Divisibility Tests and Tricks 1: Every number is always divisible by 1.

2: Look at the last digit. If it is even (0, 2, 4, 6, 8) then the number is divisible by two. For example, 24 is divisible by 2 since the last digit is 4, but 25 is not since the last digit is 5. Why does this work? First, a bit of notation (this notation is very commonly used): a|b means “a divides b”, or b is divisible by a. For example, it is true that 3 | 6 but it is not true that 4 | 6. Now to the actual proof: Let our number be x. Then write x = A B. Here, we are not using multiplication; instead this is bad notation for x = 10A + B where B is a one digit number. For example, if x = 31415 then A = 3141 and B = 5. Then the statement “x is divisible by 2” is equivalent to 2 | x, or 2 | 10A + B, which is equivalent to 2 | B since 10A is automatically divisible by 2 (2 | 10) and hence does not affect divisibility of the whole number. Therefore x is divisible by 2 if and only if its last digit, B, is divisible by 2.

3: Sum all of the digits. The number is divisible by 3 if and only if this sum is divisible by 3. For example, the number 271828 has a sum of digits of 2 + 7 + 1 + 8 + 2 + 8 = 28, so it is not divisible by 3. Why does this trick work? First we need the basic ideas of modular arithmetic. Let a mod b denote the remainder when b is divided by a. In particular, b is divisible by a when a mod b = 0. For example, 42 is divisible by 7, so 42 mod 7 = 0. We write b ≡ c mod a when b and c have the same remainder when divided by a. For example, 27 mod 5 = 2, and 27 ≡ 22 ≡ 2 ≡ 3141592 mod 5. The symbol ≡ is called “congruent” and is traditionally used. For more information, go to http://en.wikipedia.org/wiki/Modular_arithmetic. Now to the actual proof. From division we can see that 100 mod 3 = 1 and 10 mod 3 = 1. Then 10n mod 3 = (10 mod 3)n = 1n = 1. (This claim should also be easy to see with division: 999999 = 3 × 333333, so 1000000 mod 3 = 1.) We will only do an example to show this proof. Any other number can be substituted. We will not do this proof in general because the notation becomes rather messy. Our example will be 61803. We can see that 61803 mod 3 = (6 × 104 + 1 × 103 + 8 × 102 + 0 × 101 + 3 × 100 ) mod 3 = (6 mod 3)(104 mod 3) + (1 mod 3)(103 mod 3) + (8 mod 3)(102 mod 3) + (0 mod 3)(101 mod 3) + (3 mod 3)(100 mod 3) = (6 mod 3) + (1 mod 3) + (8 mod 3) + (0 mod 3) + (3 mod 3) = (6 + 1 + 8 + 0 + 3) mod 3 n

since 10 mod 3 = 1 for every natural number n. Therefore we have just shown that 61803 mod 3 = 0 if and only if (6 + 1 + 8 + 0 + 3) mod 3 = 0. This is equivalent to saying that 61803 is divisible by 3 if and only if the sum of its digits is divisible by 3. We can substitute any other number for 61803; the same argument works. Hence, any number is divisible by 3 if and only if the sum of its digits is also divisible by 3. 1

4: Look at the last two digits. If this two digit number is divisible by 4, then the whole number is divisible by 4. For example, 271828 ends in 28, which is divisible by 4, so 271828 is also divisible by 4. Why does this work? This case is very similar to the test for 2. Once again we use the bad notation of x = A B. Here B is two digits, so x = 100A + B. Then 4|x is equivalent to 4|100A + B, which is true if and only if 4|B since 100A is auotmatically divisible by 4 (100 = 4 × 25). Therefore a number is divisible by 4 if and only if the last two digits are also divisible by 4.

5: Look at the last digit. If it is a 5 or 0, then the number is divisible by 5. For example, 789526310 and 45654225 are both divisible by 5, but 78954 is not. Why does this work? This case is really quite similar to the case of divisibility by 2. Once again we represent x = A B (not multiplication) where B is one digit. We now have x = 10A + B. Then 5|x is equivalent to 5|10A + B, which is true if and only if 5|B since 10A is automatically divisible by 5 and is hence irrelevant. Therefore x is divisible by 5 if and only if the last digit of x is divisible by 5; this last digit must then be 0 or 5.

