Some questions on basic number theory I have a number of questions related to proofs based on basic properties in number theory. 
While I would post them as separate questions, I feel that they are similar enough in the method that should be used to prove them.
Also, please understand that I am sure these questions are very easy for some, they are not that easy to me and it would be great to have solutions explaining things.


*

*Show that the tenth digit of $3^k$ is even for all $k \ge 1$.


I don't know how to even start!


*Show that $n^5-n$ is divisible by 30 for all $n \gt 0$.


I tried to do this:
$ n^5 - n= n (n^4 -1) = n (n^2 -1) (n^2 +1)= n (n-1) (n+1)(n^2 +1 )$
We can try any base case, and then assuming it to be true for $k$ we get:
$ k^5 - k= k (k-1) (k+1)(k^2 +1 ) = 30x $ ($x$ is an integer).
Now I must prove that this is true as well for $k+1$:
$(k+1)^5-(k+1)= (k+1)((k+1)-1)((k+1)+1)((k+1)^2 +1) = (k+1)(k)(k+2)(k^2 + 2k+2)$
How do I prove this further?


*Show that there exists a positive integer $n$ such that $n!$ has exactly $1993$ zeros at the end.


No idea how to solve it.
 A: Answering the 1st part, $3^{20}$ is equal to 3,486,784,401 which has an odd tenth digit from either end. So, it is actually false.
Also for the 2nd part, to prove divisibility by 30, it is sufficient to prove divisibility of 2, 3 and 5. We can do this using $n^5-1=n(n-1)(n+1)(n^2+1)$
Out of $n$, $n-1$ and $n+1$, there has to be at least 1 multiple of 2 and 1 multiple of 3.
If $n$, $n-1$ and $n+1$ are all not multiples of 5, then $n=5k+2$ or $n=5k+3$ for some natural no. $k$. By substituting the values in the equation, $n^2+1$ happens to be a multiple of 5 in both cases.
A: For question 3:
A number $n$ ends in a $k$ zeroes, if $10^k$ divides $n$. As $n! = n*(n-1)!$, it's clear that the number of zeroes on $n!$ is not decreasing. To show that the requested $n$ exists you then need to show that the number of zeroes doesn't stop growing.
Bonus question: Why isn't there a $n$ so that $n!$ ends in exactly 11 zeroes?
A: Hint: These are questions solved via modular arithmetic. For example the second asserts that $n^5=n$ (mod $30$), or $n^4=1$.
