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  Divisibility Rules

From the Archives of the Math Forum's Internet project Ask Dr. Math - our thanks to Ethan 'Dr. Math' Magness, Steven 'Dr. Math' Sinnott, and, for the explanation of why these rules work, Robert L. Ward (Dr. Rob).

Dividing by 3

Add up the digits: if the sum is divisible by three, then the number is as well. Examples:
1. 111111: the digits add to 6 so the whole number is divisible by three.
2. 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?

From: "Dr. Math"
To: keving@ecentral.com (Kevin Gallagher)
Subject: Re: Divisibility of a number by 3

As Kevin Gallagher wrote to Dr. Math
On 5/11/96 at 21:35:40 (Eastern Time),

>I'm looking for a SIMPLE way to explain to several very bright 2nd
>graders why the divisibility by 3 rule works, i.e. add up all the
>digits; if the sum is evenly divisible by 3, then the number is as well.
>Thanks!
>Kevin Gallagher

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
Check out our web site - http://forum.swarthmore.edu/dr.math/

Dividing by 4

Look at the last two digits. If they are divisible by 4, the number is as well.
Examples:
1. 100 is divisible by 4.
2. 1732782989264864826421834612 is divisible by four also, because 12 is divisible by four.

Dividing by 5

If the last digit is a five or a zero, then the number is divisible by 5.

Dividing by 6

Check 3 and 2. If the number is divisible by both 3 and 2, it is divisible by 6 as well.
Robert Rusher writes in:

Another easy way to tell if a [multi-digit] number is divisible by six . . .
is to look at its [ones digit]: if it is even, and the sum of the [digits] is
a multiple of 3, then the number is divisible by 6.

Dividing by 7

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.
Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.

Dividing by 8

Check the last three digits. Since 1000 is divisible by 8, if the last three digits of a number are divisible by 8, then so is the whole number.
Example: 33333888 is divisible by 8; 33333886 isn't.

How can you tell whether the last three digits are divisible by 8? Phillip McReynolds answers:

If the first digit is even, the number is divisible by 8 if the last two digits are. If the first digit is odd, subtract 4 from the last two digits; the number will be divisible by 8 if the resulting last two digits are. So, to continue the last example, 33333888 is divisible by 8 because the digit in the hundreds place is an even number, and the last two digits are 88, which is divisible by 8. 33333886 is not divisible by 8 because the digit in the hundreds place is an even number, but the last two digits are 86, which is not divisible by 8.

Dividing by 9

Add the digits. If they are divisible by nine, then the number is as well. This holds for any power of three.

Dividing by 10

If the number ends in 0, it is divisible by 10.

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 = 0           SAVE 8

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

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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.

See also Divisibility by 11 in the Dr. Math archives.

Dividing by 12

Check for divisibility by 3 and 4.

Dividing by 13

Here's a straightforward method supplied by Scott Fellows:

Delete the last digit from the given number. Then subtract nine times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number.

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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.

 

The Math Forum
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8 January 1999