Welcome back to Divisibility Shortcuts! In the first lesson, you discovered that divisibility by 2, 5, and 10 comes down to a single glance at the last digit — a shortcut powered by the way base ten handles multiples of $10$. Now, in this second lesson of five, we move on to two new divisors that demand a different approach: 3 and 9. This time, the last digit alone will not be enough. Instead, you will learn a technique called the digit-sum test, a method that checks all the digits at once. By the end of this lesson, you will be able to test divisibility by 3 and 9 quickly, and you will understand the place-value reasoning that makes the trick work.

Consider the numbers $13$, $23$, $33$, $43$, $53$, $63$, $73$, , and . They all end in , yet only some of them are divisible by (specifically , , and ). The last digit is the same every time, but divisibility changes. This tells us that the last digit cannot be the deciding factor for .

The digit sum of a number is simply the total you get when you add each of its individual digits together. For example:

• The digit sum of $47$ is $4 + 7 = 11$.
• The digit sum of $583$ is $5 + 8 + 3 = 16$.

Here is the rule: a whole number is divisible by 3 if and only if the sum of its digits is divisible by 3.

The process has two stages. First, compute the digit sum. Then, check whether that sum is divisible by $3$. Let's walk through a few examples:

The rule for $9$ follows exactly the same pattern but with a higher bar: a whole number is divisible by 9 if and only if the sum of its digits is divisible by 9.

Let's revisit the same numbers and test for $9$:

Since $9$ is itself a multiple of $3$, the test for $9$ is simply a tighter version of the test for $3$. When you check a number against both rules, there are exactly three possible outcomes:

1. Divisible by both 3 and 9 — the digit sum is divisible by $9$ (which guarantees divisibility by $3$ too). Example: .

Understanding why a shortcut works keeps it from feeling like mere memorization, and once again the answer lives in place value. Consider a three-digit number written in expanded form:

We can rewrite each power of $10$ using the fact that and :

Let's put the full process together with a practical scenario. Suppose a warehouse receives a shipment of 7,938 items, and you need to know whether they can be packed evenly into crates of $3$ or crates of $9$ with nothing left over.

Step 1 — Compute the digit sum.

Here is a compact summary to keep alongside the last-digit rules from our first lesson:

In this lesson, you learned that divisibility by 3 and 9 can both be tested by computing the digit sum of a number. If that sum is divisible by $3$, the number is divisible by $3$; if the sum is divisible by $9$, the number is divisible by $9$ (and by $3$ as well). The reason this works is that each power of $10$ is just one more than a multiple of , so the leftover from every digit collects into the digit sum. These two rules now join the last-digit tests for , , and in your growing toolkit of divisibility shortcuts.