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Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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04 Feb 2014, 17:15
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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.
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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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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. Does all this make sense? Mike
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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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05 Feb 2014, 01:07
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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.



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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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05 Feb 2014, 08:45
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)
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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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05 Feb 2014, 09:38
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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 (n2) is prime, (n1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient. Answer (D)
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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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05 May 2014, 11:37
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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 (n2) is prime, (n1) 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.
Thanks in advance



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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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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). Does all this make sense? Mike
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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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05 May 2014, 13:01
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). Does all this make sense? Mike Now I get it Thank you for the explanation!!!



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Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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13 Aug 2015, 17:14
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 n2 is the prime 2?



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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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17 Aug 2015, 10:13
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 n2 is the prime 2? Yes, it is to exclude the case where (n2) 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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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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20 Oct 2016, 07:42
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 got to D. n>5 > n!+n+1/(n+1) > rewrite it: n!/(n+1) + (n+1)/(+1) n!/(n+1) +1 we need to find n!/(n+1) 1. n+2 is prime > n=9, n=11, n=15, n=17, etc. let's take few examples and identify the pattern 9! 9! contains a 2 and a 5, so definitely, it will be a multiple of 10. remainder will be 0. 11!/12 > 11! 11! contain a 3 and a 4, so when divided by 12, the remainder will be 0...same thing continues for all the options... 2. n2 prime n=7, 11, 17, etc. 7!/8  remainder is 0 11!/12  remainder is 0. same pattern statement 2 is sufficient. answer is D.



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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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30 Dec 2017, 02:38
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. VERITAS PREP OFFICIAL SOLUTION: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 (n2) is prime, (n1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient. Answer (D)
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Re: Given n>5 , when (n!+n+1) is divided by (n+1) [#permalink]
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22 Mar 2018, 10:39
Hi All, This question provides a great opportunity to TEST Values, but the real "secret" to it is in understanding how factorials and division "work." This is a relatively rare issue on the GMAT and the explanation is going to be a bit longwinded. We're told that N > 5 and we're asked for the remainder when (N! + N + 1) is divided by (N+1). Fact 1: (N+2) is prime. Let's TEST N = 9 (9! +9+1) / (9+1) = (9! + 10)/10 This can be "split" into… 9!/10 + 10/10 9! = 9x8x7x6x5x4x3x2x1 = (2x5)x(9x8x7x6x4x3x1) 9!/10 has no remainder and 10/10 has no remainder, so (9! + 10)/10 has a remainder of 0 With the limitations provided by Fact 1, the remainder will ALWAYS be 0. Here's why: Since N + 2 = prime and N > 5, then… N MUST be odd…. N+1 MUST be even….. N! MUST be even….(because there's a "2" in the sequence) N! MUST be a multiple of (N+1)….(because there's a "2" and the "odd number" that you need for 2(odd) = (N+1)) (N! + N+1)/(N+1) can always be split into this…. N!/(N+1) + (N+1)/(N+1) So… N!/(N+1) has no remainder and (N+1)/(N+1) = 1 and has no remainder. Fact 1 ALWAYS provides a remainder of 0 Fact 1 is SUFFICIENT. Fact 2: (N2) is prime Since N > 5 and Fact 2 tells us that N = odd, this creates the same circumstances as in Fact 1. Fact 2 is SUFFICIENT. Final Answer: GMAT assassins aren't born, they're made, Rich
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