no.of zeros in 1*2^2*3^3*....49^49

• count the number of 5's
  
  5^5 => 5 5's
  10^10 = 2^10*5^10 => 10 5's
  15^15 => 15 5's
  20^20 => 20 5's
  25^25 =(5^2)^25 = 5^50 => 50 5's
  30^30 => 30 5's
  35^35 => 35 5's
  40^40 => 40 5's
  45^45 -> 45 5's
  
  total no. of 5's = 5 + 10 + 15 + 20 + 50 + 30 + 35 + 40 + 45 = 250
  
  i.e. product has a term 5^250 , also no. of 2's is much more than 250
  
  no.of zeros in 1*2^2*3^3*....49^49 = 250
• 10 years agoHelpfull: Yes(22) No(4)
• ans is 250.
• 10 years agoHelpfull: Yes(3) No(3)
• 0,it does not consist any zero.
• 10 years agoHelpfull: Yes(1) No(12)
• no. of zero =20
  1*2^2*3^3.....49^49=(49!)^2
  and no of zero in 49! is =no. of 5's
  no. of 5 in 49! =49/5+49/5^2+....
  total no of 5 in 49!=10
  and no of 0's in (49!)^2=10*2=20
• 10 years agoHelpfull: Yes(1) No(3)
• Answer will be 250
• 10 years agoHelpfull: Yes(0) No(1)
• 105..
  5^5 has one 5 with which any even no will multiply to give one 0,so for 1,25,35,45...so total five 0's.and 10^10,20^20,30^30,40^40 gives 10+20+30+40=100 0's.They will multiply to produce 105 no of 0's.
• 10 years agoHelpfull: Yes(0) No(3)