Find the no of zeros in the product of 1^1*2^2*3^3*.....*49^49??

• consider the case of multiples of 5
  
  5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45
  or
  5^5 x 10^10 x 15^15 x 20^20 x (5^2)^25 x 30^30 x 35^35 x 40^40 x 45^45
  or
  5^5 x 10^10 x 15^15 x 20^20 x 5^50 x 30^30 x 35^35 x 40^40 x 45^45
  
  total zeros at the end of product 5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45 will be 5+10+15+20+50+30+35+40+45 = 250 zeros
• 11 years agoHelpfull: Yes(79) No(17)
• 2*5=10
  5^5 5 times 5
  10^10 10 times 5
  15^15 15
  ......
  25^25 50
  .....
  5+10+15+20+2*25+30+35+40+45=250
• 11 years agoHelpfull: Yes(12) No(4)
• can u explain me the logic behind taking multiples of 5. 5^5 do not contain single 0
• 11 years agoHelpfull: Yes(10) No(6)
• same as of aman but
  5+10+15.......+45
  
  gives sum 225
• 11 years agoHelpfull: Yes(7) No(9)
• Normally to count number of zeroes we need no of 5x2 pairs
  As the pair 5x2=10 generate a zero.
  In this case more number of 2's than 5.
  So it would b sufficient to count no of 5,s
  5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45
  IN 25^25 WE GET 50 5'S AS 25=5*5
  COUNT TOTAL OCCURENCES OF 5
  5+10+15+20+50+30+35+40+45= 250 ZEROES
• 9 years agoHelpfull: Yes(6) No(0)
• 10^2*20^2*30^2*40^2
  Then we get the number of zeros as = 8
  i.e eight zeros.
• 11 years agoHelpfull: Yes(3) No(48)
• Can u explain clearly
• 9 years agoHelpfull: Yes(2) No(0)
• No. Of zeros = no. Of 5
  So just count how many 5s are .
• 9 years agoHelpfull: Yes(1) No(1)