## Odd product of tens digits

The product of the tens digits of a few given perfect squares is odd. What is the units digit of the product of these perfect squares?

[SOLVED]

#### Hagen von Eitzen solved this puzzle:

Since $$(10k + n)^2 = n^2 + 20k(n + 5)$$, both the units digit and the parity of the tens digit depend only on $$n \bmod{10}$$.

Checking the numbers $$0^2, 1^2, \dots, 9^2$$ we find that only $$4^2 = 16$$ and $$6^2 = 36$$ have odd tens digits.

Since a product of integers is odd if and only if each integer is odd, every given perfect squares must have odd tens digit. So, each has its units digit as $$6$$. Since $$6 \cdot 6 \equiv 6 \pmod{10}$$, $$6^m \equiv 6 \pmod{10}$$ for all integers $$m \gt 0$$. Therefore, the units digit of the product of the squares is $$6$$.

#### Joe Wells solved this puzzle:

If we examine the squares of the digits $$0, 1, \cdots, 9$$, we see that $$4$$ and $$6$$ are the only two whose squares have odd tens digits. Further, both end with $$6$$.

Consider squaring some multi-digit number with $$4$$ as the units digit. Since $$4$$ is even, $$4$$ times the tens digit of the number yields an even number as the tens digit. But when we add $$4^2 = 16$$ to it, the units digit is $$6$$, and $$1$$ from $$16$$ is added to the even tens digit, thereby making it odd.

A similar argument holds for any multi-digit number with $$6$$ as the units digit, except that we add $$3$$ from $$6^2 = 36$$ to the tens digit. So each given number has $$6$$ as its units digit, and their product has $$6$$ as its unit digit.

#### Ankit Mandal solved this puzzle:

All natural numbers are of the form $$(10x + k)$$ where x and k are integers such that $$x \geqslant 0$$ and $$0 \leqslant k \leqslant 9$$.

\begin{align*} (10x + 0)^2 & = 100x^2 \\ (10x + 1)^2 & = 100x^2 + 20x + 1 \\ (10x + 2)^2 & = 100x^2 + 40x + 4 \\ (10x + 3)^2 & = 100x^2 + 60x + 9 \\ (10x + 4)^2 & = 100x^2 + 80x + 16 \\ (10x + 5)^2 & = 100x^2 + 100x + 25 \\ (10x + 6)^2 & = 100x^2 + 120x + 36 \\ (10x + 7)^2 & = 100x^2 + 140x + 49 \\ (10x + 8)^2 & = 100x^2 + 160x + 64 \\ (10x + 9)^2 & = 100x^2 + 180x + 81 \\ \end{align*}

We can conclude that the tens digit of the square of any natural number ending in $$4$$ or $$6$$ is always odd. In any other case, it is even.

The problem states that the product of the tens digits of a few given perfect squares is odd. Therefore, the tens digit of each square must be odd, and thus the units digit of each square must be $$6$$. The answer is $$6$$.

### Credit

This is an original puzzle from cotpi.