Divisibility by 9 via Digit Sum

The classic divisibility test for $9$—adding the digits and checking if the result is a multiple of $9$—is more than a quick mental trick; it’s a beautiful application of algebra and modular arithmetic that directly connects to the HSC Extension 1 proofs syllabus. By unpacking a three-digit number into its place-value components, we can reveal exactly why the digit sum works and then prove the equivalence in both directions. Mastering this proof will strengthen your ability to construct biconditional arguments and recognise where modular relationships simplify the reasoning.

Problem Statement

Prove that a three-digit number is divisible by $9$ if and only if the sum of its digits is divisible by $9$.

---

Hints

Let $N = 100a + 10b + c$ where $a, b, c$ are the digits.

Rewrite this as $N = 9(11a + b) + (a + b + c)$ to establish the relationship between $N$ and the digit sum.

Then prove both directions of the biconditional using this relationship.

---

Solutions

Let $N = 100a + 10b + c$ be a three-digit number with digits $a, b, c$.

Let $S = a + b + c$ be the sum of digits.

To see the link between $N$ and its digit sum, we’ll rewrite $N$ in a form that isolates multiples of $9$, leaving the digit sum as a remainder-like term.

Key Relationship:

Rewrite $N$:

$$\begin{aligned} N &= 100a + 10b + c \\ &= (99a + a) + (9b + b) + c \\ &= 9(11a + b) + (a + b + c) \\ &= 9(11a + b) + S \end{aligned}$$

From this identity we see that the difference between $N$ and $S$ is a multiple of $9$, which immediately gives us a modular equivalence. Therefore: $N - S = 9(11a + b)$, which means $N \equiv S \pmod{9}$.

With this core relationship in hand, proving the biconditional comes down to simple modular arithmetic.

Direction 1 ($\Rightarrow$): If $9|N$, then $9|S$.

If $N \equiv 0 \pmod{9}$, then from $N \equiv S \pmod{9}$, we get $S \equiv 0 \pmod{9}$.

Thus $9|S$.

Direction 2 ($\Leftarrow$): If $9|S$, then $9|N$.

If $S \equiv 0 \pmod{9}$, then from $N \equiv S \pmod{9}$, we get $N \equiv 0 \pmod{9}$.

Thus $9|N$.

Since we have shown both implications, the proof of the if-and-only-if statement is complete. $\blacksquare$

Note: This proof generalizes to all positive integers and divisibility by 9. The key is the modular relationship $N \equiv S \pmod{9}$.

---

Takeaways

• Reconstruct the full proof from the hint and the solution outline, and justify every transformation explicitly.
• Check edge cases and verify where each assumption is used in the argument.

---

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Last Resorts, HSC Trigonometry