Divisibility rules

These divisibility rules help determine when integers are divisible by particular other integers.

Divisibility Rule for 2 and Powers of 2

A number is divisible by $2^n$ if the last ${n}$ digits of the number are divisible by $2^n$ .

Divisibility Rule for 3 and 9

A number is divisible by 3 or 9 if the sum of its digits is divisible by 3 or 9, respectively. Note that this does not work for higher powers of 3. For instance, the sum of the digits of 1899 is divisible by 27, but 1899 is not itself divisible by 27.

Divisibility Rule for 5 and Powers of 5

A number is divisible by $5^n$ if the last n digits are divisible by that power of 5.

Proof

An understanding of basic modular arithmetic is necessary for this proof.

Consider, for example, the test for divisibility by $9$ :

Let $N$ be a positive integer. Then $N$ is divisible by $9$ if and only if the sum of the base-ten digits of $N$ is divisible by $9$ .

Arithmetic mod $9$ can be used to give an easy proof of this criterion:

Suppose that the base-ten representation of $N$ is

$N = a_k a_{k-1} \cdots a_2 a_1 a_0$ ,

where $a_i$ is a digit for each $i$ . Then the numerical value of $N$ is given by

$N = a_k \cdot 10^k + a_{k-1} \cdot 10^{k-1} + \cdots + a_1 \cdot 10^1 + a_0 \cdot 10^0$ .

Now we know that, since $10 - 1 = 9$ , we have $10 \equiv 1$ (mod $9$ ). So by the "exponentiation" property above, we have $10^j \equiv 1^j \equiv 1$ (mod $9$ ) for every $j$ .

Therefore, by repeated uses of the "addition" and "multiplication" properties, we can write

$a_k \cdot 10^k + a_{k-1} \cdot 10^{k-1} + \cdots + a_1 \cdot 10^1 + a_0 \cdot 10^0 \equiv a_k \cdot 1 + a_{k-1} \cdot 1 + \cdots + a_1 \cdot 1 + a_0 \cdot 1$ (mod $9$ ).

Therefore, we have

$N \equiv a_k + a_{k-1} + \cdots + a_1 + a_0$ (mod $9$ ).

That is, $N$ differs from the sum of its digits by a multiple of $9$ . It follows, then, that $N$ is a multiple of $9$ if and only if the sum of its digits is a multiple of $9$ .

Since we also have $10 \equiv 1 \pmod 3$ , the same proof also gives the divisibility rule for 3. The proof fails for 27 because $10 \not\equiv 1 \pmod{27}$ .

Divisibility Rule for 11

A number is divisible by 11 if the alternating sum of the digits is divisible by 11.

Proof

$10\equiv -1\pmod{11}$ . Since the number is in base 10, as we assume, the digits would be of powers of 10, or in mod 11, powers of -1. The number 3806 thus becomes:

$3\cdot (-1)^3+8\cdot (-1)^2+0\cdot (-1)^1+6\cdot (-1)^0=-3+8-0+6\equiv 0\pmod{11}$ .

Divisibility Rule for 7

Rule 1: Partition $n$ into 3 digit numbers from the right ( $d_3d_2d_1,d_6d_5d_4,\dots$ ). If the alternating sum ( $d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots$ ) is divisible by 7, then the number is divisible by 7.

Rule 2: Truncate the last digit of ${n}$ , and subtract twice that digit from the remaining number. If the result is divisible by 7, then the number is divisible by 7. This process can be repeated for large numbers.

Divisibility Rule for 13

See rule 1 for divisibility by 7. A number is divisible by 13 if the same specified sum is divisible by 13.

More general note

For every prime number other than 2 and 5, there exists a rule similar to rule 2 for divisibility by 7. For a general prime $p$ , there exists some number $q$ such that an integer is divisible by $p$ if and only if truncating the last digit, multiplying it by $q$ and subtracting it from the remaining number gives us a result divisible by $p$ . Divisibility rule 2 for 7 says that for $p = 7$ , $q = 2$ . The divisibility rule for 11 is equivalent to choosing $q = 1$ . The divisibility rule for 3 is equivalent to choosing $q = -1$ . These rules can also be found under the appropriate conditions in number bases other than 10.