Image via WikipediaSo, most know how to tell if something is divisible by 2 or 5, and many know how to tell if something is divisible by 9. What about other numbers?

So, here are strategies for discovering divisibility from 2 to 10, and then we'll talk about some rarer, more surprising divisibility tricks:

Example #1: 189 -> 1 + 8 + 9 = 18, 18 is divisible by 9, so 189 is also divisible by 9

Example #2: 137781 -> 1 + 3 + 7 + 7 + 8 + 1 = 27, 27 is divisible by 9, so 137781 is also divisible by 9

(Note: If adding six numbers together in your head seems difficult, look for my next post on number-grouping tricks. Soon, adding six numbers together will be a cinch!)

Example #1: 66 -> 6 + 6 = 12, 12 is divisible by 3, so 66 is also divisible by 3

Example #2: 15309 -> 1 + 5 + 3 + 9 = 18, 18 -> 1 + 8 = 9, 9 is divisible by 3, so 18 is divisible by 3; 18 is divisible by 3 so 15309 is divisible by 3

Example #1: 72, 72 is divisible by 2 because it ends in a 2, 72 is divisible by 3 because 7 + 2 = 9 and 9 is divisible by 3, so 72 is also divisible by 6

Example #2: 30618, 30618 is divisible by 2 because it ends in an 8, 30618 is divisible by 3 because 3 + 0 + 6 + 1 + 8 = 18 and 18 is divisible by 3, so 30618 is also divisible by 6

Example #1: 124, 24 is divisible by 4 (cut 24 in half -> 12, then in half again -> 6), so 124 is divisible by 4

Example #2: 562312, 12 is divisible by 4, so 562312 is divisible by 4

Example #1: 200, cut 200 in half -> 100, cut 100 in half -> 50, cut 50 in half -> 25, because 200 cuts in half evenly three times, it is divisible by 8

Example #2: 429192, 192 is divisible by 8 (192 in half is 96, 96 in half is 48, 48 in half is 24), so 429192 is divisible by 8

Example #1: 266, take the last digit (6) and double it -> 12, subtract it from the first two digits -> 26 - 12 = 14. 14 is divisible by 7, so 266 is divisible by 7.

Example #2: 13034, take the last digit (4) and double it -> 8, subtract it from the rest of the digits -> 1303 - 8 = 1295. Is 1295 divisible by 7? Take the last digit (5), and double it -> 10, subtract it from the rest of the digits -> 129 - 10 = 119. Is 119 divisible by 7? Take the last digit (9) and double it -> 18, subtract it from the rest of the digits -> 11 - 18 = -7. -7 is divisible by 7, so 119, 1295, and 13034 are all divisible by 7. Phew!

So, the divisible by seven rule is a unique one. For all you budding mathematicians, here is why it works.

Let's take a number n that has more than one digit.

That means that it can be written like this:

10*x + y where y is the last digit and x is all the others.

Example: 147 can be written as 10*14 + 7

Remember that we said that if n is divisible by 7, then the difference between the rest of the numbers and twice the last digit would be a multiple of seven. So x - 2*y should be a multiple of 7. Let's write that as 7*z. So x - 2*y = 7*z, or x = 7*z + 2*y.

Thus, we get:

10*(7*z + 2*y) + y

Example: 378 can be written as 10*(7*3 + 2*8) + 8

Almost there. Now, it's just algebra.

10*(7*z + 2*y) + y = 70*z + 20*y + y = 70*z + 21*y = 7*(10*z + 3*y)

And that's it! n = 7*(10*z + 3*y), which means that n is divisible by 7

Glad you asked! :)

Notice the reason that 7 could be factored out of the second term. It's because we had a multiple of 10 (20 in that case) added to 1, which created a number that was divisible by 7, in that case 21. Are there other multiples of ten that, when 1 is added to it, give us a composite number?

31 and 41 don't work, because they are prime. But, 51 does work! If we could have that multiple of 10 be 50, then we could probably come up with a new divisibility rule.

