Primal Factors

$n^5-n=n(n^4-1)=n(n^2+1)(n+1)(n-1)$ and rearranging factors, $\\n^5-n=(n^2+1)(n-1)(n)(n+1)\\$ Since three of the factors are consecutive integers, the expression is divisible by 2 and by 3 and thus by 6. Also by Fermat's little theorem which states that if $p$ is a prime number, then for any integer $a$ , the number $a^p - a$ is an integer multiple of $p$ . Since $n$ is an integer and 5 is prime, $n^5-n$ is divisible by 5. Therefore the expression is divisible by 6 and 5 and thus by 30. For $n=2, \quad n^5-n=30$ . Therefore the expression cannot be divisible by any integer greater than 30. So the answer is $\boxed{30}$ .

Sorry! I got the right answer through a very naive and unsafe way.

First, I realized that if an integer B divides an integer A , then B also divides -A (the negative of A ). This way, I could think of positive n 's only, plus zero. Then, I enumerated the first few integers: 0, 1, 2, 3. From the question statement, I assumed that the answer should be one of the options. For n =zero, all options would do. For n =1, all options would do. For n =2, only 30 would do; this implies that 20, 60 and 90 would not. Still confirmed with n =3. So, the only possible solution, from those provided, is 30.

As I said, sorry! I know this is not a scientific approach. However, in this specific example, it worked!

If you substitute $2$ for $n$ , the expression becomes $2^5-2$ , which equals $30$ . We can already see here that $30$ is only divisible by one of the answer choices: $30$ . And since the problem stated that $n$ could be any integer, and we've found that only $30$ satisfies $n=2$ , then $30$ must be the correct answer.

1. proving that $n^5-n$ is divisible by $30$ means at least that this expression is divisible by $2$ , $5$ and $3$
2. the little fermat theorem obviously tells us that it's divisible by $5$
3. $n^5-n=n*(n-1)*(n+1)*(n^2+1)$ . If $n$ is even the result is even and if $n$ is odd the result is even too because $n-1$ and $n+1$ are even. So the expression is divisible by $2$
4. $n^5-n=n*(n^2-1)*(n^2+1)$ . if $n$ is divisible by $3$ the result is obvious. If not, applying the Euler theorem we have $n^{\phi(3)}\equiv 1 \pmod 3 => n^2 \equiv 1 \pmod 3$ that's mean $n^2-1$ is divisible by $3$ for any number coprime with $3$ .
5. so $2$ , $5$ and $3$ are divisor or any $n^5-n$ and $2^5-2=30$ . So $30$ is the GCD of all $n^5-n$ . We voluntary ignore the cases of $0$ and $1$ because the result of the expression is $0$

Guys, you have to be kidding. Assuming there's at least one correct answer given, just substitute, n=2. The only answer that divides 30 is 30.

if n = 2; then n5 - n = 30.

So 30 is the answer.

But, the answer, 30, doesn’t work where n=1. Therefore, isn’t the answer actually 1?

$n^5-n$ is 30 when $n=2$ and 30 is the only of answers which divides 30 so I conclude that if any of the answers work for all integers, it has to be 30.

Comment: the problem says "ANY" integer, so presumably we must consider -1,0 and 1. In these cases the expression equals 0, 1 or 2, none of which is divisable by 30 or any of the other choices.

This isn't right. What if n is -1? (-1)^5 - 1 = -1 - 1 = -2. 30 does not divide -2.

Since we're told that one of these is a solution, just let n=2 which eliminates three possibilities leaving only 30.

I enjoyed this question, got the answer correct, but the answers provided made it possible to get this question right even if a person did not know how to use Fermat's little theorem, which I had not learned in all my years of existence , including taking calculus one and two twice .

Since two to the fifth power minus one is equal to 30, one does not need to know this theorem in order to get the answer correct. If you want to make this problem harder , you should have at least two possible correct answers , at least two answers that are factors of 30. Given the nature of the serum, that the number is a multiple of two three and five, I would suggest that the answers be 5 , 6 , 10, and 30 . Or, change one of the existing answers to 10 .

ploting any value of n can be used for verification if u plot n=2 then n^5-5 =2^5-2 =30 so this result is greater than 20, where n=3 n^5-5 =240 240%30==0 so ans is 30 This calculation is enough for me