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

• ans =250
  0 will be produced when an even number be multiplied by 5
  or a num multiplied by 10
  now 10^10=10 0's
  20^20= 20 0's
  30^30=30 0's
  40 ^40=40 0's
  5^5=5 0's
  15^15=15 0's
  35^35=35 0's
  45^45=45 0's
  but 25^25 can be written as
  5^25*5^25=50 0's
  summing up 250 0's
  
• 12 years agoHelpfull: Yes(38) No(4)
• ans is 250
• 12 years agoHelpfull: Yes(17) No(6)
• I apologize for my incorrect solution as I read the question incorrectly.
  the number of zero's is 225
  
  the number of zero's can be found by finding number of 2's and 5's.
  since in this number, number of occurrences 2's will always be greater than number of occurrences of 5's.
  therefore we need to find number of 5's only.
  number of occurrences of 5's
  in 5^5 = 5
  10^10=10
  15^15=15
  20^20=20
  25^25=25
  30^30=30
  35^35=35
  40^40=40
  45^45=45
  sum=225.
  therefore 225 5's can be paired with 225 2's.
  total number of zeros = 225
  
  
• 11 years agoHelpfull: Yes(8) No(7)
• when 25^ 25 then no of 5 is 50 not 25....
  bcoz ((5^2)25)....5^50 so total is 250 exact...
• 12 years agoHelpfull: Yes(6) No(4)
• total zeros =5+10+15+20+25+30+35+40+45
  = 225
• 12 years agoHelpfull: Yes(4) No(14)
• we know that 0 will occur when 2 will be multiplied by 5 ,here 5 will be less than 2 so we will calculate power of 5
  now 10^10=10 0's
  20^20= 20 0's
  30^30=30 0's
  40 ^40=40 0's
  5^5=5 0's
  15^15=15 0's
  35^35=35 0's
  45^45=45 0's
  but 25^25 can be written as
  5^25*5^25=50 0's
  
  total ans =250
  
• 9 years agoHelpfull: Yes(4) No(0)
• we calculate no of 5 on once place 5^5+15^15+25^25+35^35+45^45+(2*5)^10+(2*5)^20+(2*5)^30+(2*5)^40=225
• 12 years agoHelpfull: Yes(2) No(11)
• 20 zeros.
  
  
  this number can be viewed as (49!)^2
  now to find the number of zeros we need to find the occurrences of 2's and 5's only.
  [] denotes the floor value.
  occurrences of 2's =[49/2]+[49/(2^2)]+[49/(2^3)]+[49/(2^4)]+[49/(2^5)]
  =24+12+6+3+1
  =46
  occurrences of 5's =[49/5]+[49/(5^2)]
  =9+1
  =10
  10 5's can be paired with 10 2's out of 46.
  therefore number of 10's in 49! = 10.
  therefore number of 10's in (49!)^2=20.
  
  total number of 0's = 20.
• 11 years agoHelpfull: Yes(1) No(6)
• neeraj kumar note this.. 25^25 = 50 zeros...therefore 250...
• 10 years agoHelpfull: Yes(1) No(1)
• the fives(5) will be less than twos(2).Hence we need to count only the fives.
  thus:5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45
  gives us:5+10+15+20+25+30+35+40+45 fives.Thus the product has 250 zeros
  
• 9 years agoHelpfull: Yes(1) No(1)
• how plz explain it clearly........
  
• 12 years agoHelpfull: Yes(0) No(1)
• it is 49!*49!
  (49/5 + 49/(5^2) )*2 = 20 zeros
• 10 years agoHelpfull: Yes(0) No(2)
• 18 we have 10*10,20*20,30*30,40*40;
  and 5*5,15*15,25*25,35*35,45*45 each 5 will result a 0(5*2=10)
  hence total zero=8+10=18
• 10 years agoHelpfull: Yes(0) No(1)
• (5^5)*(2*5)^10*(3*5)^15*(2*2*5)^20*(5*5)^25*(2*3*5)^30*(5*7)^35*(2*2*2*5)^40*(3*3*5)^45
  here we'll take only 5
  Total number of five
  (5^5) = 5
  (2*5)^10 =(2^10*5^10) 10
  (3*5)^15 =(3^15*5^15) 15
  (2*2*5)^20 =(2^20*2^20*5^20) 20
  (5*5)^25 =(5^25*5^25) 25+25
  (2*3*5)^30 =(2^30*3^30*5^30) 30
  (5*7)^35 =(5^35*7^35) 35
  (2*2*2*5)^40 =(2^40*2^40*2^40*5^40) 40
  (3*3*5)^45 =(3^45*3^45*5^45) 45
  
  total = 250
  
  Note :: most of person confused ..why (25+25) ...because when we will (5*5)^25=(5)^50
• 8 years agoHelpfull: Yes(0) No(0)