The positive integers
![x_1, \cdots , x_n](/media/m/b/0/6/b06df0b298b685bf5ea789e1d32b96e9.png)
,
![n \geq 3](/media/m/5/4/8/54807b3bf99aa939833fe57bf8d891d3.png)
, satisfy
![x_1 < x_2 <\cdots< x_n < 2x_1](/media/m/f/7/f/f7f95f1209e0322db6324ca1b7703ba6.png)
. Set
![P = x_1x_2 \cdots x_n.](/media/m/4/f/9/4f96fb3b289aaa9812480213e31aee28.png)
Prove that if
![p](/media/m/1/c/8/1c85c88d10b11745150467bf9935f7de.png)
is a prime number,
![k](/media/m/f/1/3/f135be660b73381aa6bec048f0f79afc.png)
a positive integer, and
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
is divisible by
![pk](/media/m/f/1/6/f16c66a069632ad87dc31ddab21abfab.png)
, then
%V0
The positive integers $x_1, \cdots , x_n$, $n \geq 3$, satisfy $x_1 < x_2 <\cdots< x_n < 2x_1$. Set $P = x_1x_2 \cdots x_n.$ Prove that if $p$ is a prime number, $k$ a positive integer, and $P$ is divisible by $pk$, then $\frac{P}{p^k} \geq n!.$