+0  
 
0
48
1
avatar

Let S = 1!2!...100!. Prove that there exists a unique positive integer "k" such that S/k! is a perfect square.

 Apr 21, 2020
 #1
avatar+20767 
+1

100! · 99! · 98! ·  ...  · 3! · 2! · 1!

=  [100! · 99!] · [98! · 97!]  · ...  · [4! · 3!] · [2! · 1!]

=  [100 · 99! · 99!] · [98 · 97! · 97!] · ... [4 · 3! · 3!] · [2 · 1! · 1!]

=  [100 · 99!2] · [98 · 97!2] · ... · [4 · 3!2] · [2 · 1!2]

=  [100 · 98 · 96 ·... ·  6 · 4 · 2] · [99!2 · 97!2 · ... · 3!2 · 1!2]

=  250[50 · 49 · 48 · ... · 3 · 2 · 1] · [99!2 · 97!2 · ... · 3!2 · 1!2]

=  250 · [50!] · [99!2 · 97!2 · ... · 3!2 · 1!2]

 

So, if you let k!  =  50!, the result will be a perfect square (because 250 and [99!2 · 97!2 · ... · 3!2 · 1!2] are perfect squares)

but I can't guarantee that it is unique.

 Apr 21, 2020

21 Online Users

avatar
avatar
avatar