Young Fermats · Pre-Algebra · Multiples & Divisors · 6th Grade · NYC Math Class
Multiples & Divisors — 6th Grade Math Class with 25 Practice Questions, Divisibility Rules & Hidden Answers
A complete class-ready lesson on multiples and divisors (also called factors) for 6th grade math: the seven divisibility rules every middle-schooler should own, prime vs. composite numbers, an honest introduction to GCF and LCM, and 25 practice questions with click-to-reveal step-by-step answers. Built for the SOMATH Young Fermats Pre-Algebra program (grades 5–6) on the Upper West Side of Manhattan.
This is the class where two words that sound almost identical — multiple and divisor — finally get separated in a 6th grader’s head. It’s a small vocabulary fix with a huge downstream payoff: fractions, ratios, and pre-algebra all sit on top of it. If a student can look at 24 and instantly say “its divisors are 1, 2, 3, 4, 6, 8, 12, 24 and 24 is a multiple of each of them,” they are ready for equivalent fractions, simplifying, common denominators, and GCF/LCM word problems.
Every question below has a hidden button that reveals the answer and the reasoning, so a student can practice honestly and then check their thinking. Written by the same team that teaches the Young Fermats Pre-Algebra class at SOMATH, a math-focused school on the Upper West Side of NYC run by cofounder Marcelo Ambrozio (Northwestern) and cofounder Vivianne Wright (Harvard).
- Warm-up: read the “multiple vs. divisor” theory box aloud and drill 5 quick fact checks (8 min).
- Chant multiples of 3, 4, 6, 7, 8, 9 out loud (5 min).
- Teach the divisibility rules table with a specific number for each rule (12 min).
- Students attempt the 25 questions with the answers hidden (20 min).
- Reveal answers as a group and re-teach any rule that missed 3+ students (10 min).
What’s in this lesson
- Multiples vs. divisors — the vocabulary fix
- Finding every divisor of a number
- The seven divisibility rules for 6th grade
- Prime vs. composite — and why 1 is neither
- Prime factorization — the factor tree
- GCF and LCM the honest way
- 25 practice questions
- 5 challenge word problems for early 7th graders (Q26–30)
- Answer key summary
- About SOMATH & Young Fermats
1. Multiples vs. divisors — the vocabulary fix
Two words, two directions
The confusion between multiple and divisor is the single biggest reason 6th graders get GCF/LCM problems wrong. Fix the vocabulary and half the errors disappear.
A multiple of a number is what you get when you multiply it by a whole number. Multiples go up. The multiples of 6 are:
6, 12, 18, 24, 30, 36, 42, … (infinite — they never stop)
A divisor of a number is a whole number that divides it evenly, with no remainder. Divisors are the small numbers that fit inside a bigger number. The divisors of 24 are:
1, 2, 3, 4, 6, 8, 12, 24 (finite — there are only 8 of them)
The rule that ties them together: if a is a divisor of b, then b is a multiple of a. These are the same fact told from two directions.
- 6 is a divisor of 24 ↔ 24 is a multiple of 6
- 3 is a divisor of 30 ↔ 30 is a multiple of 3
- 7 is a divisor of 49 ↔ 49 is a multiple of 7
Divisor = factor. In 6th grade, the two words mean the same thing. Textbooks sometimes prefer “factor” when they’re talking about multiplication (4 × 5 = 20, so 4 is a factor of 20) and “divisor” when they’re talking about division (20 ÷ 4 = 5, so 4 is a divisor of 20). Same list, same numbers.
2. Finding every divisor of a number
The pairs method
To find every divisor of a number, list divisors in pairs that multiply to that number. Start at 1, walk up, and stop when the pair crosses itself.
Example: divisors of 36.
1 × 36 = 36 → pair: (1, 36) 2 × 18 = 36 → pair: (2, 18) 3 × 12 = 36 → pair: (3, 12) 4 × 9 = 36 → pair: (4, 9) 5 → not a divisor (36 ÷ 5 has a remainder) 6 × 6 = 36 → pair: (6, 6) ← STOP: pair crossedRead every number that appeared: 1, 2, 3, 4, 6, 9, 12, 18, 36. That’s 9 divisors. This is the same list you’d get by testing each number 1–36 individually, but the pairs method is much faster because you only go up to about √36 = 6.
Why it works. Every divisor d of n has a partner: n ÷ d is also a divisor. So divisors come in pairs. Once you pass √n, you’re just seeing the same pairs from the other side.
