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!



  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 $\sum\limits_{k=1}^{\lfloor\log_5n\rfloor}\left\lfloor\dfrac{n}{5^k}\right\rfloor.$

  • $\begingroup$ @C.I.J.: What can you say about $2\cdot3\cdots p_n\cdot A$ ? - Not much. Why ? $\endgroup$ – Lucian Sep 25 '15 at 9:55
  • 2
    $\begingroup$ @Lucian Let's say $A=(1/p_1)+(1/p_2)+\cdots+(1/p_n).$ Then $$\underbrace{2\cdot3\cdots p_n\cdot A}_{even}=\underbrace{3\cdots p_n}_{odd}+\underbrace{2\cdot5\cdots p_n+\cdots+2\cdot3\cdots p_{n-1}}_{even}$$ $\endgroup$ – CIJ Sep 25 '15 at 11:41
  • $\begingroup$ @C.I.J.: Thank you ! $($As silly at it sounds, I would have never guessed it in a million years !$)$ $\endgroup$ – Lucian Sep 25 '15 at 11:48
  • $\begingroup$ @Lucian Don't worry, I think I should include that in my answer $\endgroup$ – CIJ Sep 25 '15 at 11:52

Let p1,...,pn be different prime numbers. Can the sum of their reciprocals be an integer? If A = 1/p1 + ... + 1/pn, then A.p1....pn = p2....pn + p1.p3....pn + ... Let us pick any prime, pi. It divides the left side and is contained in every summand of the right side, except for the one where pi is omitted. Moreover this product of the other primes is not divisible by it. So it divides every summand but one: so it does not divide the right side. Contradiction. So the sum of reciprocals of different primes (not forcibly consecutive ones) can not be an integer.


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