Divisibility Rules

Divisibility Rules

How can you tell whether a number is divisible by another number (leaving no remainder) without actually doing the division? Why do "divisibility rules" work?

3	If the sum of the digits is divisible by three, the number is also
4	If the last two digits are divisible by 4, the number is also
5	If the last digit is a 5 or a 0, the number is divisible by 5.
6	If the number is divisible by both 3 and 2, it is also divisible by 6.
7	Take the last digit, double it, and subtract it from the rest of the number; if the answer is divisible by 7 (including 0), then the number is also
8	If the last three digits are divisible by 8, the number is also
9	If the sum of the digits is divisible by 9, the number is also. *This holds for any power of 3
10	If the number ends in 0, it is divisible by 10.
11	Subtract the sum of the even digits from the sum of the odd digits; if the difference, including 0, is divisible by 11, the number is also.
12	If the number is divisible by both 3 and 4, it is also divisible by 12.
13	Delete the last digit from the number, then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number.

Why do these 'rules' work?

Dividing by 3

Add up the digits: if the sum is divisible by three, then the number is as well. Examples:

111111: the digits add to 6 so the whole number is divisible by three.
87687687. The digits add up to 57, and 5 plus seven is 12, so the original number is divisible by three.

Why does the 'divisibility by 3' rule work?

The only way that I can think of to explain this would be as follows:

Look at a 2 digit number: 10a+b=9a+(a+b). We know that 9a is divisible by 3, so 10a+b will be divisible by 3 if and only if a+b is.

Similarly, 100a+10b+c=99a+9b+(a+b+c), and 99a+9b is divisible by 3, so the total will be iff a+b+c is.

This explanation also works to prove the divisibility by 9 test. It clearly originates from modular arithmetic ideas, and I'm not sure if it's simple enough, but it's the only explanation I can think of.

Doctor Darren, The Math Forum

Dividing by 4 and 8

Since 100, 1000 and so on are multiples of 4, it follows (as for 2) that a number is divisible by 4 if the number represented by its last two digits is a multiple of 4. Example: 3728 is a divisible by 4 because 28 is.

Powers of 10 from 1000 on are divisible by 8, therefore it follows that a number is divisible by 8 if the number represented by its last three digits is a multiple of 8. Example: 3728 is a divisible by 8 because 728 is.

Dividing by 11

Let's look at 352, which is divisible by 11; the answer is 32. 3+2 is 5; another way to say this is that 35 -2 is 33.

Now look at 3531, which is also divisible by 11. It is not a coincidence that 353-1 is 352 and 11 x 321 is 3531.

Here is a generalization of this system. Let's look at the number 94186565.

First we want to find whether it is divisible by 11, but on the way we are going to save the numbers that we use: in every step

we will subtract the last digit from the other digits, then save the subtracted amount in order. Start with

	9418656 - 5=	9418651	SAVE 5
Then	941865 - 1 =	941864	SAVE 1
Then	94186 - 4 =	94182	SAVE 4
Then	9418 - 2 =	9416	SAVE 2
Then	941 - 6 =	935	SAVE 6
Then	93 - 5 =	88	SAVE 5
Then	8 - 8=	1	SAVE 8

Now write the numbers we saved in reverse order, and we have 8562415, which multiplied by 11 is 94186565.

Here's an even easier method, contributed by Chis Foren:

Take any number, such as 365167484.

Add the 1,3,5,7,..,digits.....3 + 5 + 6 + 4 + 4 = 22

Add the 2,4,6,8,..,digits.....6 + 1 + 7 + 8 = 22

If the difference, including 0, is divisible by 11, then so is the number.

22 - 22 = 0 so 365167484 is evenly divisible by 11.

Other Numbers

And here's a more complex method that can be extended to other formulas:

1	1 (mod 13)
10	-3 (mod 13) (i.e., 10 - -3 is divisible by 13)
100	-4 (mod 13) (i.e., 100 - -4 is divisible by 13)
1000	-1 (mod 13) (i.e., 1000 - -1 is divisible by 13)
10000	3 (mod 13)
100000	4 (mod 13)
1000000	1 (mod 13)

Call the ones digit a, the tens digit b, the hundreds digit c, ..... and you get:

a - 3*b - 4*c - d + 3*e + 4*f + g - .....

If this number is divisible by 13, then so is the original number.

You can keep using this technique to get other formulas for divisibility for prime numbers. For composite numbers just check for divisibility by divisors.