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M23-12

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M23-12  [#permalink]

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New post 16 Sep 2014, 01:18
1
8
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

52% (01:20) correct 48% (01:24) wrong based on 155 sessions

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Re M23-12  [#permalink]

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New post 16 Sep 2014, 01:18
1
Official Solution:


First notice that \(n^2-n=n(n-1)\).

(1) \(n\) is divisible by 11. If \(n=11\), then \(n(n-1)=11*10\) and the answer is NO but if \(n=11*12\), then \(n(n-1)=11*12*(11*12-1)\) and the answer is YES. Not sufficient.

(2) \(n\) is divisible by 19. If \(n=19\), then \(n(n-1)=19*18\) and the answer is NO but if \(n=19*12\), then \(n(n-1)=19*12*(19*12-1)\) and the answer is YES. Not sufficient.

(1)+(2) From above we have that \(n\) is divisible by \(11*19=209\). Now, if \(n=11*19\), then \(n(n-1)=11*19*(209-1)=11*19*208\) and the answer is NO (since \(11*19*208\) is not divisible by 3, then it's not divisible by \(12=3*4\) either) but if \(n=11*19*12\), then \(n(n-1)=11*19*12*(11*19*12-1)\) and the answer is YES. Not sufficient.


Answer: E
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Re: M23-12  [#permalink]

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New post 25 Nov 2016, 00:48
Can you explain how you know that n(n−1)=11∗12∗(11∗12−1) is divisible by 12?
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Re: M23-12  [#permalink]

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New post 25 Nov 2016, 00:57
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Re: M23-12  [#permalink]

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New post 07 May 2018, 19:24
The explanation is good. This is a high quality question.
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Re: M23-12  [#permalink]

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New post 05 Jul 2018, 20:18
My try is by doing with statement.
1. for being divisible by 11 let try with number 11. 11^2-11=110/12 which is not divisible. so first statement cannot be true.
2.for being divisible by 19 let try with number 19. 19^2-19=342/12 which is not divisible. so 2nd statement cannot be true.

now let combine both 11*19
for being divisible by 11*19 let try with LCM of both number 209. 209^2-209=209(209-1)/12= (209*208)/12 since none of these 208 and 209 is divisible by 3 thus t is not divisible by 12 too. so even combining these two wont solve problem.
so E is only ans left

LET ME KNOW IF I AM WRONG.
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Re: M23-12  [#permalink]

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New post 05 Jul 2018, 20:33
1
aravsha wrote:
My try is by doing with statement.
1. for being divisible by 11 let try with number 11. 11^2-11=110/12 which is not divisible. so first statement cannot be true.
2.for being divisible by 19 let try with number 19. 19^2-19=342/12 which is not divisible. so 2nd statement cannot be true.

now let combine both 11*19
for being divisible by 11*19 let try with LCM of both number 209. 209^2-209=209(209-1)/12= (209*208)/12 since none of these 208 and 209 is divisible by 3 thus t is not divisible by 12 too. so even combining these two wont solve problem.
so E is only ans left

LET ME KNOW IF I AM WRONG.



You are wrong the way you have written..
in DS, a statement is sufficient if it can give a definite answer - definite YES or Definite NO
here you have just worked on getting a NO as the answer. you have to further check if it is true that there is some case where it will be div by 12 too
if there is a case , it is insufficient then only
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: M23-12 &nbs [#permalink] 05 Jul 2018, 20:33
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