ardz24 wrote:Does the prime number p divide n!?
(1) Prime number p divides n! + (n+2)!.
(2) The prime number p divides (n+2)!/n!.
What's the best way to determine which statement is sufficient?
(1) Prime number p divides n! + (n+2)!.
n! + (n+2)! = n! + n!*(n + 1)*(n +2) = n![1 + (n + 1)*(n +2)]
Case 1: Say p = 2 and n = 2, then n![1 + (n + 1)*(n +2)] = 2![1 + (2 + 1)*(2 +2)] = 2!*13 = 26.
We see that n! = 2! = 4 is divisible by p = 2. The answer is Yes.
Case 2: Say p = 13 and n = 2, then n![1 + (n + 1)*(n +2)] = 2![1 + (2 + 1)*(2 +2)] = 2!*13 = 26.
We see that n! = 2! = 4 is NOT divisible by p = 13. The answer is No.
No unique answer. Insufficient.
(2) The prime number p divides (n+2)!/n!.
(n+2)!/n! = [n!*(n+1)*(n+2)] / n! = (n+1)*(n+2).
Thus, the statement states that the prime number p divides (n+1)*(n+2).
Case 1: Say p = 2 and n = 2, then (n + 1)*(n +2) = (2 + 1)*(2 +2) = 12.
We see that n! = 2! = 4 is divisible by p = 2. The answer is Yes.
Case 2: Say p = 3 and n = 2, then (n + 1)*(n +2) = (2 + 1)*(2 +2) = 12.
We see that n! = 2! = 4 is NOT divisible by p = 3. The answer is No.
No unique answer. Insufficient.
(1) and (2) combined:
From Statement 2, we see that p divides (n + 1)*(n +2), thus it must not divide [(n + 1)*(n +2) + 1], a consecutive integer since ](n + 1)*(n +2)] and [(n + 1)*(n +2) + 1] are co-prime to each other. Co-primes do not share any common factor other than 1.
Take few examples: 2 and 3; 17 and 18; 22 and 23, etc.
Since p does not divide [(n + 1)*(n +2) + 1], and from Statement 1, we know that p divides n![1 + (n + 1)*(n +2)], thus p must divide n!. Sufficient.
The correct answer:
C
Hope this helps!
-Jay
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