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Re M2701
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16 Sep 2014, 00:26
Official Solution: (1) \(p\) is a factor of \((n+2)!n!\). If \(n=2\) then \((n+2)!n!=22\) and for \(p=2\) the answer will be YES, but for \(p=11\) the answer will be NO. Not sufficient. (2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\). Simplify: \(\frac{(n+2)!}{n!}=(n+1)(n+2)\). Now, if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) the answer will be YES, but for \(p=3\) the answer will be NO. Not sufficient. (1)+(2) \((n+2)!n!=n!((n+1)(n+2)1)\). Now, \((n+1)(n+2)1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are coprime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it cannot be a factor of \((n+1)(n+2)1\). Thus, in order for \(p\) to be a factor of \(n!*((n+1)(n+2)1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient. Answer: C
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Bunuel wrote: Official Solution:
(1) \(p\) is a factor of \((n+2)!n!\). If \(n=2\) then \((n+2)!n!=22\) and for \(p=2\) the answer will be YES, but for \(p=11\) the answer will be NO. Not sufficient. (2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\). Simplify: \(\frac{(n+2)!}{n!}=(n+1)(n+2)\). Now, if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) the answer will be YES, but for \(p=3\) the answer will be NO. Not sufficient. (1)+(2) \((n+2)!n!=n!((n+1)(n+2)1)\). Now, \((n+1)(n+2)1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are coprime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it cannot be a factor of \((n+1)(n+2)1\). Thus, in order for \(p\) to be a factor of \(n!*((n+1)(n+2)1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C hello, one small doubt. After combining (1) and (2)  will "p" always be 1 ??



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Re: M2701
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15 Mar 2015, 18:07
minhaz3333 wrote: Bunuel wrote: Official Solution:
(1) \(p\) is a factor of \((n+2)!n!\). If \(n=2\) then \((n+2)!n!=22\) and for \(p=2\) the answer will be YES, but for \(p=11\) the answer will be NO. Not sufficient. (2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\). Simplify: \(\frac{(n+2)!}{n!}=(n+1)(n+2)\). Now, if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) the answer will be YES, but for \(p=3\) the answer will be NO. Not sufficient. (1)+(2) \((n+2)!n!=n!((n+1)(n+2)1)\). Now, \((n+1)(n+2)1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are coprime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it cannot be a factor of \((n+1)(n+2)1\). Thus, in order for \(p\) to be a factor of \(n!*((n+1)(n+2)1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C hello, one small doubt. After combining (1) and (2)  will "p" always be 1 ?? hi, p will always not be 1and is not one in this case too.... combined , it tells us that the first has been broken down to n!(n+1)(n+2) and second (n+1)(n+2)1... now (n+1)(n+2) and (n+1)(n+2)1 will have only 1 as factor as these two are consecutive... p is prime and cannot be 1.. so p can be a factor of (n+1)(n+2)1 in (2) but will be a factor of n! in (1)
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Re: M2701
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22 Jul 2016, 20:36
Hi Bunuel , In the steps below, (1)+(2) (n+2)!−n!=n!((n+1)(n+2)−1). Now, (n+1)(n+2)−1 and (n+1)(n+2) are consecutive integers. Could you please let me know, why did you discard the n! here, and have taken (n+1)(n+2)−1 , instead of n!((n+1)(n+2)−1). Where am I missing? Thanks!



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Re: M2701
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23 Jul 2016, 01:45
Bunuel wrote: Official Solution:
(1) \(p\) is a factor of \((n+2)!n!\). If \(n=2\) then \((n+2)!n!=22\) and for \(p=2\) the answer will be YES, but for \(p=11\) the answer will be NO. Not sufficient. (2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\). Simplify: \(\frac{(n+2)!}{n!}=(n+1)(n+2)\). Now, if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) the answer will be YES, but for \(p=3\) the answer will be NO. Not sufficient. (1)+(2) \((n+2)!n!=n!((n+1)(n+2)1)\). Now, \((n+1)(n+2)1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are coprime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it cannot be a factor of \((n+1)(n+2)1\). Thus, in order for \(p\) to be a factor of \(n!*((n+1)(n+2)1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C great explanation Bunuel question too:)



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I used the same logic for 1) and 2) combined, but a bit different for 1) and 2) separated:
1) p is a factor of (n+2)!  n! and this means that p is a factor of the following expression: (n+2)!  n! = n!*(n+1)*(n+2)  n! = n! * [(n+1)*(n+2)  1] Then it means that p is a factor of either n! or [(n+1)*(n+2)  1], but not both of them. Actually, I don't know how to prove that it can't be a factor of both, but the expression is pretty weird and my gut feeling is that the restriction is that these number are somehow coprimes... Any thoughts?
2) p is a factor of (n+2)!/n! and this means that p is a factor of the following expression: (n+2)!/n! = [n!*(n+1)*(n+2)]/n! = (n+1)*(n+2) This means that p is a factor of either (n+1) or (n+2), which are coprime, thus it will give two different answers for the question.
3) Combined, as Bunuel mentions, the only way p to be a factor of both expression is to be the factor of n!, otherwise it should be the factor of coprimes and this is not feasible.



