Theorem

A number is congruent to its digit sum modulo $9$

Arithmetic • Number System • Divisibility

Formula

N \equiv (sum of the digits of N) (mod 9)

For any positive integer

N

with digits

d_{k} \dots d_{1} d_{0}

N \equiv d_{k} + \dots + d_{1} + d_{0} (mod 9)

, because

10 \equiv 1 (mod 9)

and so every power of

10

\equiv 1

. The same congruence holds modulo

3

How to use it

1Repeat the digit sum until one digit remains; that digital root is the remainder modulo $9$ , except that a root of $9$ means remainder $0$ .
2Use it to reduce a huge base before raising it to a power modulo $9$ .
3It is the source of the familiar tests: $9 ∣ N$ exactly when $9$ divides the digit sum, and $3 ∣ N$ exactly when $3$ does.

What is the remainder when

86, 419

is divided by

9

Step-by-step

Answer

1

Watch out

Only moduli

9

and

3

work this way, because only for those is

10 \equiv 1

. Modulo

11

the digits must be taken with alternating signs, and for any other modulus the digit sum says nothing at all.

Builds on

Also known as

digit sum, casting out nines, divisibility by 9, digital root, digit sum modulo 9