Divisibility by 9: A Digit-Sum Shortcut

Here’s a classic puzzle with a beautiful answer rooted in modular arithmetic:

Problem: Devise a simple rule to test if any number is divisible by 9, and prove why it works.

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The Rule

A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

For example:

• 729 → $7 + 2 + 9 = 18$, and $18$ is divisible by 9 → ✅
• 1234 → $1 + 2 + 3 + 4 = 10$, and $10$ is not divisible by 9 → ❌

But why does this work?

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Step-by-Step Proof (Base 10)

Let a number $N$ be written in decimal as:

$N = a_k \cdot 10^k + a_{k-1} \cdot 10^{k-1} + \cdots + a_1 \cdot 10 + a_0$

We want to analyze $N \mod 9$. Notice:

$10 \equiv 1 \pmod{9},\quad 10^n \equiv 1^n \equiv 1 \pmod{9}$

So:

$N \equiv a_k + a_{k-1} + \cdots + a_1 + a_0 \pmod{9}$

That is, $N$ and the sum of its digits are congruent modulo 9.

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Therefore:

• $N \equiv \text{digit sum} \mod 9$
• So, $N \equiv 0 \mod 9 \iff \text{digit sum} \equiv 0 \mod 9$

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Final Answer

A number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Reference

• [1] Divisibility Rules and Modular Arithmetic