Digital Root of 3 Consecutive Numbers ending in Multiple of 3

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Theorem

Let $n$, $n + 1$ and $n + 2$ be positive integers such that $n + 2$ is a multiple of $3$.

Let $m = n + \paren {n + 1} + \paren {n + 2}$.


Then the digital root of $m$ is $6$.


Proof

Let $n + 2$ be expressed as $3 r$ for some positive integer $r$.

Then:

\(\displaystyle m\) \(=\) \(\displaystyle \paren {3 r - 2} + \paren {3 r - 1} + 3 r\)
\(\displaystyle \) \(=\) \(\displaystyle 9 r - 3\)
\(\displaystyle \) \(=\) \(\displaystyle 9 \paren {r - 1} + 6\)


The result follows from Digital Root of Number equals its Excess over Multiple of 9.

$\blacksquare$


Historical Note

According to David Wells in his Curious and Interesting Numbers, 2nd ed. of $1997$, this result is due to Iamblichus Chalcidensis.


Sources