A two digit number is 18 less than the sum of the squares of its digits How many such numbers

suppose the digit 10x+y.
so, 10x+y=x^2+y^2-18
=>(x-6)(x-4)=(y-7)(-y-6)
so conddition satisfied for x=6,4 and y=7 so no of digit 2. 67,47.
8 years agoHelpfull: Yes(18) No(4)

(2) 2
63, 82
8 years agoHelpfull: Yes(0) No(6)
Ans: Option 1
Take N = 10a+b.
Given that, 10a+b+18 = (a+b)2
for a = 1 to 9, the L.H.S. will be, 28+b, 38+b, 48+b,.....,108+b.
As LHS is perfect square for the values of b = 1 to 9, only 28+b, 48+b, 58+b, 78+b can be equal to 36, 49, 64, 81 for b = 8, 1, 6, 3 respectively. But only 78+b = 81 for b = 3 So only one such number is possible. I.e, 63
8 years agoHelpfull: Yes(0) No(10)
(a+b)^2 -ab = a^2 + b^2 -ab
ie; accrding to question , xy = (x^2 -+y^2) - 18 => can be written as
18 = (x^2 +y^2)-xy which is same as the first equation
so, this is only possibe with 52 and 25
hence.. 2 is solution
8 years agoHelpfull: Yes(0) No(5)

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