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# Given n>5 , when (n!+n+1) is divided by (n+1)

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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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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 (n-2) is prime, (n-1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient.

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Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

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04 Feb 2014, 18:33
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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.

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
1
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 May 2014, 11:37
1
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.

Hi Karishma,

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.

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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, 12:37
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NAL9 wrote:
Hi Karishma,

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.

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,

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.

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 n-2 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
3
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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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. n-2 prime
n=7, 11, 17, etc.
7!/8 - remainder is 0
11!/12 - remainder is 0.

same pattern
statement 2 is sufficient.

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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 (n-2) is prime, (n-1) and (n+1) must be even and hence composite. Remainder must be 0. Sufficient.

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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 long-winded.

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: (N-2) 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.

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Re: Given n>5 , when (n!+n+1) is divided by (n+1)  [#permalink]

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