Divisibility Rules — Quick Tests for 2, 3, 4, 5, 6, 8, 9, and 10

By MathPal Editorial Team·Published Apr 17, 2026

Grade: 6 | Topic: Arithmetic

What You Will Learn

A number is divisible by another number when the division leaves no remainder. In this lesson you will learn shortcut tests — called divisibility rules — for 2, 3, 4, 5, 6, 8, 9, and 10. These rules let you check divisibility in your head, which is extremely helpful when simplifying fractions, finding factors, and working with LCM and GCF.

Theory

Why Divisibility Rules Matter

Imagine you need to simplify $\frac{126}{198}$. Before you can divide numerator and denominator by a common factor, you first need to figure out which numbers divide evenly into 126 and 198. Divisibility rules give you a fast way to test candidates without long division.

The Rules

Divisor	Rule	Quick Check
2	The last digit is even (0, 2, 4, 6, or 8)	34 ends in 4 — divisible
3	The sum of the digits is divisible by 3	$5+1+6 = 12$ and $12 \div 3 = 4$ — yes
4	The last two digits form a number divisible by 4	732: $32 \div 4 = 8$ — yes
5	The last digit is 0 or 5	485 ends in 5 — divisible
6	The number is divisible by both 2 and 3	132 is even and $1+3+2=6$ (div by 3) — yes
8	The last three digits form a number divisible by 8	5120: $120 \div 8 = 15$ — yes
9	The sum of the digits is divisible by 9	$7+2+9 = 18$ and $18 \div 9 = 2$ — yes
10	The last digit is 0	450 ends in 0 — divisible

How the Rules for 3 and 9 Work

The rule for 3 works because of how our base-10 number system behaves. Consider a three-digit number with digits $a$, $b$, $c$:

$100a + 10b + c = 99a + 9b + (a + b + c)$

Since $99a + 9b$ is always divisible by 3, the entire number is divisible by 3 exactly when $a + b + c$ is divisible by 3. The same reasoning extends to 9 (since 99 and 9 are both divisible by 9).

Combining Rules — The Rule for 6

The rule for 6 is a combination: a number must pass both the test for 2 (even last digit) and the test for 3 (digit sum divisible by 3). This works because $6 = 2 \times 3$ and 2 and 3 share no common factor.

Worked Examples

Example 1: Is 345 divisible by 3?

Add the digits: $3 + 4 + 5 = 12$.

Check: $12 \div 3 = 4$ with no remainder.

Yes, 345 is divisible by 3. And indeed $345 \div 3 = 115$.

Example 2: Is 528 divisible by 6?

Step 1 — Test for 2: The last digit is 8, which is even. Passes.

Step 2 — Test for 3: Digit sum is $5 + 2 + 8 = 15$. Since $15 \div 3 = 5$, the digit sum is divisible by 3. Passes.

Since 528 passes both tests, yes, 528 is divisible by 6. Check: $528 \div 6 = 88$.

Example 3: Is 4,736 divisible by 8?

Look at the last three digits: 736.

Divide: $736 \div 8 = 92$ with no remainder.

Yes, 4,736 is divisible by 8.

Example 4: Is 2,853 divisible by 9?

Add the digits: $2 + 8 + 5 + 3 = 18$.

Check: $18 \div 9 = 2$. The digit sum is divisible by 9.

Yes, 2,853 is divisible by 9. And $2853 \div 9 = 317$.

Common Mistakes

Mistake 1: Confusing the rules for 3 and 9

❌ "The digit sum of 123 is 6. Since 6 is divisible by 3, the number is divisible by 9."

✅ For divisibility by 9, the digit sum must be divisible by 9, not just by 3. Here, $1 + 2 + 3 = 6$. Since 6 is not divisible by 9, the number 123 is not divisible by 9 (though it is divisible by 3).

Mistake 2: Checking only one condition for the rule of 6

❌ "135 has a digit sum of 9, which is divisible by 3, so 135 is divisible by 6."

✅ The rule for 6 requires divisibility by both 2 and 3. The number 135 is odd (last digit is 5), so it fails the test for 2. Therefore 135 is not divisible by 6.

Mistake 3: Using the wrong digits for the rule of 4

❌ "In 5,316 the first two digits 53 give $53 \div 4 = 13$ R 1, so not divisible by 4."

✅ The rule for 4 checks the last two digits, not the first two. The last two digits are 16, and $16 \div 4 = 4$. So 5,316 is divisible by 4.

Practice Problems

1. Is 7,290 divisible by 5?

Show Answer

Yes. The last digit is 0, so 7,290 is divisible by 5. ($7290 \div 5 = 1458$)

2. Is 462 divisible by 6?

Show Answer

Test for 2: last digit is 2 (even) — passes. Test for 3: $4 + 6 + 2 = 12$ and $12 \div 3 = 4$ — passes. Since both tests pass, yes, 462 is divisible by 6. ($462 \div 6 = 77$)

3. Is 1,572 divisible by 9?

Show Answer

Digit sum: $1 + 5 + 7 + 2 = 15$. Since $15 \div 9 = 1$ R 6, the digit sum is not divisible by 9. So no, 1,572 is not divisible by 9.

4. Is 3,816 divisible by 8?

Show Answer

Last three digits: 816. $816 \div 8 = 102$. Since there is no remainder, yes, 3,816 is divisible by 8.

5. Which of these numbers are divisible by 4: 212, 530, 748, 914?

Show Answer

Check the last two digits of each:

• 212: $12 \div 4 = 3$ — yes
• 530: $30 \div 4 = 7$ R 2 — no
• 748: $48 \div 4 = 12$ — yes
• 914: $14 \div 4 = 3$ R 2 — no

The numbers divisible by 4 are 212 and 748.

Summary

• Divisibility rules are mental shortcuts to test whether one number divides evenly into another.
• Rules for 2, 5, and 10 only look at the last digit.
• Rules for 3 and 9 use the digit sum.
• The rule for 4 checks the last two digits; the rule for 8 checks the last three digits.
• The rule for 6 combines the rules for 2 and 3 — both must pass.
• These tests are especially useful when simplifying fractions, finding GCF and LCM, and factoring.

Related Topics

• Fractions — Complete Guide — use divisibility rules to simplify fractions faster.
• LCM and GCF — divisibility rules help you identify common factors and multiples.
• Simplifying Fractions — apply divisibility tests to reduce fractions to lowest terms.

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