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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
04 Feb 2014, 17:33
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guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1) , what is the remainder ?
(1) (n+2) is a prime number.
(2) (n−2) is a prime number.
This is a truly brilliant question, and I am very happy to help.
When when (n!+n+1) is divided by (n+1), (n + 1) obviously divides evenly into the (n + 1), so the question is: what is the remainder when we divide (n!) by (n+1)?
Well, if n > 5, then the only odd numbers available are prime. Whether (n + 2) or (n - 2) is prime, that means that n must be odd. If n is odd, (n + 1) is even, and therefore is product of 2 and something smaller than n. That means, both factors of (n + 1) must be contained in (n!), so (n + 1) divides evenly into (n!) with a remainder of zero. Each statement is sufficient on its own. OA = (D).
For example, let n = 21. Both n - 2 = 19 and n + 2 = 23 are prime. We know that (n + 1) = 22 is even: thus, 22 = 2*11, and of course, (21!) must contain every factor from 1 to 21, so it will include both a factor of 2 and factor of 11. Therefore, (21!) is necessarily a multiple of 22, and were we to divide, we would get a remainder of zero.
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
05 Feb 2014, 00:07
mikemcgarry wrote:
guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1) , what is the remainder ?
(1) (n+2) is a prime number.
(2) (n−2) is a prime number.
This is a truly brilliant question, and I am very happy to help.
When when (n!+n+1) is divided by (n+1), (n + 1) obviously divides evenly into the (n + 1), so the question is: what is the remainder when we divide (n!) by (n+1)?
Well, if n > 5, then the only odd numbers available are prime. Whether (n + 2) or (n - 2) is prime, that means that n must be odd. If n is odd, (n + 1) is even, and therefore is product of 2 and something smaller than n. That means, both factors of (n + 1) must be contained in (n!), so (n + 1) divides evenly into (n!) with a remainder of zero. Each statement is sufficient on its own. OA = (D).
For example, let n = 21. Both n - 2 = 19 and n + 2 = 23 are prime. We know that (n + 1) = 22 is even: thus, 22 = 2*11, and of course, (21!) must contain every factor from 1 to 21, so it will include both a factor of 2 and factor of 11. Therefore, (21!) is necessarily a multiple of 22, and were we to divide, we would get a remainder of zero.
Does all this make sense? Mike
It sure does makes sense. I just supplemented values for n in the equation to get to the answer. I hope that's okay while solving such questions.
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
05 Feb 2014, 07:45
Quote:
It sure does makes sense. I just supplemented values for n in the equation to get to the answer. I hope that's okay while solving such questions.
doing this way will not always work and might cause you to make false assessment (if something is true for a given value of n does not mean it is true for all n) _________________
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
05 Feb 2014, 08:38
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guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1), what is the remainder?
(1) (n+2) is a prime number.
(2) (n−2) is a prime number.
I would like to weigh in here since it is one of my little creations! The concept I wanted to test here was that if (n+1) is prime, it will not be a factor of n! because (n+1) cannot be broken down into smaller factors. On the other hand, a composite number can be broken down into 2 smaller factors and both will be included in n!. Here (n+1) is always even so it can definitely be broken down into two factors: 2 and n/2. In any case, n! will have both 2 and n/2 as factors (simultaneously).
Here is the detailed explanation:
Given (n! + n + 1)
\(\frac{n! + (n+1)}{n+1} = \frac{n!}{n+1} + 1\)
Now if n! is divisible by (n+1), then the remainder is 0. If not, then we don’t know the remainder. If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1). If (n+1) is prime, n! does not contain (n+1) and hence is not divisible by (n+1).
Statement I: If (n+2) is prime, (n+1) must be even and hence composite (recall that there is only one prime even number i.e. 2). Remainder must be 0. Sufficient.
Statement II: If (n-2) is prime, (n-1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient.
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
05 May 2014, 10:37
1
This post received KUDOS
VeritasPrepKarishma wrote:
guerrero25 wrote:
Given n>5 , when (n!+n+1) is divided by (n+1), what is the remainder?
(1) (n+2) is a prime number.
