Digital Root of 3 Consecutive Numbers ending in Multiple of 3

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:

\(\ds m\)

\(=\)

\(\ds \paren {3 r - 2} + \paren {3 r - 1} + 3 r\)

\(\ds \)

\(=\)

\(\ds 9 r - 3\)

\(\ds \)

\(=\)

\(\ds 9 \paren {r - 1} + 6\)

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

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $6$
• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $6$