Difference between revisions of "Divisibility rules"

Revision as of 09:25, 19 July 2006

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 $9$ .

Divisibility Rule for 11

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

Example Problems

2006 AMC 10B Problem 25

Resources

Books

The AoPS Introduction to Number Theory by Mathew Crawford.

Classes

AoPS Introduction to Number Theory Course

@@ Line 13: / Line 13: @@
-== Divisibility Rule for 9 ==
-A number is divisible by 9 if the sum of its digits is divisible by 9.
 === Proof ===
 An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.
@@ Line 43: / Line 41: @@
 That is, <math>N</math> differs from the sum of its digits by a multiple of <math>9</math>.  It follows, then, that <math>N</math> is a multiple of <math>9</math> if and only if the sum of its digits is a multiple of <math>9</math>.
 == Divisibility Rule for 11 ==

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