Divisibility rules

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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 $2^n$.

Proof

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.

Proof

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

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.

Proof

Rule 2: Truncate the last digit of ${n}$, and double that digit, subtracting the rest of the number from the doubled last digit. If the absolute value of the result is a multiple of 7, then the number itself is.

Proof

Divisibility Rule for 11

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

Proof


Divisibility Rule for 13

Multiply the last digit by 4 and add it to the rest of the number. This process can be repeated for large numbers, as with the second divisibility rule for 7.

Proof


More general note for primes

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.

More general note for composites

A number is divisible by $N$, where the prime factorization of $N$ is $p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}$, if the number is divisible by each of $p_1^{e_1}, p_2^{e_2},\ldots, p_n^{e_n}$.

Example

Is 55682168544 divisible by 36?

Solution

First, we find the prime factorization of 36 to be $2^2\cdot 3^2$. Thus we must check for divisibility by 4 and 9 to see if it's divisible by 36.

Since the last two digits, 44, of the number is divisible by 4, so is the entire number.

To check for divisibility by 9, we look to see if the sum of the digits is divisible by 9. The sum of the digits is 54 which is divisible by 9.

Thus, the number is divisible by both 4 and 9 and must be divisible by 36.

Example Problems


Resources

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See also