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M27-01

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M27-01  [#permalink]

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New post 16 Sep 2014, 01:26
2
23
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

31% (02:36) correct 69% (02:11) wrong based on 178 sessions

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Re M27-01  [#permalink]

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New post 16 Sep 2014, 01:26
1
13
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 co-prime, 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
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M27-01  [#permalink]

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New post 15 Mar 2015, 17:39
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 co-prime, 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 ?? :o
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New post 15 Mar 2015, 19:07
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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 co-prime, 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 ?? :o


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)
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New post 27 Aug 2016, 00:56
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 co-primes... 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 co-prime, 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 co-primes and this is not feasible.
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Re: M27-01  [#permalink]

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New post 22 Nov 2017, 23:42
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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.
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New post 23 Nov 2017, 00:06
sumanainampudi wrote:
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.

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Re: M27-01  [#permalink]

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New post 02 Aug 2018, 10:11
Bunuel
Great question !
I always wonder who you make these ques :-o :grin: :thumbup:
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Re: M27-01  [#permalink]

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New post 10 Sep 2018, 07:42
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For those of us who didn't catch the (n+1)(n+2) - 1 can't be a factor, there's another approach for combining statements:

Rearraging a bit, we have [(n+1)(n+2)(n!) - (n!)]/p = (n+1)(n+2)(n!)/ p - (n!) / p= an integer by (1).

By (2), we know that (n+1)(n+2)/p is an integer, so we have:

[(n+1)(n+2)(n!)/ p ]- [(n!) / p]= Integer * (n!) - n!/p = Integer. The only way that n!/p subtracted from an integer * n! can be an integer is if n!/p is an integer- which implies that n!/p = integer, meaning p is a factor of n!.
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New post 07 Jun 2019, 03:22
Hi!

I don;t understand where (n+2)!−n!=n!((n+1)(n+2)−1) came from is the combined solution
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New post 07 Jun 2019, 03:30
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New post 08 Jun 2019, 05:39
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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 \(\frac{(n+2)!}{n!}\)


Target question: Is p a factor of n!

Statement 1: \(p\) is a factor of \((n+2)!-n!\)
\((n+2)!=(n+2)(n+1)(n)(n-1)(n-2)...(3)(2)(1)\)
And \(n!=(n)(n-1)(n-2)...(3)(2)(1)\)
So, we can factor the expression to get: \(p\) is a factor of \(n![(n+2)(n+1)-1]\)
So, it COULD be the case that p is a factor of n!, in which case, the answer to the target question is YES, p IS a factor of n!
Or, it COULD be the case that p is a factor of [(n+2)(n+1)-1], in which case, the answer to the target question is NO, p is NOT a factor of n!

ASIDE: If we let n = 2, and p = 2, then we can see that p IS a factor of n!
If we n = 2, and p = 11, then we can see that p is NOT a factor of n!, but it IS a factor of [(n+2)(n+1)-1]

Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: \(p\) is a factor of \(\frac{(n+2)!}{n!}\)
Simplify the expression to get: \(p\) is a factor of \((n+2)(n+1)\)
Let's test some values that satisfy statement 2:
Case a: n = 2 and p = 2. Here, (n+2)(n+1) = (2+2)(2+1)=12, so 2 (aka p) is a factor of (n+2)(n+1). In this case, the answer to the target question is YES, p IS a factor of n!
Case b: n = 2 and p = 3. Here, (n+2)(n+1) = (2+2)(2+1)=12, so 3 (aka p) is a factor of (n+2)(n+1). In this case, the answer to the target question is NO, p is NOT a factor of n!
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 tells us that \(p\) is a factor of \((n+2)(n+1)\)
Notice that, if \(p\) is a factor of \((n+2)(n+1)\), then \(p\) is NOT a factor of \((n+2)(n+1)-1\)

Statement 1 tells us that \(p\) is a factor of \(n![(n+2)(n+1)-1]\)
Since we know that \(p\) is NOT a factor of \((n+2)(n+1)-1\), it MUST be true that \(p\) is a factor of \(n!\)
The answer to the target question is YES, p IS a factor of n!

Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Re: M27-01   [#permalink] 08 Jun 2019, 05:39
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