Divisibility Rules

What “divisible” means

One whole number is divisible by another when you can divide them and nothing is left over. The number $20$ is divisible by $5$, because $20 \div 5 = 4$ with a remainder of $0$. The number $23$ is not divisible by $5$, because $23 \div 5 = 4$ with a remainder of $3$; that leftover $3$ is what disqualifies it. Divisibility is all about the remainder being zero, nothing else.

The same fact can be stated three ways, and reading them as interchangeable will save you confusion later:

• $20$ is divisible by $5$.
• $5$ is a factor (also called a divisor) of $20$.
• $20$ is a multiple of $5$.

Each sentence says that $20$ can be cut into a whole number of equal groups of $5$, here four groups, leaving none behind.

The remainder is the only thing worth watching, so it is worth seeing rather than trusting. Below you choose how many dots there are and how many go in each row. Dots that cannot complete a row are drawn hollow, and those hollow dots are the remainder.

Set the count to $10$ and step the row width from $2$ up to $6$. Exactly two of those five widths leave nothing hollow, and each of the other three leaves a remainder you can count on the picture.

The one idea behind most rules

There is a single move underneath all of these tests: break the number into pieces along its columns. Then throw away the pieces that are obviously multiples of your divisor, and test only what is left.

Try it on $28$ with a divisor of $4$. Split $28$ into $20 + 8$. Each piece is a multiple of $4$ on its own, since $20 = 4 \times 5$ and $8 = 4 \times 2$. Push them back together and you are holding $5$ fours alongside $2$ fours, which is $7$ fours in total. So $28 = 4 \times 7$, and $4$ divides it with nothing over.

The numbers $20$ and $8$ did no special work there. Their counts of fours, $5$ and $2$, could simply be added because both pieces were measured in fours to begin with.

Watch that happen again with a divisor of $3$. Both $18$ and $12$ are multiples of $3$, since $18$ is six threes and $12$ is four threes. Put them together and you are holding six threes alongside four threes, which is ten threes. So $18 + 12 = 30$ is a multiple of $3$ as well. Take them apart instead, and six threes minus four threes leaves two threes. So $18 - 12 = 6$ is a multiple of $3$ too.

Nothing there depended on the numbers being $18$ and $12$. Two piles counted in the same unit can be added or subtracted just by adding or subtracting how many units each one holds. Write $d$ for that shared unit, and $m$ and $n$ for the two counts. What you get is the distributive property you have already met, read in reverse:

So a number that divides two others also divides their sum and their difference. Every rule for $2$, $5$, $10$, $3$, $9$, $4$, and $8$ below uses this once. We split the number into two parts: a big part that is plainly a multiple of $d$, contributing remainder $0$, and a small leftover. Then the whole question rests on that leftover alone. The rule for $6$ works differently: it does not split the number at all, and instead combines two of these rules, as the last section of the lesson shows.

Divisibility by 2, 5, and 10: read the last digit

These three are the easiest tests, and they share one reason: every column except the ones is a multiple of $10$.

Take $836$ again. The $800$ and the $30$ combine into $830$, which is $83$ tens, a clean multiple of $10$. And $10$ is $2 \times 5$, so those $83$ tens are a multiple of $2$ and a multiple of $5$ as well. All three divisions therefore come out even on the $830$ part alone. Only the $6$ is still waiting to be judged.

Nothing in that split was special to $836$. The tens, hundreds, thousands, and all columns to the left are made of tens, so any whole number, however long, can be written as

So whatever decides divisibility by $2$, $5$, or $10$ has to live in the last digit alone. That is because everything else was already a multiple of all three.

A number whose last digit is even is called an even number, and any other whole number is odd. So “even” and “divisible by $2$” are two names for the same property, and you check both by glancing at the final digit.

Divisibility by 3 and 9: add the digits

For $3$ and $9$ the last digit is useless: $12$ is divisible by $3$ but $32$ is not, and both end in $2$. The right test is the well-known one, add every digit and check that sum. The digits of $531$ add to $5 + 3 + 1 = 9$, which is divisible by both $3$ and $9$, so $531$ is too. This looks like a coincidence, but it comes out of place value as neatly as the last-digit rules. You only have to notice one thing about the numbers $10, 100, 1000,$ and so on.

