Divisibility Tests — Definition & Examples

Can we divide evenly by small numbers?

The three most common divisibility tests are for $2$, $3$, and $5$.

The Number Two$\\$ A number is divisible by $2$ if it is an even number. These numbers are the ones that end in a $0$, $2$, $4$, $6$, or $8$. This last digit is called the units digit. We only have to look at the units digit this time. The rest of the digits, no matter how many there are, don't matter at all!

The Number Three$\\$ A number is divisible by $3$ if the sum of its digits are divisible by $3$.

For example, take the number $51$. If you add its digits you get $$5+1=6$$ $6$ is divisible by $3$ so therefore $51$ is divisible by $3$. If you check on a calculator you see that $$51\div3=17$$

The Number Five$\\$ A number is divisible by $5$ if its last digit is either $5$ or $0$.

For example, take the number $34{,}910$. Its last digit is $0$ so therefore it is divisible by $5$.

Some Other Tests

There are divisibility tests the other numbers as well. $4$ through $11$ are below for those curious.

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The Number Four$\\$ A number is divisible by $4$ if its last two digits are divisible by $4$.

For example, take $3{,}728$. Its last two digits are $28$. $$28\div4=7$$ Therefore $3{,}728$ is divisible by $4$.

The Number Six$\\$ A number is divisible by $6$ if it is even and the sum of its digits is divisible by $3$.

For example, take $342$. $342$ is even and the sum of its digits are $$3+4+2=9$$ $$9\div3=3$$ Therefore $342$ is divisible by $6$.

The Number Seven$\\$ Seven is a little tricky.

1. First, take the last digit off the number.
2. Second, double that digit.
3. Third, subtract that doubled digit from the remaining digits. If the resulting number is divisible by $7$ then the original number is divisible by $7$. If you can't yet tell, do the test again on your new number.

For example, take the number $3{,}374$.

1. First, take off the last digit, $4$.
2. Now, double it, $4\cdot2=8$.
3. Then, subtract that number from the remaining digits, $337-8=329$.

Since I don't know if $329$ is divisible by $7$, I do the test again.

1. Take the last digit, $9$.
2. Double it, $9\cdot2=18$.
3. Subtract, $32-18=14$. And $14\div7=2$ so therefore $3{,}374$ and $329$ are divisible by $7$.

The Number Eight

A number is divisible by $8$ if its last three digits are divisible by $8$.

For example, take $2{,}880$. Its last three digits are $880$. $$880\div8=110$$ Therefore $2{,}880$ is divisible by $8$. If you still can't tell or the number is $3$ digits or less, you can also use the method for $2$ four times or the method for $4$ twice.

The Number Nine$\\$ A number is divisible by $9$ if the sum of its digits is divisible by $9$.

For example, take the number $3{,}447$. $$3+4+4+7=18$$ $$18\div9=2$$ Therefore $3{,}447$ is divisible by $9$.

The Number Ten$\\$ A number is divisible by $10$ if its last digit is a $0$.

For example, take the number $684{,}820$. Its last digit is $0$ so therefore it is divisible by $10$.

The Number Eleven$\\$ $11$ is another tricky one but fun. Take the digits and alternate between subtracting and adding. If the final result is divisible by $11$ then the original number is divisible by $11$.

For example, take the number $55{,}902$. Now, alternate between subtracting and adding (note: you always have to start with subtraction). $$5-5+9-0+2=11$$ And $11\div11=1$ so therefore $55{,}902$ is divisible by $11$. Also note: if you get a $0$ as your end result or a negative number that is divisible by $11$ it still counts as being divisible by $11$.

(Video) Common Divisibility Tests

by American Public University

This is a video by American Public University. This one summarizes common divisibility rules you may see.

$\large{\text{Summary}}$

The video breaks numbers down into groups based on similar divisibility checks.

You check divisibility of $2$, $5$, and $10$ by checking the last digit of the number. As a result, the video groups these three tests into one category. If the number is even, it is divisible by $2$. If the last digit is $5$ or $0$, it is divisible by $5$. If the last digit is $0$, it is divisible by $10$. Any positive number might be divisible by more than one of these. As a result I encourage you to try all three tests.

Divisibility of $3$ and $9$ is found by adding up all the digits in a number. If the sum of the digits is divisible by $3$, the number is divisible by $3$. If the sum is divisible by $9$, the number is divisible by $9$. Every number divisible by $9$ is also divisible by $3$. This is because $9$ has $3$ as a factor.

For $4$ and $8$, you have to check multiple digits on the end of a number. If the last two digits of a number are divisible by $4$, then the original number is divisible by $4$. If the number made by the last three digits of a number are divisible by $8$, the number is divisible by $8$. If the number doesn't have that many digits, just use all of them. Pretend there are zeros at the front to get the required number of digits. This won't change the value of the number!

