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

 Jun 18, 2021
 #1
avatar+2097 
+1

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

 

=^._.^=

 Jun 18, 2021
 #2
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0

THANKS!

Guest Jun 18, 2021
 #3
avatar+120930 
+1

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   !!!!

 

 

cool cool cool

CPhill  Jun 18, 2021
 #4
avatar+2097 
0

Aww, thank you. :))

 

=^._.^=

catmg  Jun 19, 2021

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