Find a triple (b, p, q) of positive integers such that p!/(2^b*q!) = 1 * 3 * 5 ... * 51

I'm going to guess that p = 51, since it goes up to 51.

That means 2^b * q! = 2 * 4 * 6... * 50 = 2^25 * 1 * 2 *... * 25.

b = 25, q = 25, p = 51

=^._.^=

THANKS!

I like that identity you used that

2*4*6*8....... = 2^(number of terms) * (number of terms)!

I'm learning all kinds of stuff from you today !!!!

Aww, thank you. :))