Divisibility
Rules
Master the shortcut tests that let you instantly determine whether one integer divides evenly into another: no long division required.
SSAT Upper Level: Quantitative
What Is Divisibility?
An integer \( a \) is divisible by another integer \( d \) (where \( d \neq 0 \)) if there exists an integer \( q \) such that:
For example, \( 36 \div 4 = 9 \) exactly, so 36 is divisible by 4. But \( 36 \div 5 = 7 \) remainder \( 1 \), so 36 is not divisible by 5.
Why Divisibility Rules Matter
★ Speed
Check divisibility in seconds without a calculator: critical on the timed SSAT.
★ Simplification
Quickly reduce fractions and find GCFs by identifying common factors.
★ Problem Solving
Many SSAT questions ask "which value is a factor of..." or "which expression is always divisible by..."
A Visual Way to Think About It
Rules for \( 2 \), \( 5 \), and \( 10 \)
Divisible by 2
The last digit is even: 0, 2, 4, 6, or 8.
Example: 1,374 ends in 4 → divisible.
Example: 1,374 ends in 4 → divisible.
Divisible by 5
The last digit is 0 or 5.
Example: 2,835 ends in 5 → divisible.
Example: 2,835 ends in 5 → divisible.
Divisible by 10
The last digit is 0.
Example: 5,290 ends in 0 → divisible.
Example: 5,290 ends in 0 → divisible.
Rules for \( 3 \) and \( 9 \)
Divisible by 3
The sum of all digits is divisible by 3.
Example: \( 528 \) → \( 5+2+8 = \) 15 → \( 15 \div 3 = 5 \) ✓
Example: \( 528 \) → \( 5+2+8 = \) 15 → \( 15 \div 3 = 5 \) ✓
Divisible by 9
The sum of all digits is divisible by 9.
Example: \( 738 \) → \( 7+3+8 = \) 18 → \( 18 \div 9 = 2 \) ✓
Example: \( 738 \) → \( 7+3+8 = \) 18 → \( 18 \div 9 = 2 \) ✓
Rules for \( 4 \) and \( 6 \)
Divisible by 4
The last two digits form a number divisible by 4.
Example: \( 1{,}316 \) → last two digits 16 → \( 16 \div 4 = 4 \) ✓
Example: \( 1{,}316 \) → last two digits 16 → \( 16 \div 4 = 4 \) ✓
Divisible by 6
The number is divisible by both 2 and 3.
Example: \( 114 \) → even ✓ and \( 1+1+4=6 \) ✓ → divisible by 6
Example: \( 114 \) → even ✓ and \( 1+1+4=6 \) ✓ → divisible by 6
Rules for \( 8 \) and \( 11 \)
Divisible by 8
The last three digits form a number divisible by 8.
Example: \( 5{,}120 \) → last three digits 120 → \( 120 \div 8 = 15 \) ✓
Example: \( 5{,}120 \) → last three digits 120 → \( 120 \div 8 = 15 \) ✓
Divisible by 11
Alternating sum of digits is divisible by 11 (including 0).
Example: \( 9{,}174 \) → \( 9 - 1 + 7 - 4 = \) 11 ✓
Example: \( 9{,}174 \) → \( 9 - 1 + 7 - 4 = \) 11 ✓
Digit-Sum vs. Last-Digit Tests
| Digit-Sum Rules | Last-Digit(s) Rules | |
|---|---|---|
| Divisors | \(3\), \(9\), \(11\) | \(2\), \(4\), \(5\), \(8\), \(10\) |
| What you check | Sum (or alternating sum) of all digits | Only the last 1, 2, or 3 digits |
| Why it works | \( 10 \equiv 1 \pmod{3} \) and \( 10 \equiv 1 \pmod{9} \) | \( 10^k \) is divisible by \( 2^k \) and \( 5^k \) |
| Speed | Fast: add digits | Fastest: glance at ending |
How to Apply Divisibility Rules
1
Identify the divisor. Which number are you testing divisibility by?
2
Pick the matching rule. Is it a last-digit rule (\(2\), \(4\), \(5\), \(8\), \(10\)) or a digit-sum rule (\(3\), \(9\), \(11\))? Or a combo (\(6\))?
3
Extract the digits you need. For last-digit rules, focus on the ending. For digit-sum rules, add all digits.
4
Test the small result. Is that smaller number divisible by the divisor? If yes, the original number is too.
5
For composite divisors (\(12\), \(15\), \(18\)...): Factor them and test each prime-power factor separately.
