The square of 13 is 169, which has the digit 6 in the tens place. The square of another number has the digit 7 in the tens place. What are the possible values for the digit in the units place of this square?
Suppose the number is $10a + b$, with $b$ being a digit. When we square it, we get

$$
(10a + b)^2 = 100a^2 + 20ab + b^2
$$

which has three terms: $100a^2$, $20ab$, and $b^2$.

The first term ends in 00, while the second term ends in an even number followed by zero. For the tens digit to be 7, which is odd, it is necessary that the tens digit of $b^2$ be odd, which only happens when $b = 4$ or $b = 6$. In each of these cases, $4^2 = 16$ and $6^2 = 36$, the units digit of the square is 6.

## Related Problem

Is there a perfect square number consisting only of the digits 0 and 6?