number of zeros in 1*2^2*3^3*4^4*....49^49

• In this problem, there are more 2's than 5's. so let us count the number of 5's.
  
  5^5 contributes 5 5's
  10^10 contributes 10 5's
  15^15 -> 15 5's
  20^20 -> 20 5's
  25^25 -> 5^50 that is 50 5's
  30^30-> 30 5's
  35^ 35-> 35 5's
  40^40-> 40 5's
  45^45 -> 45 5's
  
  so number of 5's = 5 + 10 + 15 + 20 + 50 + 30+ 35+40+45 = 250
  
  so there are 250 0's.
• 9 years agoHelpfull: Yes(14) No(0)
• total no. of 5's = 5 + 10 + 15 + 20 + 50 + 30 + 35 + 40 + 45 = 250
  
  
• 10 years agoHelpfull: Yes(2) No(0)
• @kumari manisha:how do u calculate the contribution of 5???pls explain
• 9 years agoHelpfull: Yes(1) No(0)
• no of zeros=(5+10+15+20+25+30+35+40+45)=225
• 9 years agoHelpfull: Yes(1) No(2)
• when u add 5+10+15....45 u will get 225 u will not get 250 can u explain how u got 250
• 9 years agoHelpfull: Yes(1) No(1)
• Ans is 250
• 9 years agoHelpfull: Yes(1) No(0)