Basic Divisibility Rules

It is easy to check the divisibility by some of the common numbers, so that you can find out if any given number is divisible by that number without actually dividing them.

Remember that $0$ is considered as divisible by any number, because when you divide $0$ by any other number it always leaves a remainder of $0$.

Divisibility by 2:

If the last digit of any number is divisible by $2$ , the number is divisible by $2$. If the last digit is not divisible by $2$, the number is not divisible by $2$.

Examples:

$532$ is divisible by $2$.

$29830$ is divisible by $2$. (Remember, zero is always divisible by $2$).

$892837$ is NOT divisible by 2.

Divisibility by 4:

If the number formed by the last two digits is divisible by $4$, the number is divisible by $4$, otherwise it is not.

Examples:

$293824$ is divisible by $4$. (Last two digits = $24$, which is divisible by $4$).

$293418$ is NOT divisible by $4$ (Last two digits = $18$, which is not divisible by $4$).

Divisibility by 3:

The divisibility of $3$ can be checked simply by adding the digits of a the given number. If the sum of the digits is divisible by $3$, then the number itself is divisible by $3$. We will take some example:

Is the number $2397$ divisible by $3$?

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The sum of the digits is $2 + 3 + 9 + 7 = 21$. We know that $21$ is divisible by $3$. So $2397$ is divisible by $3$. We check this:

$2397 \div 3 = 799$ and leaves a remainder of zero.

One more example. Is the number $92831$ divisible by $3$?

The sum of the digits is $9 + 2 + 8 + 3 + 1 = 23$, which is not divisible by $3$. So, the number $92831$ is not divisible by $3$.

Let's check this.

$92831 div 3 = 30943$ and leaves a remainder of $2$.

Divisibility by 9:

The divisibility by $9$ is done the same way as divisibility by $3$. If the sum of all the digits of the given number is divisible by $9$, then the number itself is divisible by $9$, otherwise not. We will take a couple of examples to see how this works.

Let us take the number 35487 and sum up all the digits.

So we get $3 + 5 + 4 + 8 + 7 = 27$. We know that $27$ is divisible by $9$, so this number is divisible by $9$.

We will take one more number as example. Let us try $298302$. Adding up the digits, we get:

$2 + 9 + 8 + 3 + 0 + 2 = 24$, and we know that $24$ is not divisible by $9$.

Therefore, $298302$ is not divisible by $9$. Indeed, when you divide $298302$ by $9$ you get $6$ as a remainder.

Divisibility by 11:

Divisibility by $11$ can be checked easily by calculating the sums of alternate digits and then taking the difference of the two sums. If the difference is divisible by $11$, then the number is divisible by $11$. Remember that $0$ is divisible by any number, so if the difference is $0$, then also the number is divisible by $11$.

Let us take the example of $267839$.

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We will split up the set of alternate digits using the red and blue colors to mark alternate digits:.

We have:

$\color{blue}{2}\raise{0.2em}{\color{red}{6}}\color{blue}{7}\raise{0.2em}{\color{red}{8}}\color{blue}{3}\raise{0.2em}{\color{red}{9}}$

We will add up all the blue digits to get $2 + 7 + 3 = 12$,

now let us add up all the red digits to get $6 + 8 + 9 = 23$

The difference of $23$ and $12$ is $11$. We know that $11$ is divisible by $11$. So the number $267839$ is divisible by $11$.

We will take one more small example to understand this better. Let us take the number $121$.

Like before we separate out the alternate digits, like before, and we get:

$\color{blue}{1}\raise{0.2em}{\color{red}{2}}\color{blue}{1}$

$1 + 1 = 2$

the other group has only one digit, which is $2$. The difference is $2 - 2 = 0$. So, $121$ is divisible by $11$.

Let us try with $439382$. We mark the alternate digits and get:

$\color{blue}{4}\raise{0.2em}{\color{red}{3}}\color{blue}{9}\raise{0.2em}{\color{red}{3}}\color{blue}{8}\raise{0.2em}{\color{red}{2}}$

Adding up one group we get:

$4 + 9 + 8 = 21$

Adding up the other group we get:

$3 + 3 + 2 = 8$

The difference of the sums is $21 - 8 = 13$. This is not divisible by $11$, so the our number $439382$ is also not divisible by $11$.

If you try the long division, you will see that $439382$ leaves a remainder of $9$ when divided by $11$.