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]

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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.

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