What is the no of occurance of 5 in the given series121122123124125126.........356

• No. of 5 betw. 121-221= 20
  No. of 5 betw. 222-322= 20
  No. of 5 betw. 323-356= 11
  so total no. of occurance of 5 will be 20+20+11=51
• 10 years agoHelpfull: Yes(13) No(1)
• 51.First consider 125-200 the number of occurrences are 18. 200-300 are 20. and 13
  ..18+20+13=51
• 10 years agoHelpfull: Yes(5) No(2)
• Ans: 51
  On considering the individual terms of the series as 3-digit no.s
  The series looks like.... 121 122 123 124 125........356
  i.e. the series is just the no.s from 121 to 356.
  On moving from 121 to 356,no. of occurrences of 5 = 51
  (counting the 5s in units place and 10s places)
• 10 years agoHelpfull: Yes(4) No(1)
• 52 is the answer
  
• 10 years agoHelpfull: Yes(2) No(4)
• 51 is the answer
• 10 years agoHelpfull: Yes(1) No(0)
• from 121-200:125,135,145,150,151,...155,...159,165,175,185,195= 18 5's
  from 200-300: 20 5's
  from 300-356:305,315,325,335,345,350,351,352,353,354,355,356=12 5's
  ans= 40
  
• 10 years agoHelpfull: Yes(1) No(0)
• from 120 t0 200 18 5's are there,200 to 300 20 5's next 305,315,325,335,345,350,351,352,353,354,355,356 13 5's so total 20+18+13=51
• 10 years agoHelpfull: Yes(1) No(0)
• 36 as it is from 121 then 122 then to 129 one 5 in each set so in this way there is a 5 in each of the sets.again in the set 151 to 159 there are all 5,s,so counting in total there are 36 5's in the set.
  
• 10 years agoHelpfull: Yes(0) No(3)
• It will be 47
  
• 10 years agoHelpfull: Yes(0) No(2)
• 5 in 1 to 100 = 20 (10in 05,15,25,...95(Unit digit) + 10 in 50,51,52,53....59(ten's place))
  101 to 200 = 20
  201 to 301 = 20
  =60
  +1(of 305)+1(315)+1(325)+1(335)+1(345)+2(355)+1(356)=8
  =68
• 10 years agoHelpfull: Yes(0) No(3)
• 121 to 200 - 18 5's
  201 to 300 - 20 5's
  301 to 356 - 13 5's
  so total 51
• 10 years agoHelpfull: Yes(0) No(0)