Example 1: Is \( 2{,}748 \) Divisible by 3, 4, and 6?
Example 2: Testing \( 53{,}064 \) for Divisibility by 8
1
Rule for 8: Check the last three digits of the number.
2
Extract: The last three digits of \( 53{,}064 \) are 064, which equals 64.
3
Test: \( 64 \div 8 = 8 \). That divides evenly.
4
Conclusion: ✓ \( 53{,}064 \) is divisible by 8.
Example 3: Is \( 7{,}293{,}516 \) Divisible by 11?
Testing Composite Divisors: \(12\), \(15\), \(18\)
For a composite divisor, factor it into coprime parts and test each separately. The number must pass all tests.
\(12\) = \( 3 \times 4 \)
Test divisibility by 3 (digit sum) and by 4 (last two digits).
\(15\) = \( 3 \times 5 \)
Test divisibility by 3 (digit sum) and by 5 (last digit is 0 or 5).
\(18\) = \( 2 \times 9 \)
Test divisibility by 2 (even) and by 9 (digit sum divisible by 9).
Common Mistakes to Avoid
✕ Mixing Up 3 and 9
Digit sum divisible by 3 does not guarantee divisibility by 9. Test: \( 123 \) → \( 1+2+3=6 \). Divisible by 3, but NOT by 9.
✕ Wrong Digits for 4 and 8
For 4, check last two digits. For 8, check last three. Students often check only the last digit for both.
✕ Alternating Sum Sign Errors
For 11: start with + from the leftmost digit and strictly alternate. Getting the signs wrong flips the result.
Complete Divisibility Rules Table
| Divisor | Rule | Quick Example |
|---|---|---|
| 2 | Last digit is even | 46: ends in 6 ✓ |
| 3 | Digit sum ÷ 3 | 123: \( 1+2+3 = 6 \) ✓ |
| 4 | Last 2 digits ÷ 4 | 732: \( 32\div4 = 8 \) ✓ |
| 5 | Last digit is 0 or 5 | 985: ends in 5 ✓ |
| 6 | Div. by 2 AND 3 | 312: even & \( 3+1+2=6 \) ✓ |
| 8 | Last 3 digits ÷ 8 | 1,024: \( 024\div8 = 3 \) ✓ |
| 9 | Digit sum ÷ 9 | 819: \( 8+1+9 = 18 \) ✓ |
| 10 | Last digit is 0 | 470: ends in 0 ✓ |
| 11 | Alternating sum ÷ 11 | 1,001: \( 1-0+0-1 = 0 \) ✓ |
Question 1 of 5
Which of the following numbers is divisible by both 3 and 4?
612: Digit sum = \( 6+1+2 = 9 \), divisible by 3 ✓. Last two digits = 12, and \( 12 \div 4 = 3 \), divisible by 4 ✓. So 612 is divisible by both. (Note: 726 is divisible by 3 but NOT by 4, since \( 26 \div 4 = 6.5 \).)
Question 2 of 5
What is the alternating sum of the digits of \( 8{,}294 \), and is it divisible by 11?
Starting from the left: \( 8 - 2 + 9 - 4 = 11 \). Since 11 is divisible by 11, 8,294 is divisible by 11.
Question 3 of 5
If \( n \) is a positive integer and \( 5n + 15 \) is divisible by 5, which of the following must also be divisible by 5?
Since \( 5n + 15 = 5(n + 3) \), it is always divisible by 5 for any integer \( n \): the divisibility tells us nothing special about \( n \). So we need an expression that is always divisible by 5. \( 10n + 30 = 10(n + 3) = 2 \times 5 \times (n + 3) \), which is always divisible by 5.
Question 4 of 5
A four-digit number has the form \( 3{,}4\_8 \). For the number to be divisible by 8, which digit could fill the blank?
For divisibility by 8, check the last three digits: \( 4\_8 \). Try each option: \( 408 \div 8 = 51 \) ✓ (exact). Testing others: \( 438 \div 8 = 54.75 \), \( 458 \div 8 = 57.25 \), \( 478 \div 8 = 59.75 \). Only 0 works.
Question 5 of 5
Which of the following is not divisible by 9?
Digit sums: \( 4{,}518 \to 4+5+1+8 = 18 \) ✓, \( 7{,}281 \to 7+2+8+1 = 18 \) ✓, \( 5{,}765 \to 5+7+6+5 = 23 \) (\( 23 \div 9 \neq \text{integer} \)) ✕, \( 3{,}609 \to 3+6+0+9 = 18 \) ✓. So 5,765 is NOT divisible by 9.