Divisibility tests: rules for 2, 3, 4, 5, 9, 10 and 25

Divisibility tests let us check whether a number is divisible by a given number without a remainder, without using long division. For 2, 5 and 10, the last digit is enough; for 4 and 25, the last two digits; and for 3 and 9, the sum of the digits.

Below you will find explanations of why each rule works, step-by-step examples and checks that help you avoid choosing the wrong rule.

First, a term: if the number $a$ is divisible by $b$ without a remainder, we call $a$ a multiple of $b$. For example, $18$ is a multiple of $3$ because $18 = 3 \cdot 6$. Divisibility tests quickly answer precisely this question: "Is this number a multiple of the given number?"

Divisibility tests for 2, 5 and 10

All three tests look only at the last digit.

Divisibility by 2: if the number ends in an even digit ($0, 2, 4, 6, 8$). For example, $358$ ends in $8$, so it is divisible by 2, whereas $471$ ends in an odd digit, so it is not.

Divisibility by 5: if the number ends in $0$ or $5$. The numbers $815$ and $2\,340$ are divisible, but $802$ is not.

Divisibility by 10: if the number ends in $0$. This is essentially a combination of the tests for 2 and 5, since $10 = 2 \cdot 5$.

Why is the last digit enough? Any natural number can be written in the form $10k + r$ (where $r$ is the last digit), and the $10k$ part is always divisible by 2, 5 and 10. So everything depends on $r$.

Divisibility tests for 3 and 9

Both of these tests use the sum of the digits.

Divisibility by 3: if the sum of the digits is divisible by 3, the number itself is also divisible by 3.

Divisibility by 9: if the sum of the digits is divisible by 9, the number is divisible by 9.

For example, for $714$ we have $7 + 1 + 4 = 12$. The number $12$ is divisible by 3, so $714$ is too: $714 = 3 \cdot 238$. But $12$ is not divisible by 9, so $714$ is not divisible by 9 either.

An important connection: every multiple of 9 is also a multiple of 3, but the converse is not always true.

Divisibility tests for 4 and 25

Now we use the last two digits, because $100$ is divisible by both 4 and 25 — the part of the number in the hundreds and higher places does not affect the test.

Divisibility by 4: if the number formed by the last two digits is divisible by 4. For example, the last two digits of $1\,936$ are $36$, and $36 = 4 \cdot 9$, so $1\,936$ is divisible by 4.

Divisibility by 25: if the number ends in $00$, $25$, $50$ or $75$. For example, $4\,725$ ends in $25$, so it is divisible by 25.

A number ending in $00$ (for example, $1\,300$) is divisible by 4, 25 and, of course, 100: $1\,300 = 4 \cdot 325 = 25 \cdot 52$.

Worked examples

Example 1. Determine which of the following numbers divide $4\,725$: 2, 3, 4, 5, 9, 10, 25.

Solution. The last digit, $5$, is odd, so the number is not divisible by 2 or 10, but it is divisible by 5. The sum of the digits is $4 + 7 + 2 + 5 = 18$; $18$ is divisible by 9, so the number is a multiple of both 3 and 9. Its last two digits are $25$, so it is divisible by 25, but because $25$ is not divisible by 4, the number is not divisible by 4. Check: $4\,725 = 9 \cdot 525 = 25 \cdot 189$. Answer: it is divisible by 3, 5, 9 and 25.

Example 2. Which of the following numbers is $2\,340$ a multiple of: 2, 3, 4, 5, 9, 10 and 25?

Solution. The number ends in $0$, so it is divisible by 2, 5 and 10. The sum of the digits is $2 + 3 + 4 + 0 = 9$, so it is also divisible by 3 and 9. Its last two digits are $40$; $40 = 4 \cdot 10$, so it is divisible by 4. But $40$ is not in the list $00, 25, 50, 75$, so the number is not divisible by 25. Check: $2\,340 = 4 \cdot 585 = 9 \cdot 260$. Answer: it is a multiple of all of them except 25.

Example 3. For a four-digit number of the form $\overline{73x2}$ to be a multiple of 9, what must the digit $x$ be?

Solution. The sum of the digits is $7 + 3 + x + 2 = 12 + x$. This sum must be divisible by 9. Since $x$ is a digit, the value of $12 + x$ ranges from $12$ to $21$, and the only number in this range divisible by 9 is $18$. Therefore, $x = 6$. Check: $7\,362 = 9 \cdot 818$. Answer: $x = 6$.

Example 4. How many natural numbers from $10$ to $100$, including both endpoints, are multiples of 25?

Solution. Such a number must end in $25$, $50$, $75$ or $00$. In this range, they are $25$, $50$, $75$ and $100$. Answer: $4$.

Independent practice

For the number $\overline{58x6}$ to be divisible by 9, what must the digit $x$ be? Short answer: $5+8+x+6=19+x$; the multiple of 9 in this range is $27$, so $x=8$.

Common mistakes

Using the digit sum to test divisibility by 4. The digit-sum test works only for 3 and 9. For example, the digits of $26$ have a sum of $8$, which is divisible by 4, but $26$ itself is not divisible by 4. For 4, always look at the last two digits.

Checking only the last digit for divisibility by 4. The number $14$ ends in $4$, but the division $14 : 4$ leaves a remainder. One digit is not enough — you need two.

Thinking "If a number is divisible by 3, it is also divisible by 9." The number $21$ is a multiple of 3 ($2 + 1 = 3$), but it is not divisible by 9. The reasoning works only in the opposite direction: a number divisible by 9 is also divisible by 3.

Testing divisibility by 25 only by checking whether the number ends in $5$. The number $135$ is divisible by 5, but its last two digits are $35$, so it is not a multiple of 25.

Summary and next step

• For 2, 5 and 10, check only the last digit.
• For 4 and 25, check the number formed by the last two digits.
• For 3 and 9, use the sum of the digits; do not apply this method to 4.
• Where possible, confirm the result using division or multiplication.
• Explore the connection between divisibility and common divisors or take topic-based tests.