Test for divisibility by 13

There are simple rules for telling whether a number is divisible by 2, 3, 4, 5, and 6.

• A number is divisible by 2 if its last digit is divisible by 2.
• A number is divisible by 3 if the sum of its digits is divisible by 3.
• A number is divisible by 4 if the number formed by its last two digits is divisible by 4.
• A number is divisible by 5 if its last digit is divisible by 5.
• A number is divisible by 6 if it is divisible by 2 and by 3.

There is a rule for divisibility by 7, but it's a little wonky. Let's keep going.

• A number is divisible by 8 if the number formed by its last three digits is divisible by 8.
• A number is divisible by 9 if the sum of its digits is divisible by 9.
• A number is divisible by 10 if its last digit is 0.

There's a rule for divisibility by 11. It's a little complicated, though not as complicated as the rule for 7. I describe the rule for 11 in the penultimate paragraph here.

A number is divisible by 12 if it's divisible by 3 and 4. (It matters here that 3 and 4 are relatively prime. It's not true, for example, that a number is divisible by 12 if it's divisible by 2 and 6.)

But what do you do when you get to 13?

Testing divisibility by 7, 11, and 13

We're going to kill three birds with one stone by presenting a rule for testing divisibility by 13 that also gives new rules for testing divisibility by 7 and 11. So if you're trying to factor a number by hand, this will give a way to test three primes at once.

To test divisibility by 7, 11, and 13, write your number with digits grouped into threes as usual. For example,

11,037,989

Then think of each group as a separate number - e.g. 11, 37, and 989 - and take the alternating sum, starting with a + sign on the last term.

989 - 37 + 11

The original number is divisible by 7 (or 11 or 13) if this alternating sum is divisible by 7 (or 11 or 13 respectively).

The alternating sum in our example is 963, which is clearly 9*107, and not divisible by 7, 11, or 13. Therefore 11,037,989 is not divisible by 7, 11, or 13.

Here's another example. Let's start with

4,894,498,518

The alternating sum is

518 - 498 + 894 - 4 = 910

The sum takes a bit of work, but less work than dividing a 10-digit number by 7, 11, and 13.

The sum 910 factors into 7*13*10, and so it is divisible by 7 and by 13, but not by 11. That tells us 4,894,498,518 is divisible by 7 and 13 but not by 11.

Why this works

The heart of the method is that 7*11*13 = 1001. If I subtract a multiple of 1001 from a number, I don't change its divisibility by 7, 11, or 13. More than that, I don't change its remainder by 7, 11, or 13.

The steps in the method amount to adding or subtracting multiples of 1001 and dividing by 1000. The former doesn't change the remainder by 7, 11, or 13, but the latter multiplies the remainder by -1, hence the alternating sum. (1000 is congruent to -1 mod 7, mod 11, and mod 13.) See more formal argument in footnote [1].

So not only can we test for divisibility by 7, 11, and 13 with this method, we can also find the remainders by 7, 11, and 13. The original number and the alternating sum are congruent mod 1001, so they are congruent mod 7, mod 11, and mod 13.

In our first example, n = 11,037,989 and the alternating sum was m = 963. The remainder when m is divided by 7 is 4, so the remainder when n is divided by 7 is also 4. That is, m is congruent to 4 mod 7, and so n is congruent to 4 mod 7. Similarly, m is congruent to 6 mod 11, and so n is congruent to 6 mod 11. And finally m is congruent to 1 mod 13, so n is congruent to 1 mod 13.

Related posts

• Divisibility rules in hexadecimal
• Fermat's factoring trick
• How long does it take to find large primes?

[1] The key calculation is

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