Fun factorial type problem.

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

Because
x!(x+1)!=(x!)2(x+1)x! (x+1)! = (x!)^2 (x+1)
It follows that
x=1100x!\prod_{x=1}^{100}x!
=y=150((2y1)!(2y)!)= \prod_{y=1}^{50} \left( (2y-1)! (2y)! \right)
=(y=150(2y1)!)2×y=150(2y)=\left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times \prod_{y=1}^{50} (2y)
=(y=150(2y1)!)2×250×50!= \left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times 2^{50} \times 50!
=(225×y=150(2y1)!)2×50!= \left(2^{25} \times \prod_{y=1}^{50} (2y-1)! \right)^2 \times 50!
The rest is child's play ;)
 
Last edited:
Here's my solution (click to reveal):-

Because
x!(x+1)!=(x!)2(x+1)x! (x+1)! = (x!)^2 (x+1)
It follows that
x=1100x!\prod_{x=1}^{100}x!
=y=150((2y1)!(2y)!)= \prod_{y=1}^{50} \left( (2y-1)! (2y)! \right)
=(y=150(2y1)!)2×y=150(2y)=\left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times \prod_{y=1}^{50} (2y)
=(y=150(2y1)!)2×250×50!= \left(\prod_{y=1}^{50} (2y-1)!\right)^2 \times 2^{50} \times 50!
=(225×y=150(2y1)!)2×50!= \left(2^{25} \times \prod_{y=1}^{50} (2y-1)! \right)^2 \times 50!
The rest is child's play ;)
Perfect.
 

A fun extension to this problem :D

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?
 
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