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Difference between revisions of "2006 JBMO Problems/Problem 1"
Line 11: Line 11:
  
 
Let us define set <math>S = \{1, 2, 3 ... n-1\}</math>
 
Let us define set <math>S = \{1, 2, 3 ... n-1\}</math>
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 +
<math>Case 1: q > p</math>
  
 
First let's note that <math>p, q \in S</math>
 
First let's note that <math>p, q \in S</math>
−
 
−
 
−
<math>Case 1: q > p</math>
 
  
 
Now, all multiples of <math>p</math> from <math>p.1</math> to <math>p.(q-1) \in S</math>
 
Now, all multiples of <math>p</math> from <math>p.1</math> to <math>p.(q-1) \in S</math>

Revision as of 22:10, 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 $(n-1)!$ for any composite number $n>4$ which will cover the case of problem statement.

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

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

$Case 1: q > p$

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

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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