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.")