Number Theory Prime Reciprocals never an integer

Question

I'm in number theory and I currently have these problems assigned as homework. I've looked through the sections containing these problems and I've solved/proved most of the other problems, but I can't figure these ones out.

1. For $n>1$, show that every prime divisor of $n!+1$ is an odd integer that is greater than $n$.

2. Assuming that $p_n$ is the $n$th prime number, show that the sum $\frac{1}{p_1}+\frac{1}{p_2}+...+\frac{1}{p_n}$ is never an integer.

3. How many zeroes end 1,111! ?

Thanks in advance!

Answer 1

Hint:

1. Note that if $n>1$ then $n\geqslant2$ and so $n!=n(n-1)\cdots2\cdot1$ is even. Also note that $\gcd(n!,n!+1)=1.$ It follows that if $p$ is a prime factor of $n!+1$ then it has to be coprime with $n!$ and so $p>n.$

2. Suppose that for some $n$ it is an integer. Call it $A.$ Then $$\begin{aligned}2\cdot3\cdots p_n\cdot A&=3\cdot5\cdots p_n+2\cdot5\cdot7\cdots p_n+\cdots+2\cdot3\cdots p_{n-1}\\\\&=3\cdot5\cdots p_n+2\cdot(5\cdot7\cdots p_n+\cdots+3\cdots p_{n-1}).\end{aligned}$$ For $A$ integer, what can you conclude?

3. The number of terminal zeros of $n!$ is equal to the largest power of $5$ that divides $n!,$ which is $\sum\limits_{k=1}^{\lfloor\log_5n\rfloor}\left\lfloor\dfrac{n}{5^k}\right\rfloor.$