Applying All Divisibility Rules

Applying All Divisibility Rules | CodeSignal Learn

Introduction

2

Your Complete Divisibility Toolkit

0, 2, 4, 6, 8

A Systematic Approach

2

Worked Examples: Running All Seven Tests

0

When a Simpler Rule Passes but a Stricter One Does Not

4

Real-World Application: Splitting a Delivery

\{2, 3, 4, 5, 6, 9, 10\}

Conclusion and Next Steps

2

Previous Lesson

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Congratulations — you have reached the final lesson of Divisibility Shortcuts! Over the previous four lessons you built a complete toolkit of quick tests: last-digit checks for $2$ , $5$ , and $10$ ; digit-sum checks for $3$ and $9$ ; a last-two-digits check for $4$ ; and a combined two-step check for $6$ . Each rule let you spot a specific divisor without performing long division. Now it is time to bring everything together and practice running the full set of tests on a single number, so you can quickly identify every divisor from $\{2, 3, 4, 5, 6, 9, 10\}$ that applies.

Think of all seven rules as tools on a belt. Each one answers a single yes-or-no question about a number, and each one takes only a few seconds. The real power appears when you use them all at once on the same number, because in just a minute or two you can learn a great deal about how that number can be split into equal groups.

Here is the full set for easy reference:

Divisor	Quick Test
2	Last digit is even ( $0, 2, 4, 6, 8$ )
3	Digit sum is divisible by $3$
4	Last two digits form a number divisible by $4$
5	Last digit is $0$ or $5$
6	Passes both the test for $2$ and the test for $3$
9	Digit sum is divisible by $9$
10	Last digit is $0$

Notice that several tests share the same starting step. The checks for $2$ , $5$ , and $10$ all look at the last digit, while $3$ , $9$ , and $6$ all use the digit sum. That overlap is good news: a little bit of work goes a long way.

When you need to test a number against all seven divisors, it helps to follow a consistent order rather than jumping around randomly. Here is one efficient sequence:

Look at the last digit. This single digit immediately tells you about $2$ , $5$ , and $10$ .
Look at the last two digits. This tells you about $4$ .
Add up all the digits. The digit sum tells you about $3$ and $9$ .
Combine results. If the number passed both the $2$ test and the $3$ test, it is also divisible by $6$ .

By grouping the tests this way, you avoid repeating work. You check the last digit once and get three answers; you compute the digit sum once and get two or three answers, since the $6$ result depends on the $3$ result plus the $2$ result you already have.

Flowchart showing the four-step systematic approach for running all seven divisibility tests in order

Let's apply the full scan to the number 2,340.

Step 1 — Last digit. The last digit is $0$ . Because $0$ is even, the number is divisible by $2$ . Because the last digit is $0$ , it also qualifies for $5$ and $10$ . Three checks done in a single glance.

Step 2 — Last two digits. The last two digits form $40$ . Since $40 \div 4 = 10$ with no remainder, the number is divisible by $4$ .

Step 3 — Digit sum. Adding the digits gives $2 + 3 + 4 + 0 = 9$ . Since $9$ is divisible by both $3$ and $9$ , the number passes both tests.

Step 4 — Combine. The number passed the tests for $2$ and $3$ , so it is also divisible by $6$ .

Result: $2{,}340$ is divisible by all seven divisors — $2, 3, 4, 5, 6, 9,$ and $10$ . That is a remarkably clean number for splitting into equal groups!

Now let's try a number that is less cooperative: 1,578.

Step 1 — Last digit is $8$ . The digit $8$ is even, so the number is divisible by $2$ . It is neither $0$ nor $5$ , so the number is not divisible by $5$ or $10$ .

Step 2 — Last two digits: $78$ . Dividing gives $78 \div 4 = 19$ remainder $2$ , so the number is not divisible by $4$ .

Step 3 — Digit sum: $1 + 5 + 7 + 8 = 21$ . Since $21 \div 3 = 7$ , the number is divisible by $3$ . However, $21 \div 9 = 2$ remainder $3$ , so it is not divisible by $9$ .

Step 4 — Combine. The number passes both the $2$ test and the $3$ test, so it is also divisible by $6$ .

Result: $1{,}578$ is divisible by $2, 3,$ and $6$ only. Compared with $2{,}340$ , far fewer group sizes work here — a useful thing to know before you start trying to divide.

As you work through these tests, you will notice that some rules are related but not equivalent. Two pairs are especially important to keep straight.

Divisibility by 2 versus 4. Every number divisible by $4$ is automatically divisible by $2$ , because $4 = 2 \times 2$ . However, the reverse is not true. Consider $38$ : the last digit is $8$ (even), so it passes the test for $2$ , but $38 \div 4 = 9$ remainder $2$ , so it fails the test for $4$ . Being even means the number has at least one factor of $2$ ; divisibility by $4$ requires two factors of $2$ .

Divisibility by 3 versus 9. Similarly, every number divisible by $9$ is automatically divisible by $3$ , since $9 = 3 \times 3$ . But a number can pass the $3$ test and still fail the $9$ test. Take $51$ : the digit sum is $5 + 1 = 6$ , which is divisible by $3$ but not by $9$ . Divisibility by $3$ needs one factor of $3$ ; divisibility by $9$ needs two.

In short, the stricter rule always includes the simpler rule, but passing the simpler rule does not guarantee the stricter one. Keep this asymmetry in mind whenever you see a number that is even but not divisible by $4$ , or whose digit sum is a multiple of $3$ but not of $9$ .

Nested oval diagrams showing that multiples of 4 are a subset of multiples of 2, and multiples of 9 are a subset of multiples of 3

Imagine you manage a small shop and a shipment of 504 items arrives. You want to know which group sizes from $\{2, 3, 4, 5, 6, 9, 10\}$ would split the delivery into perfectly equal bundles with nothing left over.

Last digit: $4$ . Even → divisible by $2$ . Not $0$ or $5$ → not divisible by $5$ or $10$ .
Last two digits: $04$ (i.e., $4$ ). $4 \div 4 = 1$ → divisible by $4$ .
Digit sum: $5 + 0 + 4 = 9$ . Divisible by $3$ and by $9$ .
Combine: Divisible by both $2$ and $3$ → divisible by $6$ .

So $504$ is divisible by $2, 3, 4, 6,$ and $9$ . You could split that shipment into bundles of $2$ ( $252$ bundles), $3$ ( $168$ ), $4$ ( $126$ ), $6$ ( $84$ ), or $9$ ( $56$ ). Bundles of $5$ or $10$ would leave items left over — and you figured all of that out in under a minute, with no calculator needed.

Illustration of a shop stockroom showing 504 boxes being split into valid equal-sized bundles for divisors 2, 3, 4, 6, and 9, with leftover boxes shown for 5 and 10

In this lesson you learned how to run all seven divisibility tests on a single number in one efficient pass. By checking the last digit first (for $2$ , $5$ , $10$ ), then the last two digits (for $4$ ), then the digit sum (for $3$ and $9$ ), and finally combining the $2$ and $3$ results (for $6$ ), you can quickly identify every applicable divisor without performing a single long division. You also saw that passing a simpler rule — like divisibility by $2$ — does not automatically guarantee its stricter relative, divisibility by $4$ , and the same goes for $3$ and $9$ .

Now it is time to put the full toolkit to work! The practice exercises ahead will have you sorting stock quantities, filling in divisibility tables, and analyzing real delivery invoices — so you can prove to yourself just how fast and reliable these shortcuts really are.