Question

A two digit number is 18 less than the sum of the squares of its digits. how many such numbers are there?

Answer 1

Let the number be  a b.    Here  a is between 1 & 9.   b is between 0 & 9.
   Its value is 10 a + b.  Value is between 10 & 99.

Given   10 a + b = (a² + b²) - 18    -- (1)
  So      a² + b² - 18 ≥ 10         ie.,  a² + b² ≥ 28
             a² + b² - 18 ≤ 99        ie.,  a² + b² ≤ 117

Now from (1) we get:
       10 a - a² = b² - b - 18
      a (10 - a) = b² - b - 18
LHS is always positive.  So RHS = b² - b -18 > 0.

Let's find the roots of above quadratic:   = [1 + √(1+72) ]/2
                  or approximately :   4.75 &  - 3.75

For the quadratic b² - b - 18 to be positive,  b > 4.75.

Hence, b = 5, 6, 7 , 8 , or 9.
Let us check it now from eq (1).

              a² - 10 a + (b² - b - 18) = 0
          Roots:  5 + √[25 - (b² - b - 18) ]

As a is real and non-negative integer, 
            b² - b - 18 ≤ 25

Checking with b = 5, 6, 7, 8, 9, we get that
                 b = 7 is the only value.
Then   a = 5 + 1 = 4 or 6. 

There are two such numbers.    47 and 67.

Answer 2

Answer: 2

Step-by-step explanation:

Let the number be 10a+b

$a^{2} + b^{2}-18 = 10a+b\\ 10a-a^{2} = b^{2}-b-18\\ a(10-a) = b^{2}-b-18$

$b^{2}-b-18$ always positive as a(10-a) is positive.

So, b = 5,6,7,8 or 9

Now, from the equation $a^{2}+b^{2} -18 = 10a+b$, $b^{2}-b-18 \leq 25$

b=7 can only satisfy this.

Then, a = 4 or 6

So, 47 and 67 are the possibilities.