In this question A^B means A raised to the power of B Start with the integers from 1 to 10^2012. Replace each of them by the sum of its digits to get a string of 10^2012 numbers. Keep doing this until you get 10^2012 single digit numbers. Let m be the number of 1's and n be the number of 2's. Then m - n

• We divide all the numbers into groups of 9. (1 to 9), (10 to 18).....
  Now when we divide each term in the first group by 9, we get 1, 2, 3 ..0 as remainders. Now digit sum is nothing but finding remainder when a number is divided by 9. So the last term, 10^2012 gives remainder 1 when divided by 9. So there is one "1" extra than 2's.
• 8 years agoHelpfull: Yes(8) No(26)
• please any one ans this question corectly....
  
• 8 years agoHelpfull: Yes(3) No(6)
• We divide all the numbers into groups of 9. (1 to 9), (10 to 18).....
  Now when we divide each term in the first group by 9, we get 1, 2, 3 ..0 as remainders. Now digit sum is nothing but finding remainder when a number is divided by 9. So the last term, 10^2012 gives remainder 1 when divided by 9. So there is one "1" extra than 2's.
  
• 8 years agoHelpfull: Yes(1) No(8)
• We divide all the numbers into groups of 9. (1 to 9), (10 to 18).....
  Now when we divide each term in the first group by 9, we get 1, 2, 3 ..0 as remainders. Now digit sum is nothing but finding remainder when a number is divided by 9. So the last term, 10^2012 gives remainder 1 when divided by 9. So there is one "1" extra than 2's.
  ans- 1
• 7 years agoHelpfull: Yes(0) No(9)