In mathematics an automorphic number (sometimes referred to as a circular number) is a number whose square "ends" in the same digits as number itself. Thus $5$ is automorphic since $25$ ends in $5$. Similarly, $625^2 = 390625$, so $625$ is also automorphic.
Given an integer $N$. How could we find the sum of all $N$ digit numbers which are automorphic? Obviously I am not looking for brute-forcing.
You may also like to take a look at what I call the inspiration problem. My solution for that is not much of mathematically rigorous, as the option were given and I knew that $25$ is automorphic. I simply subtracted from the options, I got $74,75,76$ and $77$. Again as only $1,5$ and $6$ are one digits automorphic numbers, so $74$ and $77$ could be discarded. Also any number ending with $5$ when squared and we take a modulo by $100$ gives $25$ so $75$ rejected. Hence $76$ is the other number.
However, I guess, this is not exactly a mathematical approach, since if the options are not there I don't think I could had solved it, also having more digits makes things more complicated. Hence,I was thinking if there is any other techniques for this?