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SHSAT Math • Number Properties and Integers

Divisibility Rules — Use divisibility rules to evaluate a number.

Learn quick shortcuts to tell if a number divides evenly — no long division needed.

Explore the lesson 12 min read
SHSAT illustration with a school, math function, and reading book
Inside this lesson 10 sections 5 questions 9 figures

SECTION 1

Historical Context & Motivation

Have you ever looked at a big number like 4,572 and wondered, "Can I split this evenly into groups of 3?" People have been asking questions like that for thousands of years. Long before calculators existed, mathematicians invented clever shortcuts called divisibility rules. These rules let you check whether one number divides into another without doing the full division.

~300 BCE • Euclid's Elements

The Greek mathematician Euclid wrote about even and odd numbers. He showed that every whole number is either divisible by 2 or not — one of the earliest divisibility ideas.

~500 CE • Indian Mathematicians

Scholars like Aryabhata developed the Hindu-Arabic numeral system (the digits 0–9 we use today). This place-value system made digit-based divisibility tricks possible.

1202 • Fibonacci Spreads the System

Italian mathematician Fibonacci published Liber Abaci, bringing Hindu-Arabic numerals to Europe. Merchants quickly adopted divisibility shortcuts for trading and accounting.

1801 • Gauss & Modular Arithmetic

Carl Friedrich Gauss published Disquisitiones Arithmeticae, which explained the math behind remainders. His work gave divisibility rules a formal mathematical foundation.

Today, divisibility rules are still super useful — especially on timed tests like the SHSAT. They help you simplify fractions, find factors, and solve problems faster. The big question is: how can you tell if a number is divisible by 2, 3, 4, 5, 6, 8, 9, or 10 just by looking at its digits?

SECTION 2

Core Principles & Definitions

Before we dive into the rules, let's make sure we understand the key vocabulary. When we say a number is divisible by another number, we mean it divides evenly with no remainder. For example, 12 is divisible by 3 because 12 ÷ 3 = 4 exactly. But 13 is not divisible by 3 because 13 ÷ 3 = 4 with a remainder of 1.

1

Divisibility

A number a is divisible by b if a ÷ b gives a whole number with zero remainder.
2

Factor

A factor is a number that divides evenly into another. If 15 ÷ 5 = 3, then 5 is a factor of 15.
3

Multiple

A multiple is the result of multiplying a number by a whole number. 15 is a multiple of 5 because 5 × 3 = 15.
4

Remainder

The remainder is what's left over after dividing. In 17 ÷ 5 = 3 R 2, the remainder is 2.
✦

KEY TAKEAWAY

Think of divisibility like splitting a bag of candy evenly among friends. If you have 12 pieces and 3 friends, everyone gets exactly 4 pieces — no leftovers. That means 12 is divisible by 3. If you had 13 pieces, one piece would be left over, so 13 is not divisible by 3. Divisibility rules are shortcuts that let you figure this out without actually dividing.

SECTION 3

Visual Explanation — The Divisibility Rules Chart

The diagram below shows the most important divisibility rules you need for the SHSAT. Each rule tells you exactly which digits to look at and what to check. Study this chart — it's your cheat sheet!

Divisibility Rules at a Glance÷ 2Last digit is even(0, 2, 4, 6, or 8)÷ 3Sum of all digitsis divisible by 3÷ 4Last two digitsare divisible by 4÷ 5Last digit is0 or 5÷ 6Divisible by BOTH2 AND 3÷ 8Last three digitsare divisible by 8÷ 9Sum of all digitsis divisible by 9÷ 10Last digit is0÷ 7 (Bonus)Double last digit,subtract from restQuick Memory TipRules for 2, 5, and 10 → check the LAST digit onlyRules for 3 and 9 → ADD all the digits together
This chart groups all eight divisibility rules by color. Notice that the rules for 2, 5, and 10 only require you to look at the last digit, while the rules for 3 and 9 require you to add all the digits together.

