Can't see an easy way into this...
Take 1!x 2!x3! x......100!
Which factorial can you remove ( say divide by) so the result is a square number?
Any thoughts?
You are referring to removing a single factorial? In order that something be a square, every prime divisor must occur an even number of times. How many multiples of 2 occur in each of those factorials? How many factors of 3? ....
Yes a single factorial. So if it was 1!x2!x3!x4! , removing the 2! would lead to a square number because it would become 1!x3!x3!x2x2 which is square.
This is the way I have been thinking about it but got stuck .
$\displaystyle 100!99! = 100 (99!)^2$
$\displaystyle 98!97! = 98 (97!)^2$
etc.
Multiply the above to get
$\displaystyle 1!2!3!\text{...}\text{..}100! = 2^{50}50!m^2$
(m an integer)
so
$\displaystyle \frac{1!2!3!\text{...}\text{..}100! }{50!}$ is a perfect square