find the number of zeros in the product 1^1+2^2.......+49^49?

• we have to count no.of 5's and no.of 10's
  5^5=5
  10^10=10
  15^15=15
  20^20=20
  25^25=(5*5)^2
  30^30=30
  35^35=35
  40^40=40
  45^45=45
  i.e total=5+10+15+20+50+30+35+40+45=250 zeros
• 11 years agoHelpfull: Yes(17) No(1)
• 10^10,20^20,30^30,40^40 are the numbers which contains zero.
  so,the number of zeros are: 10+20+30+40=100.
• 11 years agoHelpfull: Yes(14) No(9)
• qn should be find the number of zeros in the product 1^1*2^2*3^3....48^48*49^49
  count the no. of 5's in 1^1*2^2*3^3....48^48*49^49
  5^5*10^10*15^15*20^20*25^25*30^30*35^35*40^40*45^45(only these terms consists 5)
  =5^5*5^10*5^15*..............*5^45
  so no. of 5's in 1^1*2^2*3^3....48^48*49^49=5+10+15+20+25+25+30+35+40+45( as 25^25=5^50)
  =5(1+2+3+........+9)+25
  =5*45 +25=250
  no. of zZEROES=250
• 11 years agoHelpfull: Yes(9) No(1)
• we have to count no. of 5's so count powers of these as...
  5^5..+10^10+15^15.....45^45
  =>sum of powers 5+10+15+20+25+25+30+35+40+45=250
  25 is two times cz 25^25=(5^2)^25=>5^50.
  
• 11 years agoHelpfull: Yes(5) No(0)
• Sorry guys the solution by ABIRA is correct.
  250 is ans in case 1^1*2^2*3*3.....
• 11 years agoHelpfull: Yes(3) No(1)
• this is not the correct qsn. . it should be as
  1^1*2^2*3^3........49^49.
  
  then no of zeros are = 250.
• 11 years agoHelpfull: Yes(2) No(0)
• 250 is the answer...
• 11 years agoHelpfull: Yes(1) No(1)
• formula used=n(n+1)(2n+1)/6
  ans=40425
• 11 years agoHelpfull: Yes(0) No(14)
• To each reducing power 2 zeros will be added.. so 49x2=98 zeros..
• 11 years agoHelpfull: Yes(0) No(4)