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Re: M2701
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06 Sep 2016, 09:09
Bunuel wrote: Official Solution:
(1) \(p\) is a factor of \((n+2)!n!\). If \(n=2\) then \((n+2)!n!=22\) and for \(p=2\) the answer will be YES, but for \(p=11\) the answer will be NO. Not sufficient. (2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\). Simplify: \(\frac{(n+2)!}{n!}=(n+1)(n+2)\). Now, if \(n=2\) then \((n+1)(n+2)=12\) and for \(p=2\) the answer will be YES, but for \(p=3\) the answer will be NO. Not sufficient. (1)+(2) \((n+2)!n!=n!((n+1)(n+2)1)\). Now, \((n+1)(n+2)1\) and \((n+1)(n+2)\) are consecutive integers. Two consecutive integers are coprime, which means that they don't share ANY common factor but 1. For example, 20 and 21 are consecutive integers, thus the only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it cannot be a factor of \((n+1)(n+2)1\). Thus, in order for \(p\) to be a factor of \(n!*((n+1)(n+2)1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.
Answer: C Hi Bunuel, Can n = 1?
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Re: M2701
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20 Sep 2016, 02:52
Aren't these questions unnecessarily difficult to be a GMAT question?



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Re: M2701
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29 Dec 2016, 11:16
LOVE the question. But what are the chances of something like this showing up Bunuel? IT took me a good 10 mins to solve this :/



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Re: M2701
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31 Oct 2017, 02:32
Can N=1??



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Re: M2701
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22 Nov 2017, 22:42
Bunuel wrote: If \(n\) is a positive integer and \(p\) is a prime number, is \(p\) a factor of \(n!\)?
(1) \(p\) is a factor of \((n+2)!n!\)
(2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\) I solved it in this way stmt 1 : p is fact of ((n+2)(n+1)1) * N! P can be a factor of ((n+2)(n+1)1) or P can be a factor of n! Stmt 2 : P is afactor of (n+2)(n+1)*n!/n! p is a factor of (n+2)(n+1) Combined as p is a factor of (n+2)(n+1) , p cannot be factor of ((n+2)(n+1)1). So P is a factor of n!. Can someone please validate this approach.
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Re: M2701
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22 Nov 2017, 23:06



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Re: M2701
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02 Aug 2018, 09:11
Bunuel Great question ! I always wonder who you make these ques



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Re: M2701
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06 Sep 2018, 18:57
Bunuel wrote: sumanainampudi wrote: Bunuel wrote: If \(n\) is a positive integer and \(p\) is a prime number, is \(p\) a factor of \(n!\)?
(1) \(p\) is a factor of \((n+2)!n!\)
(2) \(p\) is a factor of \(\frac{(n+2)!}{n!}\) I solved it in this way stmt 1 : p is fact of ((n+2)(n+1)1) * N! P can be a factor of ((n+2)(n+1)1) or P can be a factor of n! Stmt 2 : P is afactor of (n+2)(n+1)*n!/n! p is a factor of (n+2)(n+1) Combined as p is a factor of (n+2)(n+1) , p cannot be factor of ((n+2)(n+1)1). So P is a factor of n!. Can someone please validate this approach. _______________ That's correct. I get this approach, but shouldn't Statement 2 yield the answer as B if she got to "P is a factor of (n+2)(n+1), hence it's not a factor of n!?



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Re: M2701
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10 Sep 2018, 06:42
For those of us who didn't catch the (n+1)(n+2)  1 can't be a factor, there's another approach for combining statements:
Rearraging a bit, we have [(n+1)(n+2)(n!)  (n!)]/p = (n+1)(n+2)(n!)/ p  (n!) / p= an integer by (1).
By (2), we know that (n+1)(n+2)/p is an integer, so we have:
[(n+1)(n+2)(n!)/ p ] [(n!) / p]= Integer * (n!)  n!/p = Integer. The only way that n!/p subtracted from an integer * n! can be an integer is if n!/p is an integer which implies that n!/p = integer, meaning p is a factor of n!.










