Divisibility

Tricks

Tricks and shortcuts for testing divisibility:

2

We know that if the last digit in a number is divisible by 2, the entire number is divisible by 2. This stems from the fact that 2 divides 10 perfectly. Thus, every number can be split into a two components: the first component is an integer multiple of 10, the second component is the last digit of the number.

Example: 416

416 = 41 * 10 + 6

We know 2 will divide 41 * 10 perfectly, because 2 divides 10. Thus, the only question remaining is whether 2 divides the last digit, 6, which it does. So, 2 divides 416.

3

We know that if the digit sum of a number is divisible by 3, the number itself is divisible by 3. If the sum of the digits forms a large number, its digits can be added together in a like fashion.

Example: 15,230,127

15,230,127 => 1 + 5 + 2 + 3 + 0 + 1 + 2 + 7 = 21 
21 => 2 + 1 => 3

The sum of the digits is 3, therefore 3 divides 15,230,127.

4

If the last two digits of a number are divisible by 4, the entire number is divisible by 4. The mechanism for this is similar to the mechanism for the divisibility tests for 2 and for 5. Because 4 divides 100 perfectly, any number can be split into its hundreds and greater component, and its tens and ones digits. 4 will evenly divide the hundreds and greater component, since it perfectly divides 100. The only remaining thing to determine is whether 4 divides the last 2 digits.

5

If a number ends in 0 or 5, it is divisible by 5. This is due to the fact that 5 divides 10 perfectly, like 2, so any number can be split into its ones value (which 5 may or may not divide) and everything else (which 5 will divide).

9

We know that if the digit sum of a number is divisible by 9, the number itself is divisible by 9.

11

To test for divisibility by 11:

• Add the numbers in the even positions (second, fourth, and so on).
• Add the digits in the odd positions (first, third, and so on)
• If the difference between the sums is 0 or a multiple of 11, the number is divisible by 11.
• Otherwise, it is not a multiple of 11.

Example: 181,182,375

181,182,375 => (1+1+8+3+5) - (8+1+2+7) = 18 - 18 = 0

Therefore, 181,182,375 is divisible by 11.

Proof/Explanation

This introduces the term modulus - similar to remainder. 18 has a remainder of 7 when divided by 11, and 18 = 7 mod 11.

The digits 0, 1, 2, 3, ..., 9 are, of course, 0, 1, 2, ..., 9 mod 11

0, 10, 20, ..., 90 are, by actual division, 0, 10, 9, ..., 2 mod 11 = 0, -1, -2, ..., -9 mod 11

0, 100, 200, ... 900 are again 0, 1, 2, ..., 9 mod 11 and so on

The sum of two given numbers has the same modulus as the sum of the moduli of the numbers. For a given number N:

N = a + 10 b + 100 c + 1000 d + ...

Now expressing that mod 11:

N mod 11 = a mod 11 + 10 b mod 11 + 100 c mod 11 + 1000 d mod 11 + ... = a mod 11 - b mod 11 + c mod 11 - d mod 11 + ... = a mod 11 + c mod 11 + ... - ( b mod 11 + d mod 11 + ... )

Since a, b, c, d, are the digits of N, starting from right to left, the test must be correct.