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Does the prime number p divide n!?
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20 Dec 2010, 09:10
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Does the prime number p divide n!? (1) Prime number p divides n! + (n+2)!. (2) The prime number p divides (n+2)!/n!.
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Re: Number System
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20 Dec 2010, 10:30
surendar26 wrote: Does the prime number p divide n!?
1. Prime number p divides n! + (n+2)!. 2. The prime number p divides (n+2)!/n!. The question should be: is the prime number p a factor of n!, where n is a nonnegative integer. (Basically the question asks whether \(p\leq{n}\)) (1) p is a factor of \(n!+(n+2)!\) > \(p\) is a factor of \(n!(1+(n+1)(n+2))\). Now, if \(p=3\) and \(n=0\) (\(n!+(n+2)!=1+2=3\)) then the answer will be NO: \(p=3\) is not a factor of \(n!=0!=1\). But if \(p=2\) and \(n=2\) (\(n!+(n+2)!=2+24=26\)) then the answer will be YES: \(p=2\) is a factor of \(n!=2!=2\). Two different answer, hence not sufficient. (2) p is a factor of \(\frac{(n+2)!}{n!}\) > \(p\) is a factor of \((n+1)(n+2)\). Now, if \(p=2\) and \(n=0\) then the answer will be NO but if \(p=2\) and \(n=2\) then the answer will be YES. Not sufficient. (1)+(2) Now, \((n+1)(n+2)\) and \((n+1)(n+2)+1\) 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 only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)+1\), thus in order \(p\) 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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Re: Number System
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15 Apr 2011, 21:15
Actually there is a pattern in divisibility. Let's say a+b is divisible by c then we are not sure if a is divisible by c. But if b is divisible by c then a is divisible by c. Hence combining 1) + 2) works. I don't have see a better to solve this in 60 secs. But will love to hear from others.
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Re: Number System
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15 Apr 2011, 23:13
Understood y Statement 1 and statement 2 are eliminated. But didn't understood how Both statements are sufficient. Anyone pls help.
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Re: Does the prime number p divide n!?
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08 Oct 2017, 03:50
surendar26 wrote: Does the prime number p divide n!?
(1) Prime number p divides n! + (n+2)!. (2) The prime number p divides (n+2)!/n!. P divides N only if p<=n from 1 p divides n! [1+(n+1)(n+2)] thus p divides either n! or divides [1+(n+1)(n+2)) ... insufficient from 2 p divides (n+1) (n+2) ... no mention whether it divides n! or not ... insuff from both 1+(n+1)(n+2) and ( n+1) ( n+2) are co primes , i.e. no common divisor except 1 and since from 1 p divides n! [1+(n+1)(n+2)] and divides (n+1)(n+2) then it must divide n! ... suff C



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Re: Does the prime number p divide n!?
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08 Oct 2017, 17:22
yezz wrote: surendar26 wrote: Does the prime number p divide n!?
(1) Prime number p divides n! + (n+2)!. (2) The prime number p divides (n+2)!/n!. P divides N only if p<=n from 1 p divides n! [1+(n+1)(n+2)] thus p divides either n! or divides [1+(n+1)(n+2)) ... insufficient from 2 p divides (n+1) (n+2) ... no mention whether it divides n! or not ... insuff from both 1+(n+1)(n+2) and ( n+1) ( n+2) are co primes , i.e. no common divisor except 1 and since from 1 p divides n! [1+(n+1)(n+2)] and divides (n+1)(n+2) then it must divide n! ... suff C Can u pls explain n! [1+(n+1)(n+2)] with steps? Thanks



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Re: Does the prime number p divide n!?
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08 Oct 2017, 17:43
zanaik89 wrote: yezz wrote: surendar26 wrote: Does the prime number p divide n!?
(1) Prime number p divides n! + (n+2)!. (2) The prime number p divides (n+2)!/n!. P divides N only if p<=n from 1 p divides n! [1+(n+1)(n+2)] thus p divides either n! or divides [1+(n+1)(n+2)) ... insufficient from 2 p divides (n+1) (n+2) ... no mention whether it divides n! or not ... insuff from both 1+(n+1)(n+2) and ( n+1) ( n+2) are co primes , i.e. no common divisor except 1 and since from 1 p divides n! [1+(n+1)(n+2)] and divides (n+1)(n+2) then it must divide n! ... suff C Can u pls explain n! [1+(n+1)(n+2)] with steps? Thanks Just like 5! = 5*4*3.... (N+2)! = (n+2)(n+1)(n)(n1).... Thus , (N+2)! = (n+2)(n+1)*n! So in n!+(n+2)! We can take n! As a common factor Sent from my iPhone using GMAT Club Forum mobile app



