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If n is an integer greater than 50, then the expression (n^2
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12 May 2014, 14:46
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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 IIIFor a discussion of how to use the properties of consecutive integers to unlock problems such as this, see: http://magoosh.com/gmat/2014/consecutiv ... thegmat/Mike
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Re: If n is an integer greater than 50, then the expression (n^2
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12 May 2014, 19:12
(n^22n)(n^21) = (n1)n(n+1)(n^22)
ii. Product of 3 consecutive integers is always divisible by 3 and since one of n1,n,n+1 is even => The product is divisible by 6
i. n= odd => n1 and n+1 are even, so the product is divisible by 4 n= even => n and n^22 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
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13 May 2014, 02:35
(n^22n)(n^21) = n(n2)(n1)(n+1) or (n2)(n1)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
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23 Nov 2016, 06:20
We have product of \(4\) consecutive integers \((n2)*(n1)*n*(n+1)\)
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
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14 Aug 2014, 00:05
Plugged in small, nonobvious 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
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25 Nov 2016, 15:27
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
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25 Nov 2016, 15:55
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. Pluggingin 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
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25 Nov 2016, 16:15
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
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02 Sep 2017, 09:40
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 IIIFor a discussion of how to use the properties of consecutive integers to unlock problems such as this, see: http://magoosh.com/gmat/2014/consecutiv ... thegmat/Mike \((n^2  2n)(n^2  1)\) =n(n2)(n1)(n+1) =(n2)(n1)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
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09 Jan 2020, 17:16
I came to a surprising find for such problems.
Product of n consecutive integer numbers is divisible by n. For example, take any 10 consecutive nos. then you will get at least one multiple of 10.




Re: If n is an integer greater than 50, then the expression (n^2
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09 Jan 2020, 17:16