6: For divisibility by 6, we can simply look at divisibility by 2 and 3 separately. If both are satisfied, then the number must also be divisible by 6. For example, 3|24 and 2|24; 24 is divisible by both 2 and 3. Therefore it is divisible by 6. A number such as 15 is divisible by 3 but not 2. Therefore 15 is not divisible by 6. Why does this work? It is easy to show that 6 = 2 × 3. Then if x is divisible by 6, it must have 6 as a factor. Hence it must have both 2 and 3 as factors. The fact that either is a factor does not affect the other’s divisibility, so we can test for 2 and 3 separately. (For a case where we cannot test separately, see the entry for either 8 or 12).

7: For divisibility by 7, we have a number x. Take the last digit, remove it from x. This makes x0 . Double the removed digit, subtract the doubled digit from x0 . Let this resulting number be x00 . Then x is divisible by 7 if and only if x00 is divisible by 7. Therefore, we can repeat the process of removing, doubling, and subtracting until the result is a number that is known to be either divisible or not divisible by 7; this determines whether x is divisible by 7. This divisiblilty test is relatively complicated. We look at a simple example in order to demonstrate how it works. Let our number be 31415. Then we take the last digit: 5. Double it: 10. Chop off the last digit: 3141. Subtract: 3141-10=3131. Repeat. Take the last digit: 1. Double it: 2. Chop off the last digit: 313. Subtract: 313-2=309. Repeat. Take the last digit: 9. Double it: 18. Chop off the last digit: 30. Subtract: 30-18=12. It is known that 12 is not divisible by 7, so 31415 is not either. Why does this work? Let our number be x = A B, where B is a one digit number, in our stupidly bad notation that does not represent multiplication. So x = 10A + B. Then we have a series of equivalent statements: 7|10A + B ⇔ 7|(7A + 7B) + (3A − 6B) ⇔ 7|7(A + B) + 3(A − 2B) ⇔ 7|3(A − 2B) ⇔ 7|A − 2B since subtracting a multiple of 7 or dividing by 3 does not affect divisibility by 7. But then A − 2B is exactly removing the last digit, doubling, and subtracting. Q.E.D. 2

8: Look at the last three digits. The number is divisible by 8 if and only if the last three digits is divisible by 8. For example, 78446597264 is divisible by 8 since 264 is divisible by 8. Why does this work? The proof is analogous to the case of divisibility of 2 and 4; here we have x = 1000A + B where B is three digits, and we know that 1000 = 8 × 125, so that 8|1000A + B ⇔ 8|B. We cannot just test for divisibility by 2 and 4 since divisible by 4 automatically implies divisible by 2; for example, 20 is divisible by 2 and 4 but not 8. This method fails because 2 and 4 are not relatively prime. In other words, they have a common factor (2); their greatest common divisor is bigger than 1. Whenever we attempt to test for divisibility by testing each factor individually, we must make sure that the factors are relatively prime. For more information, including another example, see the case of testing for divisibility by 12.

9: Sum the digits. The number is divisible by 9 if and only if the sum of the digits is divisible by 9. For example, the sum of digits of 784557 is 36, which is divisible by 9. Therefore 784557 is too. Why does this work? This proof is analogous to the case of divisibilty by 3. Once again we have 10n mod 9 = 1 for every natural number n, so we can use exactly the same argument as divisibility by 3 except for mod 9 instead of mod 3.

10: Look at the last digit. The number is divisible by 10 if and only if it is the last digit is 0. For example, 5494560 is divisible by 10 but 1231654 is not. Why does this work? We can think about this in two ways. First, divisibilty by 10 is equivalent to divisibility by both 2 and 5. 2 requires a last digits of 0, 2, 4, 6 or 8, and 5 requires a last digit of 0 or 5. The only overlap is a last digit of 0, so divisible by 10 requires a last digit of 0. Second, we can repeat the argument for 2 and 5 and show that the last digit must be divisible by 10 in order for number to be divisible by 10.