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Does the prime number p divide n!?
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Updated on: 17 Oct 2017, 18:58
Bunuel wrote: surendar26 wrote: Does the prime number p divide n!?
1. Prime number p divides n! + (n+2)!. 2. The prime number p divides (n+2)!/n!. The question should be: is the prime number p a factor of n!, where n is a nonnegative integer. (Basically the question asks whether \(p\leq{n}\)) (1) p is a factor of \(n!+(n+2)!\) > \(p\) is a factor of \(n!(1+(n+1)(n+2))\). Now, if \(p=3\) and \(n=0\) (\(n!+(n+2)!=1+2=3\)) then the answer will be NO: \(p=3\) is not a factor of \(n!=0!=1\). But if \(p=2\) and \(n=2\) (\(n!+(n+2)!=2+24=26\)) then the answer will be YES: \(p=2\) is a factor of \(n!=2!=2\). Two different answer, hence not sufficient. (2) p is a factor of \(\frac{(n+2)!}{n!}\) > \(p\) is a factor of \((n+1)(n+2)\). Now, if \(p=2\) and \(n=0\) then the answer will be NO but if \(p=2\) and \(n=2\) then the answer will be YES. Not sufficient. (1)+(2) Now, \((n+1)(n+2)\) and \((n+1)(n+2)+1\) 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 only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)+1\), thus in order \(p\) 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 u please explain this part [color=#ffff00]then it should be a factor of the first multiple of this expression: n![/color]
Originally posted by zanaik89 on 17 Oct 2017, 18:56.
Last edited by zanaik89 on 17 Oct 2017, 18:58, edited 1 time in total.



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Re: Does the prime number p divide n!?
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17 Oct 2017, 20:17
zanaik89 wrote: Bunuel wrote: surendar26 wrote: Does the prime number p divide n!?
1. Prime number p divides n! + (n+2)!. 2. The prime number p divides (n+2)!/n!. The question should be: is the prime number p a factor of n!, where n is a nonnegative integer. (Basically the question asks whether \(p\leq{n}\)) (1) p is a factor of \(n!+(n+2)!\) > \(p\) is a factor of \(n!(1+(n+1)(n+2))\). Now, if \(p=3\) and \(n=0\) (\(n!+(n+2)!=1+2=3\)) then the answer will be NO: \(p=3\) is not a factor of \(n!=0!=1\). But if \(p=2\) and \(n=2\) (\(n!+(n+2)!=2+24=26\)) then the answer will be YES: \(p=2\) is a factor of \(n!=2!=2\). Two different answer, hence not sufficient. (2) p is a factor of \(\frac{(n+2)!}{n!}\) > \(p\) is a factor of \((n+1)(n+2)\). Now, if \(p=2\) and \(n=0\) then the answer will be NO but if \(p=2\) and \(n=2\) then the answer will be YES. Not sufficient. (1)+(2) Now, \((n+1)(n+2)\) and \((n+1)(n+2)+1\) 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 only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)+1\), thus in order \(p\) 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 u please explain this part [color=#ffff00]then it should be a factor of the first multiple of this expression: n![/color] p is NOT a factor of (n+1)(n+2)+1. For p to be a factor of n!* ((n+1)(n+2)+1), it must be a factor of n!
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Does the prime number p divide n!?
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Updated on: 22 Oct 2017, 11:05
yezz wrote: surendar26 wrote: Does the prime number p divide n!?
(1) Prime number p divides n! + (n+2)!. (2) The prime number p divides (n+2)!/n!. P divides N only if p<=n from 1 p divides n! [1+(n+1)(n+2)] thus p divides either n! or divides [1+(n+1)(n+2)) ... insufficient from 2 p divides (n+1) (n+2) ... no mention whether it divides n! or not ... insuff from both 1+(n+1)(n+2) and ( n+1) ( n+2) are co primes , i.e. no common divisor except 1 and since from 1 p divides n! [1+(n+1)(n+2)] and divides (n+1)(n+2) then it must divide n! ... suff C I solve the problem in the same way. I will see my interpretation of result from solution of the statement1 that statement 1 has two answer but it answers the question that p divides n! So please explain that how come your answer is C not A. In my option, with above solution answer will be option A.



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Re: Does the prime number p divide n!?
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22 Oct 2017, 11:02
In 1 , we have an either or situation we can't be sure it divides n! Because equally probable it could divide ( 1+ ( n+1)+(n+2))
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Re: Does the prime number p divide n!? &nbs
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