a person starts writing all 4 digits numbers.how many times had he written the digit 2??

• 1)when 2 is at unit place=9*10*10*1
  2)when 2 is at tenth place=9*10*1*10
  3)when 2 is at hundred place=9*1*10*10
  4)when 2 is at thousand place=1*10*10*10
  so total no. of 2s=900+900+900+1000=3700
• 10 years agoHelpfull: Yes(97) No(14)
• Number of 2's at units place(1000 to 9999)=900
  Number of 2's at tenths place(1000 to 9999)=900
  Number of 2's at hundreds place(1000 to 9999)=900
  Number of 2's at thousands place(1000 to 9999)=1000
  
  therefore total number of 2's =(900+900+900+1000)=3700
• 10 years agoHelpfull: Yes(39) No(10)
• Total 4 digit numbers that we can form is (9*10*10*10)
  Total 4 digit numbers that we can form with out 2 is (8*9*9*9)
  so 4 digit nos with 2 we can form (9*10*10*10)-(8*9*9*9) = 3168
  
  In 3700 i feel there is redundancy becz nos are getting repeated ..
  correct me if i am wrong ..
• 10 years agoHelpfull: Yes(23) No(11)
• (1)At unit's digit = 10*10*9 = 900 (10 times in every 100)
  (2)At ten's digit = 10*10*9 = 900 (10 times in every 100)
  (3)At hundred's digit = 100*9 = 900 (100 times in every 1000)
  (4)At thousand's digit = 1000 (2000 to 2999)
  So total 900+900+900+1000=3700 times.....ans
• 10 years agoHelpfull: Yes(13) No(2)
• No of times a digit repeats from 0 to 99 = 300
  
  Going by that logic we have 3100 repetitions from 0 to 999
  
  Therefore
  
  from 1000 to 1999 = 3100
  from 2000 to 2999 = 3100+1000 (Since every one contains 2)
  from 3000 to 3999 = 3100
  from 4000 to 4999 = 3100
  from 5000 to 5999 = 3100
  from 6000 to 6999 = 3100
  from 7000 to 7999 = 3100
  from 8000 to 8999 = 3100
  from 9000 to 9999 = 3100
  
  Therefore it comes to (3100*10)+1000 = 32,000
  
• 10 years agoHelpfull: Yes(7) No(50)
• correct ans is: 4961.
  case 1: when no contain only one digit of 2.
  10*10*10+9*10*10*3=3700 i.e total 2's are=3700*1=3700.
  case 2: when no contain two digits of 2.
  10*10*3+9*10*3=570 i.e total 2's are=570*2=1140.
  case 3: when no contain three digits of 2.
  10*3+9=39 i.e total 2's are=39*3=117.
  case 4: when no contain four digits of 2.
  1 i.e total 2's are=1*4=4
  hence total no of 2's are = 3700+1140+117+4 = 4961.
• 10 years agoHelpfull: Yes(6) No(7)
• 2's at units place(1000 to 9999)=900
  2's at tenths place(1000 to 9999)=900
  2's at hundreds place(1000 to 9999)=900
  2's at thousands place(1000 to 9999)=1000
  2's =(900+900+900+1000)=3700
  
• 10 years agoHelpfull: Yes(5) No(3)
• When 2 is in unit place- 9*10*10*1= 900 Times.....
  When it is in tens place 9*10*1*10=900 Times.......
  When it is in hundred's place 9*1*10*10=900 Times.......
  When it is in 1000's place 1*10*10*10=1000 Times.......
  adding all the possibilities will give you 900+900+900+1000= 3700
• 8 years agoHelpfull: Yes(2) No(2)
• Answer is 3700
  from 10-19 --> 1
  from 20-29 --> 11
  from 30-39 --> 1
  from 40-49 --> 1
  from 50-59 --> 1
  from 60-69 --> 1
  from 70-79 --> 1
  from 80-89 --> 1
  from 90-99 --> 1
  
  which makes 20 from 0 to 99
  
  now please observer that similar pattern follows for
  000 - 099 -->20
  100 - 199 --> 20
  300 - 399 --> 20
  400 - 499 --> 20
  500 - 599 --> 20
  600 - 699 --> 20
  700 - 799 --> 20
  800 - 899 --> 20
  900 - 999 --> 20
  
  however for 200 to 299 we have that extra 2 as first digit for all these 100 numbers, so
  200 - 299 --> 120
  
  so
  000 - 999 -->300
  
  so for
  1000 - 1999 -->300
  3000 - 3999 -->300
  4000 - 4999 -->300
  5000 - 5999 -->300
  6000 - 6999 -->300
  7000 - 7999 -->300
  8000 - 8999 -->300
  9000 - 9999 -->300
  
  again for
  2000 - 2999 -->300 + 1000 as we have that extra 2 in the first digit for all these numbers, hence
  
  2000 - 2999 -->1300
  
  So total number of times digit 2 is used 1000 - 9999 is 3700
• 7 years agoHelpfull: Yes(1) No(1)
• WHICH IS THE CORRECT ANSWER
  
• 7 years agoHelpfull: Yes(0) No(1)
• 4digit number=1111......9999 so 2 divide so 3700 is divided by 4 digit number
• 7 years agoHelpfull: Yes(0) No(1)
• 1)when 2 is at thosand place=1*10*10*10
  2)when 2 is at hundred place=8*1*10*10(since we have already used 2 in thousand place we can pic one number from remaining 8 ,suppose if we take 9*1*10*10 it is a mistake.for example in case 1(ie when 2 is at thousand place there is a possibility of occourence of a 4 digit number 2257 which is also possible in case 2 when we consider as 9*1*10 *2 as we can take any number from 1 to 9 inthousand place which can also be a 2 )
  3)when 2 is at tens place=7*10*1*10(similar to the above explination )
  4)when 2 is in units place=6*10*10*1
  so total no. of 2s=600+700+800+1000=3100
• 7 years agoHelpfull: Yes(0) No(1)
• 1)when 2 is at unit place=9*10*10*1
  2)when 2 is at tenth place=9*10*1*10
  3)when 2 is at hundred place=9*1*10*10
  4)when 2 is at thousand place=1*10*10*10
  so total no. of 2s=900+900+900+1000=3700
• 5 years agoHelpfull: Yes(0) No(1)