Fun factorial type problem.

[Steven G]

Find m>0 and k>0 such that (1!2!3!...100!)/m! = k^2

 

[blamocur]

m and k both integers or only m ?

 

[Steven G]

m and k both integers or only m ?

m and k are both integers

 

[blamocur]

m and k are both integers

Just when I thought I solved it

 

[Cubist]

Here's my solution (click to reveal):-

Because
[math]x! (x+1)! = (x!)^2 (x+1)[/math]
It follows that
[math]\prod_{x=1}^{100}x![/math]
[math]= \prod_{y=1}^{50} \left( (2y-1)! (2y)! \right)[/math]
[math]=\left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times \prod_{y=1}^{50} (2y)[/math]
[math]= \left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times 2^{50} \times 50![/math]
[math]= \left(2^{25} \times \prod_{y=1}^{50} (2y-1)! \right)^2 \times 50![/math]
The rest is child's play

 

[Steven G]

Here's my solution (click to reveal):-

Because
[math]x! (x+1)! = (x!)^2 (x+1)[/math]
It follows that
[math]\prod_{x=1}^{100}x![/math]
[math]= \prod_{y=1}^{50} \left( (2y-1)! (2y)! \right)[/math]
[math]=\left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times \prod_{y=1}^{50} (2y)[/math]
[math]= \left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times 2^{50} \times 50![/math]
[math]= \left(2^{25} \times \prod_{y=1}^{50} (2y-1)! \right)^2 \times 50![/math]
The rest is child's play

Perfect.

 

[Cubist]

A fun extension to this problem ​

Replace the "100" by "4t" where t is positive integer. What are the values of t that will enable m and k to have two different (integer) values?