How many two digit numbers are there such that the sum of the number and the number formed by reversing the digits is a perfect square?

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: