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(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.

(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 ??

(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 ??

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)
_________________

In the steps below, (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. Could you please let me know, why did you discard the n! here, and have taken (n+1)(n+2)−1 , instead of n!((n+1)(n+2)−1). Where am I missing?

In the steps below, (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. Could you please let me know, why did you discard the n! here, and have taken (n+1)(n+2)−1 , instead of n!((n+1)(n+2)−1). Where am I missing?

Thanks!

Please re-read the solution. I just analyzed tow multiples of n!((n+1)(n+2)−1) separately: (n+1)(n+2)−1 first and the 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 \(\frac{(n+2)!}{n!}\). 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.

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.

(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.