View source for Divisibility rules/Rule for 3 and 9 proof

A [[number]] is [[divisible]] by 3 or 9 if the sum of its [[digit]]s 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 ==

''An understanding of [[Introduction to modular arithmetic | basic modular arithmetic]] is necessary for this proof.''

Arithmetic [[modular arithmetic | mod]] <math>9</math> can be used to give an easy [[proof]] of this criterion:

Suppose that the [[base numbers | base-ten]] representation of <math>N</math> is

<math>N = a_k a_{k-1} \cdots a_2 a_1 a_0</math>,

where <math>a_i</math> is a digit for each <math>i</math>.  Then the numerical value of <math>N</math> is given by

<math>N = a_k \cdot 10^k + a_{k-1} \cdot 10^{k-1} + \cdots + a_1 \cdot 10^1 + a_0 \cdot 10^0</math>.

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

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

<math>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</math> (mod <math>9</math>).

Therefore, we have

<math>N \equiv a_k + a_{k-1} + \cdots + a_1 + a_0</math> (mod <math>9</math>).

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>.

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

== See also ==
* [[Divisibility rules | Back to divisibility rules]]