How many 2s are between 1 and 1000?

• no. of 2's from 1 to 10=1
  but additional case arouses for tens n hundred places,i.e.,for no. b/w 21to30n200to300--
  for 21to30=10 additional twos in every hundred count
  for 200to 300=100 additional twos to both of the above cases
  so toal no. of 2's would b=1*(1000/10)+10*(1000/100)+100*(1000/1000)
  =100+100+100
  =300
  so, 300 would be correct n i apolozise for my just previous solution
• 13 years agoHelpfull: Yes(18) No(3)
• no. of 2's bet 1 to 100 = 20
  101-200 = 21
  201-300 = 120
  301- 400= 20
  401-500, 501-600, 601-700, 701-800, 801-900, 901-1000=20
  adding 20*8+21+120= 301
• 13 years agoHelpfull: Yes(5) No(4)
• @harish.............. u forgot to take 20........... while coming to answers
  consider only 2's in ones place (for clear cut explanation)
  - 2,12,32.....92.............with in hundred(10)
  102,112,122,132,142........with in two-hundred(10)like wise upto thousand we will gt 10*10=100..... similarly check for tens and hundredth place total is 100+100+100=300..............
• 13 years agoHelpfull: Yes(4) No(0)
• 300
  the no of 2's between 1 and 100 is 20, and same is for between 301 and 400, 401 and 500, and so on . but in case of the range 201 to 300, 100 additional 2's will be there. so the total no of 2's between 1 and 1000 will be 20*10+100=300
• 13 years agoHelpfull: Yes(4) No(0)
• 200
  any no from 1 to 9 appears 2o time from 1-100
  therefore from 1-1000 it will be 20*10=200
• 13 years agoHelpfull: Yes(3) No(3)
• 300 2's are there.... for 0-9 single digit 1*10pow(1-1)=2
  for 00-99 2digit 2*10pow(2-1)=20
  for 3digit 000-999= 3*10pow(3-1)= 300
  
• 13 years agoHelpfull: Yes(3) No(0)
• ok...sorry with the solution I gave..(Common man's common thinking caught a common mistake)
  I missed a 2 in every series 0f 100, so total no of 2's=290+10=300 !!
• 13 years agoHelpfull: Yes(2) No(1)
• for every series of 100 except that from 200-300 number of 2's=
  =(2 @ ten's place+ 2@ unit's place in a series)*9
  =(10+9)*9=171
  and that for the 200-300= (2-at 100's place)+(2 @ ten's place)+(2 @unit's place)
  =100+10+9=119
  hence the answer is =171+119=290
  //here's a common man's logic...I don't know how every one ended up with 300!!
• 13 years agoHelpfull: Yes(1) No(4)
• we use the formula 3b^2.ans is 300.if b is base 10.if we want 1 to 100 to find 2 we use 2b.
• 13 years agoHelpfull: Yes(1) No(0)
• 300 is seem the answer.........
• 13 years agoHelpfull: Yes(0) No(1)
• between 1 to 100 2 12 21 22 23 24 25 26 27 28 29 32 42 52 62 72 82 92
  total no == 18
  so between 1 to 1000 == 18 * 10 == 180
• 13 years agoHelpfull: Yes(0) No(14)
• since in 1 to 10 there could be only a single 2
  from 11 to 20 there could be there could b 2 2's
  from 21 t0 30 there could be(1(only one could occur at ones place)+9(since there exists one 2 at every tens place from 21 to 29))
  from 31 to 40 only single 2
  similarly from 41 to 50,51to60,61to70,71to80,81to90,91to100 there are total 6*1=6 2's
  so total 2.s from 1 to 100=1+2+10+2+6
  =19
  similarly from 100 to 200=19+1(additional for hundreds place)
  =20
  from 201 to 300=19 + 99(for hundreds place)=118
  while 301to400,501to600,601to700,701to800,801t0900,901to10000 r similar to 1to101 so total from 301to1000=6*(no. of twos from 1to101)
  =6*19
  =96
  hence total 2's from 1 to 1000=19+20+118+96
  =233
• 13 years agoHelpfull: Yes(0) No(12)
• 516
  as 0-9 except 2 r 9 nos.
  for no 1000- 0
  so consider 3digit nos
  3 possibilities- 1 '2' or 2 '2's or 3 '2's
  for 3digit no with only one 2, no of 2=3*(9c1)
  with two 2's, no.=3*2*{(9c1)(9c1)}
  so total=516=27+486+3
  
  
• 13 years agoHelpfull: Yes(0) No(4)
• 20 is the ans...
• 13 years agoHelpfull: Yes(0) No(3)
• 300 is right one
• 8 years agoHelpfull: Yes(0) No(0)
• 200
  In every 100 there must be 20 2's will come so from 1 to 1000 it will be
  20 *10=200
• 6 years agoHelpfull: Yes(0) No(0)
• 20
  in uint digit between 1 to 100 we get "10" 2 s numbers
  similarly ten digit between 1 to 100 we get "10" 2s numbers
  add 10+10=20
• 6 years agoHelpfull: Yes(0) No(0)
• from 1 to 100 there are 19
  so 1 to 1000 there are --190 2's
• 5 years agoHelpfull: Yes(0) No(0)