Question

When a two digit
number is added to
its reverse, the
result is 143. The
number is 3 less
than the sum of the
squares of its digits.
Calculate this
number.

Answer 1

take number as x and y
so number is 10x + y [x as tens digit and y as unit digit]
the sum of this digit with it's reverse is 143 so,
new reverse no is 10y + x
10x + y + 10y + x = 143
11x + 11y =143
x + y = 13............[1]

now sum of  squares of digits so x^2 + y^2
 number is 3 less than the sum of the squares of its digits.
 so , [x^2 + y^2 ] - 3..............[2]
   
the number and euation 2 is same so
10x + y =  [x^2 + y^2 ] - 3    
10x + y = [169 - 2xy] - 3
10x + y + 2xy = 166 ...... [3]

now [3]- 10*[1]

- 9y +2xy = 36
 y[ 2x-9] =36 put value of y from 1
13-x [2x-9] =36
2x^2 - 35x + 153 = 0
 solving this by √D = b²-4ac [a = 2, b = -35, c = 153]
                            =1225-1224
                            =1
                       √D =1
now   x= -b+√D÷2a    or        x=-b-√D÷2a
        x=9                  and      x=34/4
so we take x = 9
so y = 13 - 9 = 4
number is 10x + y = 10*9 + 4 = 94