Overview

Divisibility rules are shortcuts that let you test whether one number divides another without performing long division. On the AMC 8, these rules save time and often appear inside larger problems involving factors, multiples, prime factorization, GCD/LCM, and modular arithmetic.

Many harder problems combine several divisibility rules at once. A number divisible by $12$, for example, must satisfy both the rules for $3$ and $4$. Recognizing these patterns quickly can be a huge advantage in competitive math.

Key Ideas

• By 2: last digit is even.
• By 3: digit sum is divisible by 3.
• By 4: last two digits form a number divisible by 4.
• By 5: last digit is 0 or 5.
• By 6: divisible by both 2 and 3.
• By 7: difference between units digit doubled and remaining number is divisible by 7.
• By 8: last three digits form a number divisible by 8.
• By 9: digit sum is divisible by 9.
• By 10: last digit is 0.
• By 11: alternating digit sum is divisible by 11.

Remember: divisibility rules can be combined.

Core Skills

Applying Single Rules

Quickly check each rule without writing out division.

Example: Is $7{,}248$ divisible by $4$? Look only at the last two digits: 48. Since $48$ is divisible by $4$, the entire number is divisible by $4$.

Combining Rules

To check divisibility by 12, verify both 3 and 4. To check 15, verify both 3 and 5.

Example: Is $2{,}430$ divisible by $6$? A number is divisible by $6$ if it is divisible by both $2$ and $3$. Last digit is $0$, so it is divisible by $2$. Digit sum: 2+4+3+0=9. Since $9$ is divisible by $3$, the number is divisible by $6$.

Digit Sums/Differences

For divisibility rules such as $7$, $8$, and $11$, the sum/difference of digits may be equal to $0$. If this is the case, the rule is satisfied, as $0$ in this case is a multiple of the number.

Example: Is $189$ divisible by $7$? A number is divisible by $7$ if the difference between the units digit doubled and the rest of the number is divisible by $7$. Last digit is $9$, the rest of the number is $18$, and doubling the last digit is $18$, $18 - 18 = 0$, so it is divisible by $7$.

The Alternating Sum for 11

For 918082: $9 - 1 + 8 - 0 + 8 - 2 = 22$, which is divisible by 11.

Example: For $918082$, 9−1+8−0+8−2=22. Since $22$ is divisible by $11$, the original number is divisible by $11$. This rule is especially useful in AMC 8 digit problems.

Worked Example

Is $4{,}536$ divisible by 8?

Last three digits: 536. $536 \div 8 = 67$. Yes.

More Examples

Example 1: Divisible by 3

Is $5{,}742$ divisible by $3$?

Compute the digit sum: 5+7+4+2=18. Since $18$ is divisible by $3$, the number is divisible by $3$.

Example 2: Divisible by 9

Is $81{,}234$ divisible by $9$? Digit sum: 8+1+2+3+4=18. Since $18$ is divisible by $9$, the number is divisible by $9$.

Example 3: Divisible by 12

Is $3{,}456$ divisible by $12$? Check divisibility by both $3$ and $4$. Digit sum: 3+4+5+6=18, so the number is divisible by $3$. Last two digits: 56, which is divisible by $4$. Therefore, the number is divisible by $12$.

Example 4: Finding a Missing Digit

Find the digit $x$ such that $27x4$ is divisible by $9$. For divisibility by $9$, the digit sum must be divisible by $9$: 2+7+x+4=13+x. The next multiple of $9$ after $13$ is $18$, so 13+x=18. Thus, x=5.

Example 5: Divisibility by 11

Is $53{,}768$ divisible by $11$? Compute the alternating sum: 5−3+7−6+8=11. Since $11$ is divisible by $11$, the number is divisible by $11$.

Example 6: AMC 8 Style Logic

A number is divisible by both $2$ and $9$. Which of the following must also be true?

• divisible by $3$
• divisible by $6$
• divisible by $18$

Since divisibility by $9$ automatically implies divisibility by $3$, and the number is also divisible by $2$, it must be divisible by $6$. However, divisibility by $18$ requires divisibility by both $2$ and $9$, which are already given. Thus the number must be divisible by all three.

Example 7: Divisibility by 7

Is $9,023$ divisible by $7$?

Double the last digit ($3$) and subtract it from the remaining number ($902$):

$902 - 6 = 896$

Apply the rule again, doubling the last digit ($6$) and subtracting it from the remaining number ($89$):

$89 - 12 = 77$

$77$ is a multiple of $7$ ($7 * 11$), so $9,023$ is also a multiple of $7$.

Strategy Checklist

• Memorize the common divisibility rules.
• Use digit sums for $3$ and $9$ problems.
• Use only the last few digits for $4$ and $8$ problems.
• Combine rules carefully for composite numbers.
• For divisibility by $11$, alternate adding and subtracting digits.
• Look for shortcuts before attempting long division.

Common Mistakes

• Forgetting that divisibility by $6$ requires both $2$ and $3$.
• Using the last two digits for divisibility by $8$ instead of the last three.
• Adding digits incorrectly when testing divisibility by $3$ or $9$.
• Forgetting to alternate signs in the rule for $11$.
• Assuming divisibility by two numbers always implies divisibility by their product.

Practice Problems