Look at how the rules fall into two groups. The first group (for 2, 5, and 10) only cares about the last digit. The second group (for 3 and 9) asks you to add up all the digits. The rules for 4, 6, and 8 combine ideas from these two groups. Knowing which group a rule belongs to makes it much easier to remember.

SECTION 4

Mathematical Framework — Why the Rules Work

You might wonder: why does adding digits tell you about divisibility by 3? It's not magic — it's math! Let's see the reasoning behind the most-tested rules.

Divisibility by 2

RULE FOR 2

Anumberisdivisibleby2ifitslastdigitis0,2,4,6,or8.A number is divisible by 2 if its last digit is 0, 2, 4, 6, or 8.

Why? Every number can be written as (some amount × 10) + last digit. Since 10 is always divisible by 2, we only need to check the last digit.

Divisibility by 3

RULE FOR 3

Anumberisdivisibleby3ifthesumofitsdigitsisdivisibleby3.A number is divisible by 3 if the sum of its digits is divisible by 3.

Example: For 531, add 5 + 3 + 1 = 9. Since 9 ÷ 3 = 3 (no remainder), 531 is divisible by 3.

Divisibility by 9

RULE FOR 9

Anumberisdivisibleby9ifthesumofitsdigitsisdivisibleby9.A number is divisible by 9 if the sum of its digits is divisible by 9.

This works for the same reason as the rule for 3, but with a stricter requirement. For 531, the digit sum is 9, and 9 ÷ 9 = 1, so 531 is also divisible by 9.

Divisibility by 4

RULE FOR 4

Anumberisdivisibleby4ifitslasttwodigitsformanumberdivisibleby4.A number is divisible by 4 if its last two digits form a number divisible by 4.

Why two digits? Because 100 is divisible by 4, so any hundreds, thousands, etc. are already taken care of. Only the last two digits matter. Example: 916 → last two digits are 16. Since 16 ÷ 4 = 4, the number 916 is divisible by 4.

💡

SHSAT Tip

The rule for 6 is a combo rule: a number is divisible by 6 only if it passes both the rule for 2 (last digit is even) and the rule for 3 (digit sum divisible by 3). If it fails either one, it's not divisible by 6.

SECTION 5

Detailed Breakdown — Testing the Number 2,340

Let's put all the rules to work on a single number: 2,340. The diagram below walks through every divisibility check. This is the kind of systematic thinking that saves time on the SHSAT.

Divisibility Check: 2,340Digits: 2, 3, 4, 0 | Digit Sum: 2+3+4+0 = 9÷ 2 → Last digit = 0 (even)✓ YES — divisible by 2÷ 3 → Digit sum = 9✓ YES — 9 ÷ 3 = 3÷ 4 → Last two digits = 40✓ YES — 40 ÷ 4 = 10÷ 5 → Last digit = 0✓ YES — ends in 0÷ 6 → Div by 2 ✓ and 3 ✓✓ YES — passes both tests÷ 8 → Last 3 digits = 340✗ NO — 340 ÷ 8 = 42.5÷ 9 → Digit sum = 9✓ YES — 9 ÷ 9 = 1÷ 10 → Last digit = 0✓ YES — ends in 0Result for 2,340Divisible by: 2, 3, 4, 5, 6, 9, 10 — Not divisible by: 8On the SHSAT, you only need to apply the specific rule the question asks about.
This flowchart shows every divisibility test applied to 2,340. Notice how the rule for 6 depends on the results for 2 and 3. The only test 2,340 fails is divisibility by 8.
DivisorWhat to CheckResult for 2,340
2Last digit is even0 is even → ✓
3Digit sum divisible by 39 ÷ 3 = 3 → ✓
4Last two digits divisible by 440 ÷ 4 = 10 → ✓
5Last digit is 0 or 5Ends in 0 → ✓
6Divisible by both 2 and 3Both pass → ✓
8Last three digits divisible by 8340 ÷ 8 = 42.5 → ✗
9Digit sum divisible by 99 ÷ 9 = 1 → ✓
10Last digit is 0Ends in 0 → ✓

Divisibility results for 2,340

SECTION 6

Worked Example — SHSAT-Style Problem

Let's walk through a problem like one you might see on the actual SHSAT. Pay attention to each step — the method matters as much as the answer.

