1. ## Factorial problem help!

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?

2. ## Re: Factorial problem help!

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? ....

3. ## Re: Factorial problem help!

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 .

4. ## Re: Factorial problem help!

Originally Posted by rodders
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 .
Why remove it?

1! * 2! * 2! * 3! * 3! = 144

Btw, * is now the standard multiplication symbol.

5. ## Re: Factorial problem help!

$\displaystyle 100!99! = 100 (99!)^2$

$\displaystyle 98!97! = 98 (97!)^2$

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

6. ## Re: Factorial problem help!

Wow... cheers Thanks!