11: Take an alternating sum of digits. The number is divisible by 11 if and only if this alternating sum is divisible by 11. For example, if the number is 31415926 then the alternating sum would be +3 − 1 + 4 − 1 + 5 − 9 + 2 − 6 = −3, which is not divisible by 11. Therefore 31415926 is not either. Why does this work? This test is really quite similar to the cases with 3 and 9. From modular arithmetic we know that 10 ≡ −1 mod 11. This means that 10n ≡ (−1)n mod 11. Therefore when n is even, 10n mod 11 = 1, and when n is odd, 10n mod 11 = −1. We now use an example to demonstrate why this works. The example will be 17204. 17204 mod 11 = (1 × 104 + 7 × 103 + 2 × 102 + 0 × 101 + 4 × 100 ) mod 11 = (1 mod 11)(104 mod 11) + (7 mod 11)(103 mod 11) + (2 mod 11)(102 mod 11) + (0 mod 11)(101 mod 11) + (4 mod 11)(100 mod 11) = (1 mod 11) + (−1)(7 mod 11) + (2 mod 11) + (−1)(0 mod 11) + (4 mod 11) = (+1 − 7 + 2 − 0 + 4) mod 11 Hence, 17204 mod 11 = 0 if and only if the alternating sum of digits is divisible by 11. Q.E.D. 3

12: We test for divisibility by 12 by testing for 3 and 4 separately. For example, 84 is divisible by both 3 and 4, so it also divisible by 12. Why does this work? This works because 3 and 4 are relatively prime. We cannot test for 2 and 6 instead because they are not relatively prime; there is a common factor of 2. We can only split up into factors when the factors are relatively prime (such as 3 and 4). Otherwise, we are really testing for divisibility by the LCM (least common multiple) of the factors. In the case of 2 and 6, the LCM is 6. In the case of 4 and 6, the LCM would be 12. This is because we can write the prime factorization of the factors and remove the overlaps; the prime factor of 2 is an overlap in the case of factoring into 2 and 6. Look at the case of divisibility by 8!

13: For divisibility by 13, we have a number x. Take the last digit, remove it from x. This makes x0 . Multiply the removed digit by four, add the doubled digit to x0 . Let this resulting number be x00 . Then x is divisible by 13 if and only if x00 is divisible by 13. Therefore, we can repeat the process of removing, quadrupling, and adding until the result is a number that is known to be either divisible or not divisible by 13; this determines whether x is divisible by 13. We look at a simple example in order to demonstrate how it works. Let our number be 31415. Then we take the last digit: 5. Quadruple it: 20. Chop off the last digit: 3141. Add: 3141+20=3161. Repeat. Take the last digit: 1. Quadruple it: 4. Chop off the last digit: 316. Add: 316+4=320. Repeat. Take the last digit: 0. Quadruple it: 0. Chop off the last digit: 32. Add: 32+0=32. It is known that 32 is not divisible by 13, so 31415 is not either. Why does this work? This case is really quite similar to divisibility by 7. Let our number be x = A B, where B is a one digit number, in our stupidly bad notation that does not represent multiplication. So x = 10A + B. Then we have a series of equivalent statements: 13|10A + B ⇔ 13|(13A + 13B) − (3A + 12B) ⇔ 13|13(A + B) − 3(A + 4B) ⇔ 13|3(A + 4B) ⇔ 13|A + 4B since subtracting a multiple of 13 or dividing by -3 does not affect divisibility by 13. But then A + 4B is exactly removing the last digit, quadrupling, and adding. Q.E.D.

14: It is known that 14 = 2 × 7. Therefore we can test for 2 and 7 separately. A number is divisible by 14 if and only if it is divisible by both 2 and 7. This works because 2 and 7 are relatively prime; see the cases of divisibility by 8 and 12 for more information.

15: It is known that 15 = 3 × 5. Therefore we can test for 3 and 5 separately. A number is divisible by 15 if and only if it is divisible by both 3 and 5. This works because 3 and 5 are relatively prime; see the cases of divisibility by 8 and 12 for more information.

16: Look at the last four digits. If it is divisible by 16 then the whole number is too. For example, for the number 544689412345654288, we only need to look at 4288 to conclude that it is divisible by 16. The proof is analogous to 2, 4, 8; they are all powers of 2. This is because 10000 = 16 × 625. In general, divisibility by 2n requires looking at the last n digits since 10n = 2n × 5n . However, for huge values of n, this method might not be practical. 4