What is 51 divisible by? Well, it happens to be divisible by 17. Could we possibly get a 17 divisibility rule out of this?

Let's look at the pattern again. 7*(10*z + 3*y) is the end pattern that we are looking for. Let's replace 7 with 17. We get 17*(10*z + 3*y) => 170*z + 51*y => 170*z + 50*y + y => 10*(17*z + 5*y) + y

Does that look somewhat familiar? Check out the first equation that we got to represent the divisible by 7 rule: 10*(7*z + 2*y) + y. It looks pretty close.

So, let's try translating it back into English: If you multiply the last digit by 5 (instead of 2), and subtract it from the rest of the digits, if that number is divisible by 17, then the original number is divisible by 17.

Experiment #1: 867, multiply the last digit (7) by 5 -> 35, and subtract it from the rest of the digits -> 86 - 35 = 51. 51 is divisible by 17 (17*3), so 867 is divisible by 17.

Experiment #2: 6936, multiply the last digit (6) by 5 -> 30, and subtract it from the rest of the digits -> 693 - 30 = 663. Is 663 a multiple of 17? Multiply the last digit (3) by 5 -> 15, and subtract it from the rest of the digits -> 66 - 15 = 51. 51 is divisible by 17, so 663 and 6936 are also divisible by 17.

I think you probably know the answer to that. :) The other two that I found were 13 and 27. Can you figure out divisibility rules for those? Good luck!

So, here are strategies for discovering divisibility from 2 to 10, and then we'll talk about some rarer, more surprising divisibility tricks:

### Divisible by two:

If the number ends in 0, 2, 4, 6, or 8, it is divisible by 2.### Divisible by five:

If the number ends in 0 or 5, it is divisible by 5.### Divisible by ten:

If the number ends in 0, it is divisible by 10.### Divisible by nine:

If you add all the digits in a number together and that new number is divisible by 9, then it is also divisible by 9.Example #1: 189 -> 1 + 8 + 9 = 18, 18 is divisible by 9, so 189 is also divisible by 9

Example #2: 137781 -> 1 + 3 + 7 + 7 + 8 + 1 = 27, 27 is divisible by 9, so 137781 is also divisible by 9

(Note: If adding six numbers together in your head seems difficult, look for my next post on number-grouping tricks. Soon, adding six numbers together will be a cinch!)

### Divisible by three:

If you add all the digits in a number together and that new number is divisible by 3, then it is also divisible by 3.Example #1: 66 -> 6 + 6 = 12, 12 is divisible by 3, so 66 is also divisible by 3

Example #2: 15309 -> 1 + 5 + 3 + 9 = 18, 18 -> 1 + 8 = 9, 9 is divisible by 3, so 18 is divisible by 3; 18 is divisible by 3 so 15309 is divisible by 3

### Divisible by six:

If the number is divisible by 2 and 3, then it is divisible by 6.Example #1: 72, 72 is divisible by 2 because it ends in a 2, 72 is divisible by 3 because 7 + 2 = 9 and 9 is divisible by 3, so 72 is also divisible by 6

Example #2: 30618, 30618 is divisible by 2 because it ends in an 8, 30618 is divisible by 3 because 3 + 0 + 6 + 1 + 8 = 18 and 18 is divisible by 3, so 30618 is also divisible by 6

### Divisible by four:

If the last two digits are divisible by 4, then it is also divisible by 4. Or, if you can cut it in half twice evenly.Example #1: 124, 24 is divisible by 4 (cut 24 in half -> 12, then in half again -> 6), so 124 is divisible by 4

Example #2: 562312, 12 is divisible by 4, so 562312 is divisible by 4

### Divisible by eight:

If the last three digits are divisible by 8, then it is also divisible by 8. Or, if you can cut it in half evenly three times.Example #1: 200, cut 200 in half -> 100, cut 100 in half -> 50, cut 50 in half -> 25, because 200 cuts in half evenly three times, it is divisible by 8

Example #2: 429192, 192 is divisible by 8 (192 in half is 96, 96 in half is 48, 48 in half is 24), so 429192 is divisible by 8

### Divisible by seven:

Take the last digit, double it, and subtract it from the rest. If that number is divisible by 7, then your original number is also divisible by 7.Example #1: 266, take the last digit (6) and double it -> 12, subtract it from the first two digits -> 26 - 12 = 14. 14 is divisible by 7, so 266 is divisible by 7.

