ziyuenlau 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 (n+2)! / n!
Anyway, how to edit with the proper formula in this forum?
I have a hard time to solve this problem. Please help!
Hi ziyuenlau,
We have n: {1, 2, 3, ....}; p: {2, 3, 5, 7, ...}
We have to see if n! / p is integer or not.
Let's rephrase each statement.
S1: p is a factor of (n+2)! - n!
(n+2)! - n! = n!*(n+1)*(n+2) - n! = n!*[(n+1)*(n+2) - 1]
So we have p is a factor of n!*[(n+1)*(n+2) - 1]
Let's test this with a couple of test values.
Case 1: n = 1 and p = 5
n!*[(n+1)*(n+2) - 1] = 1!*[(1+1)*(1+2) - 1] = 5.
We see that 1! / 5 is not an integer. The answer is NO.
Case 2: n = 2 and p = 2
n!*[(n+1)*(n+2) - 1] = 2!*[(2+1)*(2+2) - 1] = 22.
We see that 2! / 2 is an integer. The answer is Yes. No unique answer.
S2: p is a factor of (n+2)! / n!
(n+2)! / n! = [n!*(n+1)*(n+2)] / n! = (n+1)*(n+2)
So we have p is a factor of (n+1)*(n+2).
Case 1: n = 1 and p = 2 or 3
(n+1)*(n+2) = (1+1)*(1+2) = 6.
We see that 1! / 2 is not an integer. The answer is NO.
Case 2: n = 2 and p = 2
(n+1)*(n+2) = (2+1)*(2+2) = 12.
We see that 2! / 2 is an integer. The answer is YES. No unique answer.
S1 and S2:
From S1, we know that n!*[(n+1)*(n+2) - 1] is a factor of p and from S2, we know that (n+1)*(n+2) is a factor of p.
Thus,
n!*[(n+1)*(n+2) - 1] / p = integer
=> n!*[
(n+1)*(n+2)] /
p - n! / p = integer; we know that
(n+1)*(n+2) /
p is an integer,
=> Integer - n! / p = Integer
=> n! / p = Integer - Integer = Integer
Thus, p is a factor of n!. Sufficient.
The correct answer:
C
Hope this helps!
Relevant book:
Manhattan Review GMAT Data Sufficiency Guide
-Jay
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