Divisibility by 9 via Digit Sum

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.
Key Relationship: Rewrite $N$ : $\begin{aligned} N &= 100a + 10b + c \\ &= (99a + a) + (9b + b) + c \\ &= 9(11a + b) + (a + b + c) \\ &= 9(11a + b) + S \end{aligned}$
Therefore: $N - S = 9(11a + b)$ , which means $N \equiv S \pmod{9}$ .
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 $9|N$ .
Both directions proven, so the biconditional holds.
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:

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