Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 46264

Question Stats:
37% (02:27) correct 63% (01:36) wrong based on 51 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 46264

Re M2701 [#permalink]
Show Tags
16 Sep 2014, 01:26
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
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 11 Nov 2014
Posts: 38
Concentration: Marketing, Finance
WE: Programming (Computer Software)

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 ??



Math Expert
Joined: 02 Aug 2009
Posts: 5911

Re: M2701 [#permalink]
Show Tags
15 Mar 2015, 19:07
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)
_________________
Absolute modulus :http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
GMAT online Tutor



Intern
Joined: 01 Aug 2013
Posts: 5

Re: M2701 [#permalink]
Show Tags
22 Jul 2016, 21:36
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!



Math Expert
Joined: 02 Sep 2009
Posts: 46264

Re: M2701 [#permalink]
Show Tags
23 Jul 2016, 00:15



Intern
Joined: 29 Oct 2014
Posts: 40

Re: M2701 [#permalink]
Show Tags
23 Jul 2016, 02:45
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:)



Current Student
Joined: 08 Jan 2015
Posts: 82

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.



Current Student
Joined: 31 Jan 2016
Posts: 24
Concentration: Finance, Statistics
GPA: 3.6

Re: M2701 [#permalink]
Show Tags
06 Sep 2016, 10:09
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?
_________________
If you like my post, please send some kudos! :D



Intern
Joined: 06 Feb 2016
Posts: 48
Location: Poland
Concentration: Finance, Accounting
GPA: 3.5

Re: M2701 [#permalink]
Show Tags
20 Sep 2016, 03:52
Aren't these questions unnecessarily difficult to be a GMAT question?



Intern
Joined: 10 Dec 2016
Posts: 24

Re: M2701 [#permalink]
Show Tags
29 Dec 2016, 12:16
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 :/



Intern
Joined: 24 Jun 2017
Posts: 6

Re: M2701 [#permalink]
Show Tags
31 Oct 2017, 03:32
Can N=1??



Manager
Joined: 29 Feb 2016
Posts: 106
Location: India
GPA: 3

Re: M2701 [#permalink]
Show Tags
22 Nov 2017, 23:42
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.
_________________
Find what you love and let it kill you. — Charles Bukowski



Math Expert
Joined: 02 Sep 2009
Posts: 46264

Re: M2701 [#permalink]
Show Tags
23 Nov 2017, 00:06










