$x$ is a positive integer with at least 2 digits, a perfect square, and its tens digit is odd. What are the possible units digits of x?

Question

I have no idea where to start; and guess-and- check is all I can get up to now.

Answer 1

We have $$(10a+b)^2=100a^2+20ab+b^2$$ where $b$ is a single digit number. So the tens digit is $2ab$ plus the carry-over from $b^2$. Since $2ab$ is even, the carry-over must be odd in order for the tens digit to be odd. The only possibilities for $b$ are therefore $b=4$ and $b=6$ as these are the only single-digit numbers whose square have an odd tens digit.

Answer 2

All you need to do is check the squares $\bmod 20$. A quick spreadsheet shows that $6$ is the only possibility, coming from squaring $04, 06, 14, 16$ Another way to see it is that if the number squared is $10a+b$ the square is $100a^2+20ab+b^2$ so you need $b^2$ to have an odd tens digit and only $4,6$ do.