Divisibility by 9
A simple rule, an interactive test, and a proof showing exactly why adding the digits preserves divisibility by 9.
Rule: A number is divisible by 9 exactly when the sum of its digits is divisible by 9.
Try it
The key observation
Key fact
10 ≡ 1 (mod 9)
Because 10 = 9 + 1.
Therefore
10² ≡ 1, 10³ ≡ 1, … (mod 9)
Every decimal place value behaves like 1 when looking only at remainders mod 9.
Why the rule works
1 · Expand by place value
N = a₀ + 10a₁ + 10²a₂ + ··· + 10ᵏaₖ
2 · Work mod 9
Since 10ᶦ ≡ 1,
N ≡ a₀ + a₁ + ··· + aₖ (mod 9)
N ≡ a₀ + a₁ + ··· + aₖ (mod 9)
3 · Conclude
The number and its digit sum have the same remainder when divided by 9.
↓
So the original number has remainder 0 mod 9 exactly when its digit sum has remainder 0 mod 9. Therefore, the test is correct.
Concrete example: 5,724
5724 = 5×1000 + 7×100 + 2×10 + 4
mod 9
5724 ≡ 5×1 + 7×1 + 2×1 + 4 = 18 ≡ 0
Therefore 5,724 is divisible by 9.
Mental shortcut
You can keep summing digits until one digit remains.
7,236 → 7+2+3+6 = 18 → 1+8 = 9
Since the final digit is 9, the original number is divisible by 9.