Whether a number is divisible by $6$ and $12$ depends on other tests. If a number is divisible by both $2$ and $3$, it is divisible by $6$ as well. This is because $6$ is a product of different prime numbers. In particular,

$$6 = 2^1 \times 3^1$$

If a number is divisible by both $3$ and $4$, it is also divisible by $12$. We could also check $2$ and $6$ instead of $3$ and $4$. The key is that we check pairs of numbers that multiply to $12$.

Why Do We Care?

These types of tests might seem to serve no real purpose at all. Fortunately, everything in math has a purpose, even if you don't know what it is. In this case we care about factorization of numbers. This is especially true for prime factorizations. Any number's factors are divisible by the factor. Thus we can use the tests to find factors of the number, and eventually the prime factors. In some cases we will have to use the tests more than once. Nonetheless, we have an alternative method to upside-down division.

Example: Let's find all the prime factors of $66$.

Solution: We could start randomly checking for prime factors. That being said, there is a strategy to make things go faster. It is usually easiest to start by checking the smallest prime numbers. The tests for the smaller prime numbers tend to be easier. Since $2$ is the smallest prime number, we'll start with $2$.

The test for $2$ is to check whether our number is even. Certainly $66$ is even, because the units digit is a $6$. So, we add $2$ to our list of prime factors and divide $66$ by $2$:

$$66 \div 2 = 33$$

We do this division to make it easier to keep looking for other prime factors. Any prime factor of $33$ will also be a prime factor of $66$, not to mention the $2$ we found earlier. The next smallest prime number is $3$. As a result we ask if $33$ is divisible by $3$. Remember the test for $3$. We want to check the sum of the digits of $33$:

$$3 + 3 = 6$$

Of course, $6$ is divisible by $3$, because $6 = 3 \times 2$. This means $3$ is a factor of $33$ (and of $66$). Once again, we divide away the new factor:

$$33 \div 3 = 11$$

You might remember that $11$ is a prime number. This means we don't need to run another test. We conclude that the prime factors of $66$ are $2, 3,$ and $11$. This completes the problem.

Introduction

Here we will talk about divisibility tests. These are short tests you can run on a specific number to find out if it's divisible by another number. (If a number $a$ is divisible by another number $b$, then $a\div b$ is a whole number.) Some numbers have short tests, like $2$ and $3$, but others have longer, more tedious ones, such as $7$. In this article, I'll go over the tests for $2,3,4,5,6,8,9,$ and $10$.

What's wonderful about these tests is they work for large numbers and small numbers. Many of them just depend on one or two digits of the numbers. As a result, there can be any number of digits in a multiple.

By applying a bunch of these tests you can even find the prime factors of a number! The number formed by multiplying all the prime factors will be a multiple of each prime factor.

Without further ado, let's get into the tests.

Checking $2,4,$ and $8$

The tests for powers of two are rather simple. In order for a number to be divisible by $2$, its last digit must be $2$, $4$, $6$, $8$, or $0$. In other words, the number must be an even number. The test for $4$ is slightly more complicated: $$\text{If the tens digit is even, the ones digit must be } 0,4,\text{or }8.$$ $$\text{If the tens digit is odd, the ones digit must be } 2\text{ or }6.$$ Another way to say this is that the last two digits of the number must be a multiple of $4$. Finally, for a number to be divisible by $8$, its last three digits have to be a multiple of $8$. In any of these cases, the original number is divisible by a power of $2$.

Checking $3,6,$ and $9$

In order for a number to be divisible by $3$, the sum of its digits must be a multiple of $3$. For example, $1{,}536$ is a multiple of $3$ because $1+5+3+6=15$, and $15$ is divisible by $3$. The test for divisibility by $9$ is similar: the sum of the digits of the number has to be a multiple of $9$. For example, $9{,}714$ isn't a multiple of $9$ because $9+7+1+4=21$, and $21$ isn't a multiple of $9$.

For a number to be divisible by $6$, it must be a multiple of $2$ and $3$. This is because $2\cdot3=6$. For example, $456$ is divisible by $6$ because $456$ is an even number and $4+5+6=15$, which is a multiple of $3$.

Checking $5$ and $10$

The last tests I'll show you are for $5$ and $10$, which are by far the simplest tests there are (besides the test for $2$, of course.) For a number to be a multiple of $5$, it has to end in a $5$ or a $0$, and for a number to be a multiple of $10$, it has to end in a $0$. An example is the number $743{,}190$, which is a multiple of $10$ because it ends in a $0$.

In this example we're going to talk about a handful of tests. These can help find the factors of any whole number we'd want.

Divisibility by $3$, $4$, $7$, and $9$

Divisibility by $3$

Example: Consider the number $1602$. We can look to find the sum of all of its digits. This sum is

$$1 + 6 + 0 + 2 = 9$$

Divisibility by $4$

Divisibility by $7$

Let's try this out.

Divisibility by $9$