0
$\begingroup$

At the back of the book which contains this problem, a hint is given to consider $$S_n\cdot2^{k-1}\cdot3\cdot5\cdot9\cdot\ldots$$ where $2^k \le n < 2^{k+1}$.

I don't know how this helps? I know that every positive integer can be written as a sum of powers of 2, but not sure if this is relevant.

$\endgroup$

marked as duplicate by user7530, Ian Coley, Pedro Tamaroff, Antonio Vargas, user127096 Mar 19 '14 at 19:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ An answer here : math.stackexchange.com/questions/2746/… $\endgroup$ – user119228 Mar 19 '14 at 18:27
  • $\begingroup$ I suspect the hint meant to have a $7$ instead of that $9$, i.e., multiply $S_n$ by a power of $2$ (that turns out to leave a single $2$ in the denominator of the product) and all odd numbers up to $n$. $\endgroup$ – Barry Cipra Mar 19 '14 at 18:49
0
$\begingroup$

Use Betrand-Chebyshev theorem : If p is the largest prime number less than or equal

to n, then p > n/2. Use this to show that p divides n! and p divides n!/k for any k not equal

to p and less or equal to n, the denominator is: n!, and the numerator is:

n!/2 + n!/3 + ... + n!/p + ... + n!/n. The numerator is not divisible by p but the

denominator is. So the number is not an integer.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.