Find the number of zeros at the end of the product of the factorials of the first 25 natural numbers.

1) 54
2) 55
3) 56
4) 75
5) 76

• the solution is very easy... we just have to see that number of 5 that comes in any number factorial...
  as from 1! to 4! their is no 5 in any factorial, 0*4=0
  from 5! to 9! their is one 5 in every factorial ,1*5=5
  from 10! to1 4! their is two 5 in every factorial,2*5=10
  from 15! to 19! their is three 5 in every factorial,3*5=15
  from 20! to 24! their is four 5 in every factorial,4*5=20
  but when 25! comes their are six fives as the number 25 will give u 2 fives, 6
  sum all answer = 56
• 9 years agoHelpfull: Yes(1) No(0)
• 25! = 6 zero
  20! to 25! each have 4 zero so total zero= 5*4 = 20
  15! to 19 each have 3 zero so total zero = 5*3 = 15
  10! to 14! each have 2 zero so total zero = 5*2 =10
  5! to 9! each have 1 zero so total zero = 5*1 = 5
  total sum = 56
• 9 years agoHelpfull: Yes(0) No(0)
• 3
  2*27=one 2
  5*11=one 5
  7*8=three 2's
  25*3= two*5's
  19*4=two*2's
• 9 years agoHelpfull: Yes(0) No(0)