Suppose n is an integer, such that the sum of the digits of n is 2, and 10^10 < n < 10^11 . The number of different values for n is:

(i) 11
(ii) 10
(iii) 9
(iv) 8

• a)11
  10000000001(first number)
  now 1 can be placed in 9 other positions to give 9 numbers
  20000000000 , one more number
  total=1+9+1=11
• 13 years agoHelpfull: Yes(14) No(0)
• 10000000001
  10000000010
  10000000100
  10000001000
  10000010000
  10000100000
  10001000000
  10010000000
  10100000000
  11000000000
  20000000000
  
  so the ans is 11
• 13 years agoHelpfull: Yes(12) No(0)
• 10^10=10000000000 this is 9 possiblity to get sum=2 and after 10^11=100000000000..total value of n is(1+9+1).
• 13 years agoHelpfull: Yes(4) No(0)
• the qstn says dat the sum of digits is 2.Lookng onto d options we get knw d no. is 11 as 1+1=2.
• 13 years agoHelpfull: Yes(3) No(2)
• 11
• 13 years agoHelpfull: Yes(2) No(2)
• 10 values are there...... which is the 11th one please xplain..
  1100000000
  1010000000
  1001000000
  1000100000
  1000010000
  1000001000
  1000000100
  1000000010
  1000000001
  2000000000
• 13 years agoHelpfull: Yes(2) No(2)
• ans (iii)9
  
  the options are
  1100000000
  1010000000
  1001000000
  1000100000
  1000010000
  1000001000
  1000000100
  1000000010
  1000000001
  
• 13 years agoHelpfull: Yes(0) No(4)
• as we know 10^10=10000000000,and 10^11=100000000000 but 10^10
• 13 years agoHelpfull: Yes(0) No(3)
• 10000000000 t0 100000000000
  ans is 10
• 13 years agoHelpfull: Yes(0) No(5)
• sum of the digits in n so from the options i take n as 11 so 1+1=2 ...and 11 also lies between 10^10 and 10^11
• 7 years agoHelpfull: Yes(0) No(0)