Different Square Difference

Let the digits of the number be $a, b$ and $c$ . The value of the number becomes $100a + 10b + c$ .

So the number minus the sum of its digits is

$100a + 10b + c - ( a + b + c)$

$= 99a + 9b$

$= 9 ( 11a + b )$

Now, as 9 is already a perfect square, we need $11a + b$ to be a perfect square. We also need its maximum value.

Remember that a and b are digits. Their maximum value will be 9.

Since $a$ is the hundreds digit, taking $a=9$ will have a greater value than taking $b=9$ .

So $11a+b$ becomes $99 + b$ .

As b is also a digit, the maximum value of b is 9. But the only value of b for which $99+b$ is a perfect square is 1. So $b=1$ .

The maximum value of c we can take is 9, again as c is a digit.

So the number is $\boxed{919}$

Let h = the hundreds digit Let t = the tens digit Let u = the units (or ones) digit

Then the number = 100h+10t+u Sum of the digits = h+t+u

The number - sum of digits) = perfect square.

100h+10t+u - (h+t+u) = perfect square

100h+10t+u - h-t-u = perfect square

99h+9t = perfect square

This shows that the solution is independent of u. Since we are looking for the largest three digit number, we will therefore take u=9

9(11h+t) = perfect square

So the perfect square must be a square root divisible by 9 So its square root must be divisible by 3.

Since V999 = 31.6... the largest 3-digit perfect square is 31² or 961. The largest number that does not exceed 31 that is divisible by 3 is 30. so 30 is a candidate for the square root, and 30² or 900 is a candidate for the perfect square we are looking for:

9(11h+t) = 900

llh+t = 100

   t = 100-11h

All digits are less than 10, so

      t < 10
100-11h < 10
   -11h < -90
      h > 8.18...

The only digit larger than 8.18... is 9, so h = 9

   t = 100-11(9) = 100-99 = 1

Checking: 919 has sum of digits 9+1+9 = 19

919-19 = 900 = 30²

Answer: 919