Question

A two ligit number is less than 18 than sum of the squars of its digits

Answer 1

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Answer:

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.