Digital Root of Square
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Theorem
Let $n^2$ be a square number.
Then the digital root of $n^2$ is $1$, $4$, $7$ or $9$.
Proof
Let $\map d n$ denote the digital root base $10$ of $n$.
From Digital Root is Congruent to Number Modulo Base minus 1, $\map d n \equiv n \pmod 9$.
So, let $n = 9 k + m$ where $1 \le m \le 9$.
Thus:
\(\ds n^2\) | \(=\) | \(\ds \paren {9 k + m}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 81 k^2 + 18 k m + m^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 9 \paren {9 k^2 + 2 k m} + m^2\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds m^2\) | \(\ds \pmod 9\) |
We enumerate the squares of the digits:
\(\ds 1^2\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 2^2\) | \(=\) | \(\ds 4\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 4\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 3^2\) | \(=\) | \(\ds 9\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 9\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 4^2\) | \(=\) | \(\ds 16\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 7\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 5^2\) | \(=\) | \(\ds 25\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 7\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 6^2\) | \(=\) | \(\ds 36\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 9\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 7^2\) | \(=\) | \(\ds 49\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 4\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 8^2\) | \(=\) | \(\ds 64\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod 9\) | |||||||||||
\(\ds 9^2\) | \(=\) | \(\ds 81\) | ||||||||||||
\(\ds \) | \(\equiv\) | \(\ds 9\) | \(\ds \pmod 9\) |
Hence the result.
$\blacksquare$