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Difference between revisions of "2006 JBMO Problems/Problem 1"

(Created page with "==Problem== If <math>n>4</math> is a composite number, then <math>2n</math> divides <math>(n-1)!</math>. == Solution == We shall prove a more stronger result that <math>4n...")
 
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== Solution ==
 
== Solution ==
  
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We shall prove a more stronger result that <math>4n</math> divides <math>(n-1)!</math> which will cover the case of the problem statement.
+
We shall prove a more stronger result that <math>4n</math> divides <math>(n-1)!</math> which will cover the case of problem statement.
  
 
Let <math>n = p.q</math> where <math>q \ge p \ge 2</math>.
 
Let <math>n = p.q</math> where <math>q \ge p \ge 2</math>.

Revision as of 22:02, 26 December 2018

Problem

If $n>4$ is a composite number, then $2n$ divides $(n-1)!$.


Solution

We shall prove a more stronger result that $4n$ divides $(n-1)!$ which will cover the case of problem statement.

Let $n = p.q$ where $q \ge p \ge 2$.

Let us define set $S = \{1, 2, 3 ... n-1\}$

First let's note that $p, q \in S$


$Case 1: q > p$

Now, all multiples of $p$ from $p.1$ to $p.(q-1) \in S$

Since $q > p => q > 2 => q-1 \ge 2$ we have that $p.2 \in S$ Also, since $p \ge 2$ we have that $2 \in S$

So, we have that $2, p.2, q \in S$, in other words, $4.p.q$ divides $(n-1)!$


$Case 2: q = p$

Now, all multiples of $p$ from $p.1$ to $p.(p-1) \in S$

Since $p > 2 => p-1 \ge 2$ we have that $p.2 \in S$

Also, since $n = p^2 > 4 => p > 2$ so we have that $2 \in S$

So, we have that $2, p.1, p.2 \in S$, in other words, $4.p^2$ divides $(n-1)!$


Thus $4n$ divides $(n-1)!$.


$Kris17$

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