no of zeroes at the end of the expression (2!)^2!+4!^4!+8!^8!+9!^9!+10!^10!+11!^11!

• (2!)^2!+4!^4! will not contain any zero
  8!^8! will contain 1*8! zeroes(since 8! will contain 1 zero so 8!^8! will contain 1*8! zeroes at the end of number)
  similarly
  9!^9! will contain 1*9! zeroes
  10!^10! will contain 2*10! zeroes(10! contains 2 zeroes at the end of no. so 10!^10! will contain 2*10! zeroes at the end of the number)
  similarly
  11!^11! will contain 2*11! zeroes
  total no. of zeroes at the end of expp will be = 8! + 9! + 2*10! + 2*11!
• 11 years agoHelpfull: Yes(6) No(0)
• did ny1 find a soln to dis qn???
• 11 years agoHelpfull: Yes(0) No(2)
• exactly it's a right way of doing a question.
• 11 years agoHelpfull: Yes(0) No(0)
• check for the numbr of fives in the equation.
  
  they are 5!+6!+7!+8!+9!+[2*(10!+11!)]
  that shuld be the answer
• 11 years agoHelpfull: Yes(0) No(0)
• last two digit of 4!^4! is 76
  and 2!^2!=4
  76+4=80
  only one zero at end of exp.
• 11 years agoHelpfull: Yes(0) No(0)