find the number of zeroes in 1^1*2^2*3^3*4^4 …. 48^48*49^49 ?
a. 250
b. 225
c. 545
d. 135

• only 2*5 gives 10
  
  find total exponent of 5 in given product
  
  5^5 =(1*5)^5 => 5^5
  10^10=(2*5)^10 => 5^10
  15^15=(3*5)^15 => 5^15
  20^20=(4*5)^20 => 5^20
  25^25=(5*5)^25 => (5^25)^2
  30^30=(6*5)^30 => 5^30
  35^35=(7*5)^35 => 5^35
  40^40=(8*5)^40 => 5^40
  45^45=(9*5)^45 => 5^45
  
  exponent of 5 in product = 5+10+15+20+(2*25)+30+35+40+45
  = 5*(1+2+...+9)+25
  = 250
  
  i.e product has a factor 5^250 & exponent of 2 is much more than 5
  
  so, number of zeroes = 250
• 10 years agoHelpfull: Yes(27) No(0)
• consider only 5,10,15,20.....45
  so add the number of 5 in these occurence
  so 5+10+15+20+50+30+35+40+45=250
  a is the answer
• 10 years agoHelpfull: Yes(13) No(1)