In general, an integer 
is divisible by 
iff the digit
sum 
is divisible by 
.
Write a positive decimal integer 
out digit by digit in the form 
. The following rules then determine if 
is divisible
by another number by examining the congruence properties
of its digits. In congruence notation, 
means that the remainder when 
is divided by a modulus 
is 
. (Note that it is always true that 
for any base.)
1. All integers are divisible by 1.
2. 
,
so 
for 
.
Therefore, if the last digit 
is divisible by 2 (i.e., is
even), then so is 
.
3. 
,
,
,
..., 
(mod 3). Therefore, if the digit sum 
is divisible
by 3, so is 
(Wells 1986, p. 48). In general, if the sum of any permutation of the digits
of 
in any order is divisible by 3, then so is 
.
4a. 
,
,
..., 
(mod 4). So if the last two digits are divisible by
4, then so is 
.
4b. If 
is, then so is 
.
5. 
,
so 
for 
.
Therefore, if the last digit 
is divisible by 5 (i.e., is
5 or 0), then so is 
.
6a. If 
is divisible by 3 and is even,
then 
is also divisible by 6.
6b. 
,
,
..., 
(mod 6). Therefore, if 
is divisible
by 6, so is 
.
The final number can then, of course, be further reduced using the same procedure.
7a. 
,
,
,
,
,
(mod 7), and the sequence then repeats. Therefore, if 
is divisible by 7, so is 
. This method was found by Pascal.
7b. An alternate test proceeds by multiplying 
by 3 and adding to 
, then repeating the procedure up through 
. The final number can then, of course, be further reduced
using the same procedure. If the result is divisible by 7, then so is the original
number (Wells 1986, p. 70).
7c. A third test multiplies 
by 5 and adds it to 
, proceeding up through 
. The final number can then, of course, be further reduced
using the same procedure. If the result is divisible by 7, then so is the original
number (Wells 1986, p. 70).
7d. Given a number, form two numbers 
and 
such that 
consists of all digits of the number except the last (units)
digit and 
is the last digit. Compute 
and repeat the procedure. Then the original number is divisible
by 7 iff the number in the last step is divisible by 7.
8. 
,
,
,
..., 
(mod 8). Therefore, if the last three digits are divisible
by 8, more specifically if 
is, then so is 
(Wells 1986, p. 72).
9. (Rule of nines). 
, 
, 
, ..., 
(mod 9). Therefore, if the digit
sum 
is divisible by 9, so is 
(Wells 1986, p. 74).
10. 
(mod 10), so if the last digit is 0, then 
is divisible by 10.
11. 
,
,
,
,
... (mod 11). Therefore, if 
is divisible
by 11, then so is 
.
12. 
,
,
,
... (mod 12). Therefore, if 
is divisible
by 12, then so is 
.
Divisibility by 12 can also be checked by seeing if 
is divisible by 3 and 4.
13. 
,
,
,
,
,
(mod 13), and the pattern repeats. Therefore, if 
is divisible by 13, so is 
.
For additional tests for 13, see Gardner (1991).
An interesting piece of English language trivia is that the word "indivisibilities" has more "i"s (in fact, seven of them) than any other common word. (Other words with seven i's include, honorificabilitudinitatibus, indistinguishabilities, indivisibilities, and supercalifragilisticexpialidocious. Phrases with eight i's include "Illinois fighting Illini" and "infinite divisibility." The English word with the most possible i's is floccinaucinihilipilification (nine i's), where "floccinaucinihilipilification" means "the action or habit of estimating as worthless.")