(2) (n−2) is a prime number.
I would like to weigh in here since it is one of my little creations! The concept I wanted to test here was that if (n+1) is prime, it will not be a factor of n! because (n+1) cannot be broken down into smaller factors. On the other hand, a composite number can be broken down into 2 smaller factors and both will be included in n!. Here (n+1) is always even so it can definitely be broken down into two factors: 2 and n/2. In any case, n! will have both 2 and n/2 as factors (simultaneously).
Here is the detailed explanation:
Given (n! + n + 1)
\(\frac{n! + (n+1)}{n+1} = \frac{n!}{n+1} + 1\)
Now if n! is divisible by (n+1), then the remainder is 0. If not, then we don’t know the remainder. If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1). If (n+1) is prime, n! does not contain (n+1) and hence is not divisible by (n+1).
Statement I: If (n+2) is prime, (n+1) must be even and hence composite (recall that there is only one prime even number i.e. 2). Remainder must be 0. Sufficient.
Statement II: If (n-2) is prime, (n-1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient.
Answer (D)
Hi Karishma,
could you please elaborate this part of your reasoning:
If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1)
I don't really see why if n+1 is even then n! will be divisible by n+1.
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
05 May 2014, 11:37
2
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NAL9 wrote:
Hi Karishma,
could you please elaborate this part of your reasoning:
If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1)
I don't really see why if n+1 is even then n! will be divisible by n+1.
Thanks in advance
Dear NAL9, I saw this, so I'll respond. I have a great deal of respect for Karishma, but I can answer this question.
Think about it this way. The number (73!) is a very big number, more than 100 digits. It is the product of every integer from 1 to 73, so it is automatically divisible by any integer from 1 to 73. Now, think about the next integer, 74. We want to know whether (73!), that 100+ digit number, will be divisible by 74. Well, notice that 74 = 2*37. Clearly (73!) is divisible by both 2 and by 37: therefore, it must be divisible by 2*37 = 74.
In much the same way, as long as we know that n is odd and that (n + 1) is even, we know that (n + 1) can be written as a product of 2 times some other number. That other integer, (n + 1)/2, will be much smaller than n, and therefore automatically will be one of the factors included in (n!). Of course, 2 is another factor also included in (n!). Therefore, (n!) would have to be divisible by the product of these two factors, and that product is (n + 1).
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
05 May 2014, 12:01
mikemcgarry wrote:
NAL9 wrote:
Hi Karishma,
could you please elaborate this part of your reasoning:
If (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1)
I don't really see why if n+1 is even then n! will be divisible by n+1.
Thanks in advance
Dear NAL9, I saw this, so I'll respond. I have a great deal of respect for Karishma, but I can answer this question.
Think about it this way. The number (73!) is a very big number, more than 100 digits. It is the product of every integer from 1 to 73, so it is automatically divisible by any integer from 1 to 73. Now, think about the next integer, 74. We want to know whether (73!), that 100+ digit number, will be divisible by 74. Well, notice that 74 = 2*37. Clearly (73!) is divisible by both 2 and by 37: therefore, it must be divisible by 2*37 = 74.
In much the same way, as long as we know that n is odd and that (n + 1) is even, we know that (n + 1) can be written as a product of 2 times some other number. That other integer, (n + 1)/2, will be much smaller than n, and therefore automatically will be one of the factors included in (n!). Of course, 2 is another factor also included in (n!). Therefore, (n!) would have to be divisible by the product of these two factors, and that product is (n + 1).
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
14 May 2015, 03:08
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Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
13 Aug 2015, 16:14
This is a great question; it tests thoroughly tests your brain. is there any significance of the n>5? Is it to exclude the case where n-2 is the prime 2?
Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
17 Aug 2015, 09:13
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ggurface wrote:
This is a great question; it tests thoroughly tests your brain. is there any significance of the n>5? Is it to exclude the case where n-2 is the prime 2?
Yes, it is to exclude the case where (n-2) could be an even prime.
If we are talking about primes greater than 2, they will definitely be odd so with either statement, (n+1) will definitely be even and greater than 2 (hence, non prime). _________________
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