Which of the following numbers is divisible by both 4 and 9: 1,224 or 1,242?

  1. 1

    Step 1 — Check 1,224 for divisibility by 4

    Look at the last two digits of 1,224. They form the number 24. Is 24 divisible by 4? Yes, because 24 ÷ 4 = 6.
    1,224 passes the test for 4 ✓
  2. 2

    Step 2 — Check 1,224 for divisibility by 9

    Add all the digits: 1 + 2 + 2 + 4 = 9. Is 9 divisible by 9? Yes, because 9 ÷ 9 = 1.
    1,224 passes the test for 9 ✓
  3. 3

    Step 3 — Conclusion for 1,224

    Since 1,224 is divisible by both 4 and 9, we already have our answer. But let's check 1,242 anyway for practice.
    1,224 is divisible by both 4 and 9 ✓
  4. 4

    Step 4 — Check 1,242 for divisibility by 4

    Look at the last two digits: 42. Is 42 divisible by 4? Let's check: 42 ÷ 4 = 10.5. That's not a whole number.
    1,242 fails the test for 4 ✗
  5. 5

    Step 5 — Final Answer

    Since 1,242 is NOT divisible by 4, it can't be divisible by both 4 and 9. The answer is 1,224.
    Answer: 1,224
💡

Time-Saving Strategy

On the SHSAT, if a question asks about divisibility by two numbers, start with the rule that's faster to check. The rule for 4 (look at last two digits) is usually quicker than the rule for 9 (add all digits). If a number fails the first test, you can skip the second test entirely!

SECTION 7

Strengths & Limitations of Divisibility Rules

Divisibility rules are powerful, but they have limits. Knowing when they work well — and when they don't — will help you use them wisely on test day.

StrengthsLimitations
No calculator needed — just look at digits or add them.No simple rule exists for 7 that's easy to remember.
Very fast for 2, 5, and 10 (just check the last digit).For very large numbers, the rule for 8 (check last 3 digits) can still require mental division.
The rule for 6 combines two simpler rules, making it easy to apply.Rules don't directly tell you the quotient — only whether division is even.
Helps you simplify fractions quickly by finding common factors.Doesn't work well for prime divisors larger than 10 (like 11 or 13).

Strengths vs. Limitations of Divisibility Rules

✦

KEY TAKEAWAY

Think of divisibility rules like a set of filters at a water plant. Each filter catches something different. The "÷ 2" filter catches odd numbers, the "÷ 5" filter catches numbers not ending in 0 or 5, and so on. If a number passes through all the filters you need, you know it's divisible. These filters are fast and reliable for divisors 2 through 10, which covers most SHSAT problems.

SECTION 8

Connection to Advanced Topics

Divisibility rules are your first step into a bigger world of number theory. Here's how the ideas you've learned connect to topics you'll see in high school and beyond.

What You Know NowWhere It Leads
Divisibility by 2, 3, 4, etc.Prime Factorization — breaking a number into its smallest factor building blocks.
Checking if remainder is 0Modular Arithmetic — a system that focuses entirely on remainders. Used in computer science and cryptography.
Finding common factors of two numbersGCF and LCM — Greatest Common Factor and Least Common Multiple, essential for simplifying fractions and solving equations.
Digit-sum rules for 3 and 9Casting Out Nines — an old technique for checking arithmetic, based on the same digit-sum idea.

From divisibility rules to advanced math

On the SHSAT specifically, divisibility rules often appear in questions about factors, multiples, and simplifying fractions. Mastering these rules now gives you a strong foundation for GCF/LCM problems, which are also tested. The same logical thinking — breaking a problem into smaller checks — will help you in algebra and beyond.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A student says that 345 is divisible by 6 because it is divisible by 3. Explain why the student is wrong.
Show answer
To be divisible by 6, a number must be divisible by both 2 and 3. Let's check: the digit sum of 345 is 3 + 4 + 5 = 12, and 12 ÷ 3 = 4, so yes, 345 is divisible by 3. However, the last digit is 5, which is odd. Since 345 is not divisible by 2, it fails the test for 6. The student forgot that the rule for 6 requires passing both conditions, not just one.

