a person starts writing all 4 digits numbers .how many times had he written the digit 2?
a)3700
b)32000
c)3200
d)37000

• ans:3700
  at 1's place=900 times
  at 10's place=900 times
  at 100's place=900 times
  at 1000's place(from 2000 to 2999)=1000 times
  total=1000+900+900+900=3700
• 10 years agoHelpfull: Yes(35) No(0)
• ans:a
  _ _ _ _
  2 (10) (10) (10)
  _ _ _ _
  (9) 2 (10) (10)
  _ _ _ _
  (9)(10) 2 (10)
  _ _ _ _
  (9)(10)(10) 2
  
  10*10*10+3(9*10*10)=3700
• 10 years agoHelpfull: Yes(29) No(3)
• No of two at unit digit is=900
  No of two at ten digit is=900
  No of two at hundred digit is=900
  No of two at thousand digit is=1000
  so 2 is repeated in (900+900+900+1000)=3700 is the ans
• 10 years agoHelpfull: Yes(5) No(1)
• numbers with 2 at unit digit=9*10*10=900,
  '' '' 2 at tens place=900,
  '' '' 2 at hundred place=900,
  '' '' 2 at thousand place=10*10*10=1000.therefore,number of times 2 is written=(900*3)+1000=3700
• 10 years agoHelpfull: Yes(4) No(0)
• The 4-Digit Numbers are from 1000 to 9999.
  Digit 2 is written in Units Place 900 times; In Ten's Place 900 times; In Hundred's Place 900 times and in Thousands Place 1000 times. So total=1000+900+900+900=3700
• 10 years agoHelpfull: Yes(3) No(2)
• Ans is 3700.
  from 1000 10 1999 consists of 300 digit 2. likewise 9*300(because 1000 to 9999)=2700
  also from 2000 t0 2999 consists extra 1000.
  therefore 2700+1000=3700
• 10 years agoHelpfull: Yes(2) No(0)
• Ans is 3700.
  from 1000 10 1999 consists of 300 digit 2. likewise 9*300(because 1000 to 9999)=2700
  also from 2000 t0 2999 consists extra 1000.
  therefore 2700+1000=3700
• 10 years agoHelpfull: Yes(1) No(0)
• 90*20+1000+900=3700
  in evry 100 no's 20 2's,so we hve 90 100's,2000-2999 we hve 1000 2's,1200-1299 2200-2299....9200-9299 we hve 900 2's...
• 10 years agoHelpfull: Yes(0) No(0)
• total=1000+900+900+900=a)3700
• 10 years agoHelpfull: Yes(0) No(0)
• 2001
  2002
  2003
  .
  .
  .
  2999
  hence digit 2 will repeat 1000 times at 4'th(1000's) place. Now
  1200
  1201
  1202
  .
  .
  1299 hence 2 repeat 100 times here.replace 1000's place(1) by 2,3,4,5,6,7,8,9.so digit 2 repeat 100*9=900 times at 100's place
  similarly for 10's place
  1020
  1021.
  .
  .
  1029
  .
  1120
  1129.
  .
  1220
  1229
  .
  .and so on
  so the total no=9*10*10=900
  similarly for 1's place
  1002
  1009
  .
  .
  .
  so the final ans=1000+900+900+900=3700
• 8 years agoHelpfull: Yes(0) No(0)
• Total number of four digit numbers =9000 (i.e 1000 to 9999 )
  We try to find the number of numbers not having digit 2 in them.
  Now consider the units place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
  Tens place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
  Hundreds place it can be selected in 9 ways (i.e 0,1,3,4,5,6,7,8,9)
  Thousands place can be selected in 8 ways (i.e 1,3,4,5,6,7,8,9) here '0' cannot be taken
  Total number of numbers not having digit 2 in it =9 x 9 x 9 x 8 =5832
  Total number of numbers having digit 2 in it = 9000-5832 =3168
  
• 8 years agoHelpfull: Yes(0) No(1)