Divisibility by 9

Rule for Divisibility by 9

A number is divisible by 9 if the sum of its digits is divisible by 9. For large numbers this rule can be applied again to the result. In addition, the final iteration will result in a 9.

Examples

A.) 2,880: 2 + 8 + 8 + 0 = 18, 1 + 8 = 9, so 9| 2,880.

B.) 3,564,213: 3+5+6+4+2+1+3=24, 2+4=6, so 9 does NOT divide 3,564,213.

Proof

The proof for the divisibility rule for 9 is essentially the same as the proof for the divisibility rule for 3.

For any integer x written as a_n· · · a₃a₂ a₁a₀we will prove that if 9|(a₀ + a₁+ a₂+ a₃ ... + a_n), then 9|x and vice versa.

First, we can state that

x = a₀ + a₁×10 + a₂×10² + a₃×10³... + a_n×10ⁿ

Next if we let s be the sum of its digits then

s = a₀ + a₁ + a₂ + a₃ + ... + a_{n .}

x - s = (a₀ - a₀) + (a₁×10 - a₁) + (a₂×10² - a₂) + ... + (a_n×10ⁿ - a_n)

= a₁(10 - 1) + a₂(10² - 1) + ... + a_n(10ⁿ - 1).

If we let b_k = 10^k - 1, then b_k = 9...9 (9 occurs k times) and b_k=9(1…1) and we can rewrite the previous equation as

x - s = a₁(b₁)+ a₂(b₂)+ ... + a_n(b_n)

It follows that all numbers b_k are divisible by 9, so the numbers a_k×b_k are also divisible by 9. Therefore, the sum of all the numbers a_k×b_k (which is x-s) is also divisible by 9.

Since x-s is divisible by 9, if x is divisible by 9, then so is s and vice versa.