SOLUTION: how many zeros will be there at the end of the product (2!)^2! � (4!)^4! � (6!)^6! � (8!)^8! � (10!) 10! ?

Algebra ->  Decimal-numbers -> SOLUTION: how many zeros will be there at the end of the product (2!)^2! � (4!)^4! � (6!)^6! � (8!)^8! � (10!) 10! ?      Log On


   

Question 898372: how many zeros will be there at the end of the product (2!)^2! � (4!)^4! � (6!)^6! � (8!)^8! � (10!) 10! ?
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
(2!)^2 ---> no zeros
(4!)^4 ---> no zeros

(6!)^(6!) = 6!^(720)
6/5 = 1 (nearest integer)

1 * 720 = 720 zeros

(8!)^(8!) = 8!^(40320)
8/5 = 1 (nearest integer)

1 * 40320 = 40320 zeros

(10!)^(10!)
10/5 = 2

2 * 10! zeros

Total: 2 * 10! + 8! + 6! = 7,297,944 zeros