 - 2n)(n^2 - 1) MUST be divisible by which of the following?
I. 4
II. 6
III. 18
(A) I only
(B) II only
(C) I & II only
(D) II & III only
(E) I, II, and III


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Source: — Problem Solving |

by raj44 » Sat May 24, 2014 11:08 pm
I'll go with C.

The given polynomial can be factorized into a product of 4 consecutive integers:

n (n-1)(n-2)(n+1). Now since N>50, plug in any value for N say 51. Therefore, the numbers are 51,50,49 and 52. The product of these numbers is for sure divisible by 4 and 6 , but 18.

This can be realized by using simpler set of 4 consecutive integers say 4,5,6,7 or 1,2,3,4.

I think the condition N>50, is just to make the problem look more complex that actually it is; the bottom line is- plug in numbers of questions of such types and then eliminate the options.

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by GMATGuruNY » Sun May 25, 2014 2:20 am
Mike@Magoosh wrote:If n is an integer greater than 50, then the expression (n^2 - 2n)(n^2 - 1) MUST be divisible by which of the following?
I. 4
II. 6
III. 18
(A) I only
(B) II only
(C) I & II only
(D) II & III only
(E) I, II, and III
(n² - 2n)(n² - 1) = n(n-2)(n+1)(n-1) = (n-2)(n-1)(n)(n+1).
(n-2)(n-1)(n)(n+1) is the product of four consecutive integers.

Of every 4 consecutive integers, at least one will be a MULTIPLE OF 3 and exactly one will be a MULTIPLE OF 4.
Thus, the product of four consecutive integers must be a MULTIPLE OF 12.
Implication:
Statements I and II must be true.
Eliminate any answer choice that does not include both I and II (A, B and D).

If n=51, then (n-2)(n-1)(n)(n+1) = 49*50*51*52 = (7*7)(2*5*5)(3*17)(2*2*13).
The resulting product is not divisible by 18.
Eliminate any remaining answer choice that includes statement III (E).

The correct answer is C.
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by Brent@GMATPrepNow » Sun May 25, 2014 3:48 am
I want to point out that Mitch is using an important property that can be summarized as follows:
The product of k consecutive integers is divisible by k, k-1, k-2,...,2, and 1
So, for example, the product of any 5 consecutive integers will be divisible by 5, 4, 3, 2 and 1
NOTE: the product may be divisible by other numbers as well, but these divisors are guaranteed.

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