Question

How many two-digit numbers have digits whose sum is a perfect square?

Answer 1

Hint: Here we will find the number of two-digit numbers whose sum is a perfect square. Firstly we will write the highest two-digit number and check the sum of it. Then by using the sum we will see what all perfect squares are less than or equal to that number. Then we will see that for each perfect square how many digits can be formed. Finally we will calculate it and get our desired answer.

Complete step by step solution:
So, starting with highest two-digit number we have is $99$ so sum of its digit is,
$ 9 + 9 = 18$
So, that means we will only consider the perfect square below 18 which are:
$\begin{gathered}
  {1^2} = 1 \\
  {2^2} = 4 \\
  {3^2} = 9 \\
  {4^2} = 16 \\
\end{gathered} $
Now, we will write all the possible ways to find the sum of the above perfect square using two digits.
Firstly, 1 can be written as,
$ 1 = 1 + 0$
Next, 4 can be written as,
$ 4 = 2 + 2 = 3 + 1 = 1 + 3$
Next, 9 can be written as,
$ 9 = 9 + 0 = 8 + 1 = 1 + 8 = 7 + 2 = 2 + 7 = 6 + 3 = 3 + 6 = 5 + 4 = 4 + 5$
Next, 16 can be written as,
$ 16 = 8 + 8 = 9 + 7 = 7 + 9$
So, we can write the numbers for each perfect square as,
$1:10 \\
  4:22,31,13 \\
  9:90,81,18,72,27,63,36,54,45 \\
  16:88,97,79 \\ $
Counting the above digit we get a total of 17.

So, there are 17 two-digit numbers whose sum of digits is a perfect square.

Note:
Perfect squares are those numbers which are formed when any number is multiplied by itself. In other words, we can say that if we find a square root of a perfect number we get an integer as the answer and not a decimal or fractional number. A number is a square only if it can be arranged that number point in a square. Sum of two consecutive triangular numbers forms a square. We should not get confused between the square and cube of a number. The number obtained is obtained by multiplying the number twice to itself.