The key fact is that each of those place values is exactly one more than a multiple of $9$:

and $9$, $99$, and $999$ are all multiples of $9$ (and of $3$). So suppose a digit $d$ sits in the tens column. That digit’s contribution $d \times 10$ breaks into $d \times 9$, which is a multiple of $9$, plus a single leftover $d$. Every column behaves the same way: it donates a multiple of $9$ and leaves one copy of its own digit. Sweep up all those leftovers and you have the digit sum.

If the digit sum is still large, add its digits again, and repeat until the total is small enough to read at a glance. For $9{,}996$ the digits give $9 + 9 + 9 + 6 = 33$, and adding once more gives $3 + 3 = 6$. Because $6$ is divisible by $3$ but not by $9$, the original $9{,}996$ is divisible by $3$ but not by $9$.

Divisibility by 4 and 8: check the last two or three digits

The rule for $3$ used digit sums, but $4$ goes back to the splitting idea, just one column deeper. The fact we need is that $100$ is a multiple of $4$, since $100 = 4 \times 25$.

Take $1{,}732$ and cut it at that point, into $1{,}700 + 32$. The $1{,}700$ piece is $17$ hundreds, and every hundred is $4 \times 25$, so those $17$ hundreds come to $4 \times 425$ exactly. That piece leaves nothing over. The whole question drops onto the $32$, and $32 = 4 \times 8$, so $1{,}732$ is divisible by $4$.

The same cut works on any number, because the hundreds, thousands, and higher part always counts a whole number of hundreds:

That part is a multiple of $100$, and therefore of $4$, so it contributes no remainder. So divisibility by $4$ depends only on the number made by the last two digits.

The very same reasoning, pushed out one more column, gives the rule for $8$. Here $1000$ is a multiple of $8$, because $1000 = 8 \times 125$, so everything from the thousands column up is a multiple of $8$. That is why divisibility by $8$ depends only on the last three digits.

Divisibility by 6: combine 2 and 3

Some numbers carry no rule of their own. Instead you build a test from the rules for their factors. The tidiest example is $6$.

Watch it happen on $42$. The last digit is even, so $2$ divides it, and $42 = 2 \times 21$. The digit sum is $4 + 2 = 6$, so $3$ divides $42$ as well. Now look at the $21$ sitting beside that $2$. It is $3 \times 7$, so the factor of $3$ was waiting inside the other half all along. Collecting the pieces gives $42 = 2 \times 3 \times 7 = 6 \times 7$.

Because $6 = 2 \times 3$, a number is divisible by $6$ exactly when it is divisible by both $2$ and $3$. You already own both tests, so you just run them together. Confirm that the last digit is even and that the digit sum is a multiple of $3$.

Here is why passing both tests is always enough, without any letters. Look at the multiples of $3$ in order, starting from $0$: $0, 3, 6, 9, 12, 15, 18, 21, 24, \ldots$. They alternate even, odd, even, odd, forever, because adding $3$ to an even number gives an odd one and adding $3$ to an odd number gives an even one. So exactly every other multiple of $3$ is even, and those are exactly the numbers passing both the $2$ test and the $3$ test at once. But the even multiples of $3$ are exactly the multiples of $6$: $0, 6, 12, 18, 24, \ldots$. Passing both tests always lands you on one of those.

This combining trick works whenever the two factors share no common factor bigger than $1$, as $2$ and $3$ do not. It does not work to build a test for $4$ by running the $2$ test twice over, because there the two factors are both $2$ rather than different numbers, and passing the $2$ test once already guarantees passing it again. For $6$ the split into a $2$ and a $3$ is clean, so the combined test is exactly right.

Putting the rules to work

Together these tests are far quicker than long division. The rules pay off most when you stack several on one number to learn a lot in a single pass. Here is the whole toolkit:

Common mistakes