how many zeroes are there at the end of the expression (2!)^2!+(4!)^4!+(8!)^8!+(9!)^9!+(10!)^10!+(11!)^11!

• each term of (8!)^8!+(9!)^9!+(10!)^10!+(11!)^11! consists lots of zeroea at end
  now last two digit of (2!)^2!+(4!)^4!=4+(24)^24=4+76=80
  hence last two digit of (2!)^2!+(4!)^4!+(8!)^8!+(9!)^9!+(10!)^10!+(11!)^11! = 80
  ANS no. of zero=1
• 10 years agoHelpfull: Yes(19) No(2)
• is the ans is 1?
• 10 years agoHelpfull: Yes(3) No(1)
• 8! and 9! have a 5 in them and inherently get a 0 (even number x 5 results in a trailing 0). 10! has two multiples of 5 in it (5 and 10) thus gets two zeroes in its end.
  
  The powers of 8! , 9! and 10! are all so big that the number of trailing zeroes are also huge. (to be specific, smallest of 8!, 9! and 10!*2 = 8! number of zeroes)
  
  But we dont need all those numbers as we only want the final number of trailing zeroes which is governed by the "smallest" numbers we are adding up.
  
  2! is 2
  2! ^ 2! is 2^2 = 4.
  
  You already have a non-zero units digit. But lets wait up and find the last term too.
  
  4! is 24
  4! ^ 4! is um.... well last two digits are all we care about
  Here's a nice link to finding out the last 2 digits of any power
  24^24 = 2^72 * 3^24
  = (2^10)^7 * 2^2 * (81)^6
  
  (last 2 digits of each expression = 24 * 4 * 81
  = 76
  
  76+4 = 80 resulting in only ONE trailing zero after the non-zero ten's digit of 8.
  
• 10 years agoHelpfull: Yes(2) No(1)