Base $10$ Divisibility Rules

A divisibility rule is a short rule or algorithm that can be used to determine (generally integral) divisibility. Here is a list of many divisibility rules based on the digits of integers in their base $10$ representation and corresponding proofs.

Throughout these proofs, let $k = a_{n} 10^{n} + a_{n - 1} 10^{n - 1} + \dots + a_{1} 10 + a_{0}$ be the base $10$ representation of a non-negative integer $k$ such that $a_{i} \in {1, \dots, 9}$ are the digits.

Each of these rules is only provided for non-negative integers, however most of them generalise to any integer. Rules that don't include that of $7$ , however in cases like this, divisibility for a negative number should be checked by verifying divisibility for its absolute value, since $m ∣ k ⟺ m ∣ - k$ .

Also note we only prove rules for divisibility by prime powers since for example checking divisibility by $12$ reduces to checking divisibility by $3$ and $4$ . The same is true for any integer which is not the power of a prime.

2

Theorem

A natural number is divisible by $2$ if and only if its last digit is divisible by $2$ .

Proof

Since $2 ∣ 10$ , we have that

k = 2 (a_{n} 5^{n} + a_{n - 1} 5^{n - 1} + \dots + a_{1} 5) + a_{0}

and therefore

k \equiv a_{0} (\mod 2)

which implies that

2 ∣ k ⟺ k \equiv 0 (\mod 2) ⟺ a_{0} \equiv 0 (\mod 2) ⟺ 2 ∣ a_{0} .

3

Theorem

A natural number is divisible by $3$ if and only if the sum of its digits is divisible by $3$ .

Proof

Consider that

\begin{aligned} k & = a_{n} (10^{n} - 1) + a_{n} + a_{n - 1} (10^{n - 1} - 1) + a_{n - 1} + \dots + a_{1} (10 - 1) + a_{1} + a_{0} \\ = (\sum_{i = 0}^{n} a_{i}) + a_{n} (10^{n} - 1) + a_{n - 1} (10^{n - 1} - 1) + a_{n - 1} + \dots + a_{1} (10 - 1) + a_{1} . \end{aligned}

From here it is clear that $10^{r} - 1 \equiv 1^{r} - 1 \equiv 0 (\mod 3)$ for all $r \geq 0$ , and therefore

k \equiv \sum_{i = 0}^{n} a_{i} (\mod 3)

which implies that

3 ∣ k ⟺ k \equiv 0 (\mod 3) ⟺ \sum_{i = 0}^{n} a_{i} \equiv 0 (\mod 3) ⟺ 3 ∣ \sum_{i = 0}^{n} a_{i} .

4

Theorem

A natural number is divisible by $4$ if and only if the two digit number formed from its last two digits is divisible by $4$ .

Proof

Since $4 ∣ 100$ given $4 \times 25 = 100$ , we have that

k = 4 \times 25 (a_{n} 10^{n - 2} + a_{n - 1} 10^{n - 3} + \dots + a_{2}) + a_{1} 10 + a_{0}

and therefore

k \equiv a_{1} 10 + a_{0} (\mod 4)

which implies that

4 ∣ k ⟺ k \equiv 0 (\mod 4) ⟺ a_{1} 10 + a_{0} \equiv 0 (\mod 4) ⟺ 4 ∣ a_{1} 10 + a_{0} .

5

Theorem

A natural number is divisible by $5$ if and only if its last digit is $0$ or $5$ .

Proof

As $5$ is a factor of $10$ , this proof follows almost identically to the divisibility rule for $2$ . In particular we have that

k = 5 (a_{n} 2^{n} + a_{n - 1} 2^{n - 1} + \dots + a_{1} 2) + a_{0}

and therefore

k \equiv a_{0} (\mod 5)

which implies that

5 ∣ k ⟺ k \equiv 0 (\mod 5) ⟺ a_{0} \equiv 0 (\mod 5) ⟺ 5 ∣ a_{0} .

7

Theorem

A natural number is divisible by $7$ if and only if the number formed by doubling the last digit and subtracting that from the truncated number is divisible by $7$ .

Since this rule is a little weirder, here is an example.

Example

$3976$ is divisible by $7$ .

Solution

We take the truncated number $397$ and subtract twice the last digit to get $397 - 6 \cdot 2 = 385$ . Repeating this, we have $38 - 5 \cdot 2 = 28$ . Again, we have $2 - 8 \times 2 = - 14$ and finally $1 - 4 \cdot 2 = - 7$ , which is clearly divisible by $7$ .

Proof

\begin{aligned} k \equiv 0 (\mod 7) \\ ⟺ & a_{n} 10^{n} + a_{n - 1} 10^{n - 1} + \dots + a_{1} 10 + a_{0} \equiv 0 (\mod 7) \\ ⟺ & 10 (a_{n} 10^{n - 1} + a_{n - 1} 10^{n - 2} + \dots + a_{1}) \equiv - a_{0} (\mod 7) \\ ⟺ & a_{n} 10^{n - 1} + a_{n - 1} 10^{n - 2} + \dots + a_{1} \equiv - 10^{- 1} a_{0} (\mod 7) \\ ⟺ & (a_{n} 10^{n - 1} + a_{n - 1} 10^{n - 2} + \dots + a_{1}) - 2 a_{0} \equiv - 10^{- 1} a_{0} - 2 a_{0} (\mod 7) \\ ⟺ & (a_{n} 10^{n - 1} + a_{n - 1} 10^{n - 2} + \dots + a_{1}) - 2 a_{0} \equiv - (10^{- 1} + 2) a_{0} (\mod 7) \\ ⟺ & (a_{n} 10^{n - 1} + a_{n - 1} 10^{n - 2} + \dots + a_{1}) - 2 a_{0} \equiv 0 (\mod 7) \end{aligned}

where this last equivalence holds because $10 \times - 2 = - 20 \equiv 1 (\mod 7)$ which implies that $10^{- 1} = - 2$ in $Z_{7}$ and subsequently $10^{- 1} + 2 \equiv 0 (\mod 7)$ .

Base 10 Divisibility Rules

2

Proof

3

Proof

4

Proof

5

Proof

7

Solution

Proof

Base $10$ Divisibility Rules