What is the value of the integer n?

(1) n!=n*(n – 1)! --> n*(n – 1)! equals to n! itself, so we have n!=n!. Useless statement.

(2) n^3 + 3n^2 + 2n is divisible by 3 --> n(n^2+3n+2)=n(n+1)(n+2): we have the product of 3 consecutive integers, which is ALWAYS divisible by 3. Useless statement.

(1)+(2) Two useless statements. Not sufficient.

Answer: E.
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I'm not really sure but isn't n!=n! will give 1 as an answer

n! = n! for any value of n, for which a factorial is defined. Consider this, if we had x = x, could we say that x = 1? No, x = x is true for all x's.
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I do NOT agree! The answer is C!
(1)The first statement rules out all negative numbers. Negative numbers cannot be factorials. (of interest, zero factorial is 1)
It is NOT SUFFICIENT b/c all non-negative integers are part of the solution set.

The second statement n(n^2+3n+2)=0
n(n+2)(n+1)=0
n=0 or -1 or -2
Also NOT SUFFICIENT to answer what n is. There are 3 possibilities for the answer.

Considering them together.
-1! is undefined. -2! is undefined. ZERO is a member of both solution sets and is the only number that is. That makes both of them together SUFFICIENT to know that n = 0!

I hope it helps! Stay cautious. Don't Leap to conclusions!
Bunuel Thanks for a great question! Kudos for me perhaps?
This should show up again so a lot of people see it. Good job Bot!

(2) says that "n^3 + 3n^2 + 2n is divisible by 3" not that n^3 + 3n^2 + 2n = 0.

Anyway all positive integers satisfy both statements together and the correct answer is E, not C.
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