Two digit integers can be expressed as $a*10 + b*1$ for some integers $a, b$ satisfying $1 \leq a, b \leq 9$. For example, $95$ is $9*10 + 5*1$.
Now, the "reverse order" integer of $a*10 + b*1$ is $b*10 + a*1$. For example, $59$ is $5*10 + 9*1$, and $95$ is $9*10 + 5*1$.
So, from the question, we want $10a + b + 10b + a = c^{2}$ for some $c$.
So, we are looking for integers $a$ and $b$ each between $1$ and $9$ so that $10(a + b) + a + b = c^{2}$ for some integer $c$, that is, $11(a + b) = c^{2}$, i.e., $a + b = c^{2}/11$. Maybe from here you can plug in all possible combinations? There would only be 81 to check. Like $a = 1, b = 1$, $a = 1, b = 2$, etc.
Answer
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, 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.