How many of number x ( x being integer) with 10<= x<= 99 are 18 more than the sum of their digits.

• Numbers from 20 to 29 ,total 10
  
  If '10a+b' is a number which is 18 more than the sum of the digits, then
  10a+b= 18+a+b ,9a=18 or a=2
  So numbers with a=2 and b=0,1,2,....., 9, can be formed with given criteria.
  Such a numbers from 10*2+b= 20, 21, 22, ........, 29 ,total 10 will be there.
• 8 years agoHelpfull: Yes(14) No(2)
• in the problem they given that sum of their digits is 18 and above it is possible when the number is 99
• 8 years agoHelpfull: Yes(4) No(1)
• 10a+b=a+b+18
  So 9a=18.
  a=2. Putting a=2, numbers are (10*2+1, 10*2+2, ..... 10*2+9) means 21,22,...29.
  So there are 10 such numbers...
• 8 years agoHelpfull: Yes(1) No(1)
• Suppose we have total amount is Rs 100
  we spent 1st rs 10% of 100 i.e rs 10
  now left amountis 100-10=rs 90 in my wallet
  again we spent 10% of rest amount 90 i.e rs 9
  so now in my wallet avail amount is 90-9= Rs81
  
• 8 years agoHelpfull: Yes(0) No(7)
• x+y+18=10x+y
  =>9x=18
  =>x=2
  numbers are 20, 21, 22,23, 24,25,26,27,28,29
  so total number of this type of number is equal to 10
• 8 years agoHelpfull: Yes(0) No(0)
• The number is 10n + m = 18 + n + m ==> 9n = 18 ==> n = 2 , m is any digit
  
  So, the numbers are 20, 21, 22,......29. ten numbers in total.
• 8 years agoHelpfull: Yes(0) No(0)
• not understood question
  
• 8 years agoHelpfull: Yes(0) No(0)
• 20 to 29.
  total 10 numbers.
• 8 years agoHelpfull: Yes(0) No(0)