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

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If n is an integer greater than 50, then the expression (n^2 [#permalink]

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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


For a discussion of how to use the properties of consecutive integers to unlock problems such as this, see:
http://magoosh.com/gmat/2014/consecutiv ... -the-gmat/

Mike :-)
[Reveal] Spoiler: OA

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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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New post 12 May 2014, 20:12
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(n^2-2n)(n^2-1) = (n-1)n(n+1)(n^2-2)

ii. Product of 3 consecutive integers is always divisible by 3 and since one of n-1,n,n+1 is even => The product is divisible by 6

i. n= odd => n-1 and n+1 are even, so the product is divisible by 4
n= even => n and n^2-2 are even, so the product is divisible by 4

iii. for the expression to be divisible by 18, the product should have 3,3,2
lets consider n = 100 and n= 101
n=100, 99*100*101*9998 => 99 has two threes and overall expression has plenty of 2's
n=101, 100*101*102*10199 => plenty of 2's but no 3 (sum of digits of 10199 = 20; not divisible by 3; hence 10199 not divisible by 3)

Therefore, i & ii, Hence C
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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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New post 13 May 2014, 03:35
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(n^2-2n)(n^2-1) = n(n-2)(n-1)(n+1) or (n-2)(n-1)n(n+1)
this represents product of 4 consecutive integers.

Out of these 4 integers, two will be even and two will be odd.

If the first term is divisible by 3 then the last term will also divisible by 3. (check by taking 4 consecutive integers as 54,55,56,57)
If the first term is not divisible by 3 then out of 4 consecutive integers, only one will be divisible by 3. (check by taking 4 consecutive integers as 52,53,54,55)

Hence 4 consecutive expressions may contain minimum 1 and maximum two integers divisible by 3.

Divisibility by 4: Product of two even number is always divisible by 4. hence expression is divisible by 4.
Divisibility by 6: Product of an even number and a number divisible by 3 will be divisible by 6.
Divisibility by 18: Product of an even number and two numbers divisible by 3 will be divisible by 18. However, if first number is not divisible by 3, there will be only 1 (not 2) number divisible by 3. Therefore we can't be sure that there will be 2 numbers divisible by 3. Hence divisibility by 18 is not sure.

The expression is divisible by 4 and 6 only . Hence C
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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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New post 14 Aug 2014, 01:05
Plugged in small, non-obvious numbers

For n = 2, result = 2*3*2 .... Divisible by 4 & 6

For n = 5, result = 5*24*23 ....... Divisible by 4 & 6

Answer = C
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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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We have product of \(4\) consecutive integers \((n-2)*(n-1)*n*(n+1)\)

The product of any \(n\) consecutive integers will be always divisible by \(n!\). In our example this will be \(4!=24=2^3*3^1\)

\(4=2^2\)

\(6=2*3\)

Only \(18 = 2*3^2\) has more facros of \(3\) than \(4!\)

Hence answer C.

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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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New post 25 Nov 2016, 16:27
Hi,

Can you help me clarify something. if the stem says that n is greater than 50, should not I use 51=n as the smallest test number.
if so, i get.... 49*50*51*52?

Your insight is appreciated. I got the wrong answer of E, but if you can help me close the gap. I got stuck in the words "if n is an integer greater than 50"

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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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New post 25 Nov 2016, 16:55
lalania1 wrote:
Hi,

Can you help me clarify something. if the stem says that n is greater than 50, should not I use 51=n as the smallest test number.
if so, i get.... 49*50*51*52?

Your insight is appreciated. I got the wrong answer of E, but if you can help me close the gap. I got stuck in the words "if n is an integer greater than 50"

Dear lalania1,

I'm the author of this question and I am happy to respond. :-)

My friend, with all due respect, it is a HUGE mistake to approach this a plug in problem. One would get absurdly large numbers if one used that method. Plugging-in numbers is not at all the best way to approach this problem. See the OE on this blog article.

Does all this make sense?
Mike :-)
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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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New post 25 Nov 2016, 17:15
Hi Mike,

Yes, I see your point. In essence, the question says "when will the condition MUST apply" for all numbers. Using the logic of consecutive integers and the solution steps you suggest I can clearly see how it works.

thanks Mike. I am about to finish your videos on Number Properties and then ready to take the 5 question quiz.

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Re: If n is an integer greater than 50, then the expression (n^2 [#permalink]

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New post 02 Sep 2017, 10:40
mikemcgarry 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


For a discussion of how to use the properties of consecutive integers to unlock problems such as this, see:
http://magoosh.com/gmat/2014/consecutiv ... -the-gmat/

Mike :-)


\((n^2 - 2n)(n^2 - 1)\)
=n(n-2)(n-1)(n+1)
=(n-2)(n-1)n(n+1)
So, its a multiple for 4 consecutive integers, which means there are two even numbers and two odd numbers. So it must be divisible by 4.
Also among the 4 consecutive numbers, there must be atleast one multiple of 3. So, it must be divisible by 6.
Now the number may or may not be divisible by 18 = 3*3*2.

Lets check for a value of n = 51
So number = 49*50*51*52

Bingo this no is not divisible by 18.
hence (I) & (II) only . Answer C
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Re: If n is an integer greater than 50, then the expression (n^2   [#permalink] 02 Sep 2017, 10:40
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