A two digit number is 18 less than the square of the sum of its digits. How many such numbers are there?
(1)1
(2)2
(3)3
(4)4

• 2 is the answer. we have to see the numbers which become perfect squares..hence we have to check only 25,36,49,64,81,100 which are greater than 18.subtracting 18 from 81 we get 63 which fullfills the condition mentioned.........
• 10 years agoHelpfull: Yes(46) No(17)
• take 63 and 82.
  xy+18=(x+y)^2.
  63+18=(6+3)^2.
  82+18=(8+2)^2.
• 10 years agoHelpfull: Yes(28) No(19)
• 10x+y=(x+y)^2-18
  hence the above eq is a second degree polynomial it has two solutions
  therefore the ans is 2
• 9 years agoHelpfull: Yes(22) No(0)
• xy+18=(x+y)^2
  5^2=25,25-18=7
  6^2=36,36-18=18 and 1+8=9
  7^2=49
  8^2=64,64-18=64
  9^2=81,81-18=63 and 6+3=9,9^2=81 so xy=63 answer is 1.
• 10 years agoHelpfull: Yes(6) No(24)
• For the above condition (a+b)^2=(10a+b)+18
  where ab is the two digit number..
  
  Possible values of (a+b)^2 are 16, 25, 36, 49, 64, 81, 100. from which 18 can be subtracted to get the possible values of the 2 digit. They are 18, 31, 56, 63, 82.
  
  By trail and error, you get the the values for above conditions, which will be 63 & 82.
  
  (6+3)^2=(63+18)=81
  (8+2)^2=(82+18)=100
• 8 years agoHelpfull: Yes(6) No(0)
• (a+b)^2=(10a+b)+18
  Target- to find perfect square and substract 18 from that and after that satisfy the above equation
  so values of perfect square for two digit - 16,25,36,49,81,100.
  16-18=-8 can't take.25-18=7 can't take as it is not two digit no. 36-18=18 can take .49-18=31 can take. 81-18=63. and for 100-18=82.
  Now corresponding values we get as 18,31,63,82.
  Now satisfy the equation (a+b)^2=(10a+b)+18 with these values.
  Only 63 and 82 is possible.
  Thanx
  
  
  
• 8 years agoHelpfull: Yes(4) No(2)
• xy+18=(x+y)^2.
  by applying the values to x and y from 1 to 9 we get these two which satisfies the given condition.
  63+18=(6+3)^2.
  82+18=(8+2)^2.
  
  you can check it by inserting this formulae in the basic c program
• 10 years agoHelpfull: Yes(3) No(2)
• 36 and 81
  cz
  36–18=18
  81–18 =63
  8+1=6+3=9
  both satisfies the condition
  
• 10 years agoHelpfull: Yes(1) No(11)
• answer is 4
  two digit nos 36,49,64,81
• 10 years agoHelpfull: Yes(1) No(21)
• ans = 2 ;
  63&81 and 82&100
• 10 years agoHelpfull: Yes(1) No(3)
• RAMANAREDDY why did u assume 63 and 82 ????
• 9 years agoHelpfull: Yes(1) No(1)
• 63 satisfies the condition
  
  63=(6+3)²–18
• 10 years agoHelpfull: Yes(0) No(6)
• @manokar what isthe logicbehind perfect square
• 9 years agoHelpfull: Yes(0) No(0)
• (a+b)^2=(10a+b)+18
  Target- to find perfect square and substract 18 from that and after that satisfy the above equation
  
• 8 years agoHelpfull: Yes(0) No(0)
• Take N = 10a+b.
  Given that, (10a+b)+18 = K2 = (a+b)2
  Given number = K2
  
  - 18 = (10a+b)
  
  That means, when we add 18 to the given number it should be a perfect square. So K2
  takes the
  
  following values. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ....
  1 to 16 are ruled out as if we subtract 18 from them, the resulting number is a single digit number.
  Now 25 - 18 = 7
  36 - 18 = 18
  49 - 18 = 31
  64 - 18 = 46
  81 - 18 = 63
  100 - 18 = 82
  121 - 18 = 103
  Now 63, 82 satisfies.
  so (2) 2 is answer
• 4 years agoHelpfull: Yes(0) No(0)