27. Find the number of zeroes in 11*22*33*44 …. 4848*4949 ?

• Plz give explanation teja...
• 10 years agoHelpfull: Yes(7) No(1)
• 11(1*2*3*.....449)=11*449!
  449! contains 109 5's , so no of zeros=109
• 10 years agoHelpfull: Yes(7) No(6)
• Question is wrong.
  correct question with solution-
  number of zeros in 1*2^2*3^3*4^4*....49^49
  
  Let us count the number of 5's.
  
  5^5 contributes 5 5's
  10^10 contributes 10 5's
  15^15 -> 15 5's
  20^20 -> 20 5's
  25^25 -> 5^50 that is 50 5's
  30^30-> 30 5's
  35^ 35-> 35 5's
  40^40-> 40 5's
  45^45 -> 45 5's
  
  so number of 5's = 5 + 10 + 15 + 20 + 50 + 30+ 35+40+45 = 250
  
  so there are 250 0's.
  
  
• 8 years agoHelpfull: Yes(7) No(0)
• ans is 250......actually the question is wrong ..the question is 1^1*2^2.....49^49
• 9 years agoHelpfull: Yes(5) No(1)
• think it as 4949!
  we need to determinr no of 5's in it
  4949/5 its quotient is 989
  989/5 its quotient is 197
  197/5 its quotient is 39
  39/5 its quptient is 7
  7/5 its quotient is 1
  so, it we all all of these quotient it will gibve answer 1231 no. of 5 in 4949!
  we have to reduce first 10 digits
  their contains only 2 no of 5's(5 and 10)
  s0, 1231-2=1229
• 10 years agoHelpfull: Yes(4) No(9)
• 11 22 33 ... 1111 1212 .... 2525 ....4848 4949
  no of 5 in
  55=1
  1010=1
  1515=1
  2020=1
  2525=2
  3030=1
  3535=1
  4040=1
  4545=1
  total=10
  or
  u can also find no of zero in 49!=10
• 9 years agoHelpfull: Yes(4) No(4)
• Number should be a mirror image:(All we need is numbers with 5 and o in the end)
  55=11*5^1
  555=111*5^1,calculate in the same way
  1010,1515,2020,2525,3030,3535,4040,4545......adding powers of 5,Total=11 As there are many 2's.So when 5 and 2 multiplies,zeroes will come.Answer=11
• 10 years agoHelpfull: Yes(2) No(0)
• 1229(ans)
  as no.of 5s in 11*22...4949=1229
• 10 years agoHelpfull: Yes(1) No(0)
• please give solution
• 10 years agoHelpfull: Yes(0) No(0)
• question is wrong because the last terms 4848 and 4949 are not divisible perfectly.
• 10 years agoHelpfull: Yes(0) No(1)
• Answer is 9
  
• 10 years agoHelpfull: Yes(0) No(0)
• Can anybody provide the answer
• 10 years agoHelpfull: Yes(0) No(0)
• plzzz xpln it clearly
  
• 10 years agoHelpfull: Yes(0) No(0)
• The answer is 1230
  lets !st find out how many zeros are there in 4949! by multiplying and dividing the expession by 10!.
  So, no of zeros in 4949! will be 1232
  the substract it by 2, ie. 1232-2=1230 (since we have 2 zeros in 10!)
  So 1230 is the correct answer...
• 9 years agoHelpfull: Yes(0) No(1)
• The actual question is
  Find the number of zeroes in
  1^1 *2^ 2 *3 ^3 *......48^48 *49^49? a) 250 b) 225 c) 545 d) 135
• 6 years agoHelpfull: Yes(0) No(0)