Welcome to Divisibility Shortcuts, the second course in our learning path! In the previous course, Foundations of Factors and Multiples, we built a solid understanding of factors, multiples, and prime versus composite numbers. Now we begin putting that knowledge to work with quick tests that tell us whether one number divides evenly into another — no long division required.

This is the first lesson of five in the course, and our focus today is on divisibility by 2, 5, and 10. By the end of this lesson, you will be able to check divisibility by any of these three numbers just by looking at a single digit. Even better, you will understand why that shortcut works.

As you may recall from the previous course, when we say a number is divisible by another number, we mean the division comes out exact with no remainder. For example, $30$ is divisible by $5$ because $30 \div 5 = 6$ exactly. In factor language, $5$ is a factor of $30$, and is a multiple of .

A whole number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8. These are the even digits. If the last digit is odd (1, 3, 5, 7, or 9), the number is not divisible by 2.

Let's try a few examples:

Notice that the size of the number does not matter. Whether you have a three-digit number or a ten-digit number, only the last digit decides divisibility by 2.

A whole number is divisible by 5 if its last digit is 0 or 5. That's it — just two possible endings.

Think about counting by fives: $5, 10, 15, 20, 25, 30, \ldots$ Every multiple of $5$ ends in either $0$ or $5$, and this pattern never breaks. So a number like is divisible by (last digit is ), while is not (last digit is ).

A whole number is divisible by 10 if its last digit is 0. This is the easiest rule to remember because our entire counting system is based on groups of ten.

Consider $580$. Its last digit is $0$, so it is divisible by $10$. What about $585$? The last digit is $5$, so it is not divisible by $10$, even though it is divisible by .

So far, the three rules may feel like facts to memorize. But there is a satisfying reason they work, and it lives in place value. In base ten, any whole number can be split into two parts: everything except the last digit (which forms a multiple of $10$) and the last digit itself. For instance:

The first part, $37 \times 10$, is always a multiple of . Since is divisible by , by , and by , that first part is automatically divisible by all three. So the entire number's divisibility comes down to just the contributed by the last digit. Let's spell this out for :

Before you head into practice, here is a compact table worth keeping in mind:

Notice the overlap one more time: divisibility by $10$ requires the last digit to satisfy both the rule for $2$ and the rule for $5$. The only digit that is both even and equal to $0$ or $5$ is itself.

In this lesson, you learned that divisibility by 2, 5, and 10 can each be tested by examining just the last digit of a number. You also explored why this works: in base ten, every digit except the last contributes a multiple of $10$, so only the final digit determines whether the number is divisible by $2$, $5$, or $10$. These three rules are fast, reliable, and will serve as building blocks for the more advanced tests coming later in this course.

Up next, you will put these shortcuts into action with a set of hands-on practice tasks — from checking stockroom quantities and sorting numbers into overlapping categories to explaining the place-value reasoning in your own words. Let's see how quickly you can spot those last digits!