The numbers $24^2 = 576$ and $56^2 = 3136$ are examples of perfect squares that have a units digits of $6.$
If the units digit of a perfect square is $5,$ then what are the possible values of the tens digit?
If the number 5 is in the units digit, then the original number that was squared must end with 5.
Let's set a variable to help us solve this question.
Let's let a be a number that is the multiple of 10. We have
\((a + 5) (a + 5) = \\ a^2 + 10a + 25\)
a^2 will always end with a 0 or two 0s, so it has no effect on the final digit
10a will always end with a 0 or two 0s, so it has no effect on the final digit
This means we always have 25.
So 2 is our answer.
Thanks! :)