Example #2: 13034, take the last digit (4) and double it -> 8, subtract it from the rest of the digits -> 1303 - 8 = 1295. Is 1295 divisible by 7? Take the last digit (5), and double it -> 10, subtract it from the rest of the digits -> 129 - 10 = 119. Is 119 divisible by 7? Take the last digit (9) and double it -> 18, subtract it from the rest of the digits -> 11 - 18 = -7. -7 is divisible by 7, so 119, 1295, and 13034 are all divisible by 7. Phew!

## What's with the divisible by seven rule?

So, the divisible by seven rule is a unique one. For all you budding mathematicians, here is why it works.

Let's take a number n that has more than one digit.

That means that it can be written like this:

10*x + y where y is the last digit and x is all the others.

Example: 147 can be written as 10*14 + 7

Remember that we said that if n is divisible by 7, then the difference between the rest of the numbers and twice the last digit would be a multiple of seven. So x - 2*y should be a multiple of 7. Let's write that as 7*z. So x - 2*y = 7*z, or x = 7*z + 2*y.

Thus, we get:

10*(7*z + 2*y) + y

Example: 378 can be written as 10*(7*3 + 2*8) + 8

Almost there. Now, it's just algebra.

10*(7*z + 2*y) + y = 70*z + 20*y + y = 70*z + 21*y = 7*(10*z + 3*y)

And that's it! n = 7*(10*z + 3*y), which means that n is divisible by 7

## Interesting. Can that be done with other numbers? What does this have to do with 17?

Glad you asked! :)

Notice the reason that 7 could be factored out of the second term. It's because we had a multiple of 10 (20 in that case) added to 1, which created a number that was divisible by 7, in that case 21. Are there other multiples of ten that, when 1 is added to it, give us a composite number?

31 and 41 don't work, because they are prime. But, 51 does work! If we could have that multiple of 10 be 50, then we could probably come up with a new divisibility rule.

What is 51 divisible by? Well, it happens to be divisible by 17. Could we possibly get a 17 divisibility rule out of this?

Let's look at the pattern again. 7*(10*z + 3*y) is the end pattern that we are looking for. Let's replace 7 with 17. We get 17*(10*z + 3*y) => 170*z + 51*y => 170*z + 50*y + y => 10*(17*z + 5*y) + y

Does that look somewhat familiar? Check out the first equation that we got to represent the divisible by 7 rule: 10*(7*z + 2*y) + y. It looks pretty close.

So, let's try translating it back into English: If you multiply the last digit by 5 (instead of 2), and subtract it from the rest of the digits, if that number is divisible by 17, then the original number is divisible by 17.

Experiment #1: 867, multiply the last digit (7) by 5 -> 35, and subtract it from the rest of the digits -> 86 - 35 = 51. 51 is divisible by 17 (17*3), so 867 is divisible by 17.

Experiment #2: 6936, multiply the last digit (6) by 5 -> 30, and subtract it from the rest of the digits -> 693 - 30 = 663. Is 663 a multiple of 17? Multiply the last digit (3) by 5 -> 15, and subtract it from the rest of the digits -> 66 - 15 = 51. 51 is divisible by 17, so 663 and 6936 are also divisible by 17.

## Wow! Will it work for other numbers?

I think you probably know the answer to that. :) The other two that I found were 13 and 27. Can you figure out divisibility rules for those? Good luck!

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