How many zeros will be there at the end of expression:-
(2!)^2!+(4!)^4!+(8!)^8!+(10!)^10!+(12!)^12!

• (2!)^2!+(4!)^4!+(8!)^8!+(10!)^10!+(12!)^12!
  
  = 2^2 + 24^24 + (8!)^8!+(10!)^10!+(12!)^12!
  
  last two digit of (8!)^8!+(10!)^10!+(12!)^12! = 0 as each has a factor 10.
  
  so, last two digit of (2!)^2!+(4!)^4!+(8!)^8!+(10!)^10!+(12!)^12! = 04 + 76 = 80
  
  last two digit is 80
  no. of zeroes at end = 1
  
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