3. The seven divisibility rules for 6th grade
Fast tests that don’t require long division
Divisibility rules let a student see a divisor without dividing. These seven cover almost every 6th grade problem.
| Divisor | Rule | Example |
|---|---|---|
| 2 | Last digit is 0, 2, 4, 6, or 8 (even) | 1,472 ends in 2 → ÷ 2 |
| 3 | Sum of the digits is divisible by 3 | 2,145: 2+1+4+5 = 12 → ÷ 3 |
| 4 | The number formed by the last two digits is divisible by 4 | 3,724: last two = 24 → ÷ 4 |
| 5 | Last digit is 0 or 5 | 8,435 ends in 5 → ÷ 5 |
| 6 | Divisible by both 2 and 3 | 2,148: even AND digit sum 15 → ÷ 6 |
| 9 | Sum of the digits is divisible by 9 | 3,258: 3+2+5+8 = 18 → ÷ 9 |
| 10 | Last digit is 0 | 7,290 ends in 0 → ÷ 10 |
Bonus rule for 8: the last three digits form a number divisible by 8. Useful but not required at the 6th grade level.
Watch out. There is no clean divisibility rule for 7 that’s worth memorizing in 6th grade — just divide.
4. Prime vs. composite — and why 1 is neither
Counting the divisors
Every whole number bigger than 1 falls into one of two buckets, based on how many divisors it has.
- Prime: exactly two divisors, which must be 1 and the number itself. Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
- Composite: more than two divisors. Examples: 4 (divisors 1, 2, 4), 6 (divisors 1, 2, 3, 6), 9 (divisors 1, 3, 9).
The number 1 is special. It has only one divisor (itself), so it doesn’t fit either bucket. In 6th grade we simply say 1 is neither prime nor composite.
2 is the only even prime. Every other even number has 2 as a divisor besides 1 and itself, so every other even number is composite. This is a favorite test question.
The 10 primes under 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Every 6th grader should be able to write these from memory.
5. Prime factorization — the factor tree
Every composite number is a unique product of primes
Prime factorization means writing a number as a product of prime numbers only. Every composite number has one and only one prime factorization (this fact is called the Fundamental Theorem of Arithmetic — a big name for a 6th grade truth).
Example: factor tree for 60. Split 60 into any pair of divisors, keep splitting until every branch ends in a prime.
60 / \ 6 10 / \ / \ 2 3 2 5Reading the leaves left to right: 60 = 2 × 3 × 2 × 5 = 2² × 3 × 5. Written with exponents (Young Fermats style): 60 = 2² · 3 · 5.
Try starting differently — you get the same answer. Start 60 = 4 × 15, then 4 = 2×2 and 15 = 3×5. Same primes, same exponents. Order can look different, but the factorization is unique.
Why it matters. Prime factorization is how you find GCF and LCM the honest way when the numbers get too big to list all divisors and multiples.
6. GCF and LCM the honest way
Greatest Common Factor (GCF)
GCF is the largest divisor that two (or more) numbers share.
Method 1 (list divisors). Works for small numbers.
Divisors of 18: 1, 2, 3, 6, 9, 18 Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24 Common divisors: 1, 2, 3, 6 Greatest common divisor: 6 → GCF(18, 24) = 6Method 2 (prime factorization). Works for any size number.
18 = 2 · 3² 24 = 2³ · 3 GCF = product of primes to the LOWEST power that appears in BOTH = 2¹ · 3¹ = 6 → GCF(18, 24) = 6Least Common Multiple (LCM)
LCM is the smallest multiple that two (or more) numbers share.
Method 1 (list multiples).
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, … Multiples of 8: 8, 16, 24, 32, 40, 48, … Smallest common multiple: 24 → LCM(6, 8) = 24Method 2 (prime factorization).
6 = 2 · 3 8 = 2³ LCM = product of primes to the HIGHEST power that appears in EITHER = 2³ · 3¹ = 24 → LCM(6, 8) = 24Quick check: for any two positive whole numbers a and b, GCF(a, b) × LCM(a, b) = a × b. Test with 6 and 8: 2 × 24 = 48 = 6 × 8. ✓
- “How many small groups of the same size?” → GCF. (Splitting apart evenly.)
- “When will two events line up again?” → LCM. (Cycles catching up.)
25 practice questions — multiples, divisors, divisibility, primes, GCF & LCM
Answers are hidden by default. Click Show answer & solution to check.
Q1. Multiple check
Is 84 a multiple of 7?
Yes.
7 × 12 = 84, so 84 is a multiple of 7. (Equivalently, 7 is a divisor of 84.)
Q2. Divisor check
Is 6 a divisor of 90?
Yes.
Rule for 6: divisible by 2 AND by 3. 90 is even (last digit 0) ✓. Digit sum 9+0 = 9, divisible by 3 ✓. So 90 ÷ 6 = 15 with no remainder.
Q3. First five multiples
List the first five multiples of 9.
9, 18, 27, 36, 45.
Multiply 9 by 1, 2, 3, 4, 5. (Note: 0 × 9 = 0 is technically the very first multiple, but 6th grade textbooks usually start counting multiples at 1×.)
