how many numbers are there between 10-99 such that the number exceeds its sum of its digits by 18?

• let the no. be xy
  then 10x+y-(x+y)=18
  on solving we get x=2 and y gets cancelled out which means that it is independent of the value of second digit in the number
  hence the all the nos starting with 2 .. therefore the nos will be 20-29 which is 10 nos in all
• 10 years agoHelpfull: Yes(50) No(4)
• here, question clearly clarifies that, the sum of their individual digits should be exceeds 18. the number should be between 10-99 .. clearly 9+9 = 18 which means it does not exceeds .. so no number between given range should satisfy given condition.. so zero is the answer. am i right ?
• 10 years agoHelpfull: Yes(6) No(1)
• the numbers will be from 20-29
  20=2+18,
  21=3+18,
  22=2+18,
  and so on upto
  29=11+18
  hence total possible values= 10
• 10 years agoHelpfull: Yes(5) No(0)
• the question is not clear. plz somebody clear the question first.
• 10 years agoHelpfull: Yes(2) No(0)
• if 10x+y-(x+y)=18 ;
  x=2
  so numbers from 20 to 29.
  in total = 10 nos.
• 10 years agoHelpfull: Yes(2) No(0)
• there are 10 possible no.s i.e from 20-29
• 10 years agoHelpfull: Yes(0) No(2)
• 1+1=2 exceed by 18 than it 20. similarly 1+2=21,1+3=22,1+4=23,1+5=24,1+6=25,1+7=26,1+8=27,1+9=28,2+9=29 hence no.is 9
  
• 10 years agoHelpfull: Yes(0) No(1)
• zero;;;jkvbfjhvb
• 10 years agoHelpfull: Yes(0) No(1)
• i think its 1 only 99 satisfy the condition i.e 9+9=18 other combination cannot exceed sum be 18
  
• 10 years agoHelpfull: Yes(0) No(2)