M27-01

Question

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!}

Bunuel · Accepted Answer

Official Solution:

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! . 
 
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!} . 
 
We 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) Simplifying the equation from (1) gives  (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, so the only common factor they share is 1. As from (2),  p  is a factor of  (n+1)(n+2) , it cannot be a factor of  (n+1)(n+2)-1 . Thus, for  p  to be a factor of  n!*((n+1)(n+2)-1) , from (1), it should be a factor of the first multiple of this expression:  n! . Sufficient.

Answer: C

SI777 · Answer

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.

minhaz3333 · Answer

hello, one small doubt. After combining (1) and (2) - will "p" always be 1 ??

chetan2u · Answer

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)

Bunuel · Answer

_______________
That's correct.

lukachiri · Answer

Bunuel Great question ! I always wonder who you make these ques

gregsz · Answer

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  /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:

-  = 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!.

amandesai17 · Answer

Hi!

I don;t understand where (n+2)!−n!=n!((n+1)(n+2)−1) came from is the combined solution

Bunuel · Answer

(n + 2)! = 1*2*3*...*n*(n + 1)(n + 2) = n!*(n + 1)(n + 2) .

Thus,  (n+2)!−n!=n!*(n + 1)(n + 2) - n! =n!((n+1)(n+2)−1)

BrentGMATPrepNow · Answer

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!  
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  , 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

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!  
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

coreyander · Answer

I think this is a  high-quality  question and I agree with explanation.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

M27-01

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