Q4. All divisors of 36
List every divisor of 36.
1, 2, 3, 4, 6, 9, 12, 18, 36 (9 divisors).
Pairs method: 1×36, 2×18, 3×12, 4×9, 6×6. Stop when the pair crosses itself at 6.
Q5. Divisibility by 3
Is 4,281 divisible by 3?
Yes.
Sum the digits: 4 + 2 + 8 + 1 = 15. 15 is divisible by 3 (5 × 3), so 4,281 is divisible by 3. Check: 4,281 ÷ 3 = 1,427.
Q6. Divisibility by 4
Is 5,318 divisible by 4?
No.
Rule for 4: the last two digits must form a number divisible by 4. Last two digits are 18, and 18 ÷ 4 = 4 remainder 2. So 5,318 is not divisible by 4.
Q7. Divisibility by 9
Is 60,732 divisible by 9?
Yes.
Digit sum: 6 + 0 + 7 + 3 + 2 = 18. 18 is divisible by 9, so 60,732 is divisible by 9. (Check: 60,732 ÷ 9 = 6,748.)
Q8. Divisibility by 6
Which of these are divisible by 6? 522, 1,050, 4,215, 8,124
522, 1,050, and 8,124 are divisible by 6. 4,215 is not.
Rule: divisible by 2 AND by 3.
• 522: even ✓, digit sum 9 ✓ → ÷ 6.
• 1,050: even ✓, digit sum 6 ✓ → ÷ 6.
• 4,215: odd (last digit 5) ✗ → not ÷ 6.
• 8,124: even ✓, digit sum 15 ✓ → ÷ 6.
Q9. Fill in the divisor
What single digit could go in the blank to make 3,4_5 divisible by 3?
0, 3, 6, or 9.
Digit sum without the blank: 3 + 4 + 5 = 12, which is already divisible by 3. So the blank digit itself must be divisible by 3 to keep the total divisible by 3. Single digits divisible by 3 are 0, 3, 6, and 9. Verify: 3,405 ÷ 3 = 1,135 ✓, 3,435 ÷ 3 = 1,145 ✓, 3,465 ÷ 3 = 1,155 ✓, 3,495 ÷ 3 = 1,165 ✓.
Q10. Prime or composite?
Classify each: 21, 29, 51, 57, 91.
29 is prime. 21, 51, 57, and 91 are composite.
• 21 = 3 × 7 → composite.
• 29: no divisor between 2 and √29 (≈ 5.4) — not divisible by 2, 3, or 5 → prime.
• 51 = 3 × 17 (digit sum 6) → composite.
• 57 = 3 × 19 (digit sum 12) → composite.
• 91 = 7 × 13 → composite. (This is the “trick” number — it looks prime because it’s not divisible by 2, 3, or 5, but 7 × 13 gets it.)
Q11. Primes under 20
List every prime number less than 20.
2, 3, 5, 7, 11, 13, 17, 19 (eight primes).
1 is not prime (only one divisor). All the other numbers under 20 have a divisor besides 1 and themselves: 4 (2), 6 (2, 3), 8 (2), 9 (3), 10 (2, 5), 12 (2, 3), 14 (2, 7), 15 (3, 5), 16 (2), 18 (2, 3).
Q12. Prime factorization of 84
Write 84 as a product of primes.
84 = 2² · 3 · 7.
Factor tree: 84 = 2 × 42 = 2 × 2 × 21 = 2 × 2 × 3 × 7. Group the twos: 2² · 3 · 7. Check: 4 · 3 · 7 = 84 ✓.
Q13. Prime factorization of 120
Write 120 as a product of primes.
120 = 2³ · 3 · 5.
120 = 2 × 60 = 2 × 2 × 30 = 2 × 2 × 2 × 15 = 2 × 2 × 2 × 3 × 5. Group the twos: 2³ · 3 · 5. Check: 8 · 3 · 5 = 120 ✓.
Q14. Count the divisors
How many divisors does 100 have? List them.
9 divisors: 1, 2, 4, 5, 10, 20, 25, 50, 100.
Pairs: 1×100, 2×50, 4×25, 5×20, 10×10. Stop when the pair repeats.
Q15. Common divisors
List the common divisors of 24 and 36.
1, 2, 3, 4, 6, 12.
Divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24. Divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. The divisors that appear in both lists are 1, 2, 3, 4, 6, 12. (Note: the largest one, 12, is the GCF.)
Q16. GCF by listing
Find GCF(18, 30).
GCF(18, 30) = 6.
Divisors of 18: 1, 2, 3, 6, 9, 18. Divisors of 30: 1, 2, 3, 5, 6, 10, 15, 30. Common: 1, 2, 3, 6. Greatest = 6.
Q17. GCF by prime factorization
Find GCF(48, 60) using prime factorization.
GCF(48, 60) = 12.