Question 2: BASIC CALCULATION

Is the number 7,236 divisible by 9? Show your work using the divisibility rule.
Show answer
To check divisibility by 9, add all the digits: 7 + 2 + 3 + 6 = 18. Now check if 18 is divisible by 9: 18 ÷ 9 = 2, which is a whole number. Since the digit sum (18) is divisible by 9, the number 7,236 is divisible by 9. You can verify: 7,236 ÷ 9 = 804 exactly.

Question 3: INTERMEDIATE

Which of these numbers is divisible by both 4 and 5: 820, 1,340, or 1,455?
Show answer
First, check divisibility by 5 (last digit must be 0 or 5): 820 ends in 0 ✓, 1,340 ends in 0 ✓, 1,455 ends in 5 ✓. All three pass. Now check divisibility by 4 (last two digits divisible by 4): 820 → 20 ÷ 4 = 5 ✓, 1,340 → 40 ÷ 4 = 10 ✓, 1,455 → 55 ÷ 4 = 13.75 ✗. So 820 and 1,340 are divisible by both 4 and 5. Notice that 1,455 fails the test for 4 even though it passes for 5. Both conditions must be met.

Question 4: APPLIED

A school has 1,368 students. The principal wants to divide them into equal teams for a field day. Can the students be divided into teams of 6 with nobody left out? What about teams of 8?
Show answer
For teams of 6, check divisibility by both 2 and 3. Last digit of 1,368 is 8 (even), so ÷ 2 ✓. Digit sum: 1 + 3 + 6 + 8 = 18, and 18 ÷ 3 = 6, so ÷ 3 ✓. Since both pass, yes, teams of 6 work (1,368 ÷ 6 = 228 teams). For teams of 8, check the last three digits: 368. Is 368 ÷ 8 a whole number? 368 ÷ 8 = 46, which is whole. So yes, teams of 8 also work (1,368 ÷ 8 = 171 teams). Both arrangements are possible with no students left over.

Question 5: CRITICAL THINKING

The number 5,2□4 has a missing digit (shown as □). What digit could go in the blank to make the number divisible by 9? Is there more than one answer?
Show answer
For divisibility by 9, the digit sum must be divisible by 9. The known digits give us 5 + 2 + 4 = 11. If we call the missing digit d, then we need 11 + d to be divisible by 9. The multiples of 9 near 11 are 9 (too small, since d must be 0–9 and 11 + 0 = 11) and 18. Setting 11 + d = 18 gives d = 7. The next multiple is 27, so 11 + d = 27 gives d = 16, which is too big (digits are 0–9). Therefore, the only answer is d = 7, making the number 5,274. Check: 5 + 2 + 7 + 4 = 18, and 18 ÷ 9 = 2 ✓. There is exactly one solution.

SUMMARY

Lesson Summary

Summary

Divisibility rules are shortcuts that let you check whether a number divides evenly by 2, 3, 4, 5, 6, 8, 9, or 10 — without doing long division. For 2, 5, and 10, you only need to look at the last digit. For 3 and 9, you add all the digits and check whether that sum is divisible by 3 or 9. For 4 and 8, check the last two or three digits. The rule for 6 is a combo — the number must pass both the test for 2 and the test for 3.

On the SHSAT, these rules save you valuable time. Use them to simplify fractions, find factors, and eliminate wrong answer choices quickly. Remember: a number that is divisible by a product (like 6 = 2 × 3) must be divisible by each of its factors. Master these rules, and you'll handle number-properties questions with confidence.

Varsity Tutors • SHSAT Math • Divisibility Rules — Use divisibility rules to evaluate a number.