Simple Problem

 

Question: Find all two-digit positive integers such that the sum of the number and the number obtained by reversing the order of digits is a perfect square.

Detailed solution: Let $xy$ be any such two-digit number (Note: Here $xy$ is just the form of a two-digit number like 25,43,etc., where $x$ and $y$ are digits).

Therefore, $xy$ can be written as $10x+y$ (in expanded form). Number obtained by reversing the order of digits of $xy$ is $yx=10y+x$.

According to question, The sum of $xy$ and the number obtained by reversing the order of digits of $xy$ i.e., $yx$ is a perfect square, which means

$(10x+y)+(10y+x)=$Perfect square

or, $11x+11y=$Perfect square

or, $11(x+y)=$Perfect square .....$(1)$

But, maximum value of $x$ and $y$ are 9 each, so maximum value of $x+y$ is $9+9=18$. Observe that, to satisfy equation $(1), x+y=11k²$, for some positive integer $k$. For $k=1,x+y=11$ and for $k>1$ i.e., for $k≥2, x+y≥44$, which is not possible. Therefore, we get that $x+y=11$ is the only choice to satisfy equation $(1)$.

Hence, $(x,y)=(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2)$ satisfies. [Note: $(0,11),(1,10),(10,1)$ and $(11,0)$ are not possible because 10 and 11 are not digits]

So, all such two-digit positive integers are $29,38,47,56,65,74,83$ and $92$

Hope you all liked it!