M27-01

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

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)

_______________
That's correct.

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

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!.

Hi!

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

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

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

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

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

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