48 = 2⁴ · 3 60 = 2² · 3 · 5 GCF = 2² · 3 = 12 (lowest power of each shared prime)Q18. LCM by listing
Find LCM(4, 6).
LCM(4, 6) = 12.
Multiples of 4: 4, 8, 12, 16, 20, 24, … Multiples of 6: 6, 12, 18, 24, … First common one: 12.
Q19. LCM by prime factorization
Find LCM(12, 18).
LCM(12, 18) = 36.
12 = 2² · 3 18 = 2 · 3² LCM = 2² · 3² = 4 · 9 = 36 (highest power of each prime)Cross-check with GCF: GCF(12, 18) = 6. GCF × LCM = 6 × 36 = 216 = 12 × 18 ✓.
Q20. GCF word problem
A teacher has 24 pencils and 36 erasers. She wants to make identical goody bags with no leftovers. What is the greatest number of bags she can make, and what goes in each bag?
12 bags — each with 2 pencils and 3 erasers.
“Split into identical groups with no leftovers” is a GCF problem. GCF(24, 36) = 12. Each bag gets 24 ÷ 12 = 2 pencils and 36 ÷ 12 = 3 erasers.
Q21. LCM word problem
Bus A leaves the depot every 15 minutes. Bus B leaves every 20 minutes. They both just left at 8:00 AM. When is the next time they leave together?
9:00 AM (60 minutes after 8:00).
“When will two cycles line up again?” is an LCM problem. LCM(15, 20): 15 = 3 · 5, 20 = 2² · 5, LCM = 2² · 3 · 5 = 60 minutes. 8:00 + 60 min = 9:00 AM.
Q22. GCF used to simplify a fraction
Simplify 42 / 56 to lowest terms.
42 / 56 = 3 / 4.
GCF(42, 56): 42 = 2 · 3 · 7, 56 = 2³ · 7. GCF = 2 · 7 = 14. Divide top and bottom by 14: 42 ÷ 14 = 3, 56 ÷ 14 = 4. Answer: 3/4. This is why GCF matters — it’s how you simplify fractions in one step.
Q23. LCM used to add fractions
Add 1/6 + 1/8. Give the answer in lowest terms.
1/6 + 1/8 = 7/24.
Common denominator = LCM(6, 8) = 24. Rewrite: 1/6 = 4/24, 1/8 = 3/24. Add: 4/24 + 3/24 = 7/24. GCF(7, 24) = 1, so 7/24 is already in lowest terms.
Q24. Missing divisor
A number has exactly these divisors: 1, 2, 3, 4, 6, 8, 12, 24. What is the number?
24.
The largest divisor of a number is always the number itself.
Q25. Common multiples on a list
List the first three positive multiples that 4 and 5 have in common.
20, 40, 60.
LCM(4, 5) = 20. Every common multiple of 4 and 5 is a multiple of the LCM. So the common multiples are 20, 40, 60, 80, … The first three are 20, 40, 60.
5 challenge word problems for early 7th graders (Q26–30)
Same tools, longer problems. Great warmups for the SHSAT and for pre-algebra placement.
Q26. Bake sale bags
A bake sale has 72 brownies, 96 cookies, and 120 cupcakes. Each treat bag must have the same number of brownies, cookies, and cupcakes, with no treats left over. What is the greatest number of bags possible, and how many of each treat go in a bag?
24 bags — each with 3 brownies, 4 cookies, and 5 cupcakes.
GCF(72, 96, 120). Prime factor each: 72 = 2³ · 3², 96 = 2⁵ · 3, 120 = 2³ · 3 · 5. GCF = 2³ · 3 = 24. Divide each: 72÷24 = 3 brownies, 96÷24 = 4 cookies, 120÷24 = 5 cupcakes.
Q27. Blinking lights
A red light blinks every 8 seconds. A blue light blinks every 12 seconds. A green light blinks every 20 seconds. All three just blinked together. How many seconds until they all blink together again?
120 seconds (2 minutes).
LCM(8, 12, 20). Prime factor: 8 = 2³, 12 = 2² · 3, 20 = 2² · 5. LCM = 2³ · 3 · 5 = 120 seconds.
Q28. Rectangle tile problem
A rectangle measures 48 cm by 60 cm. What is the largest square tile (in cm) that can tile the rectangle exactly with no cutting?
12 cm by 12 cm tiles.
The tile’s side length must be a divisor of both 48 and 60. The largest such divisor is GCF(48, 60) = 12. Check: 48/12 = 4 tiles across, 60/12 = 5 tiles down, so 20 tiles total fill the rectangle.
Q29. Prime factorization from a clue
A whole number has prime factorization 2³ · 3 · 5². How many positive divisors does it have?
24 divisors.
Rule from 7th-grade pre-algebra: if n = pa · qb · rc, the number of positive divisors is (a+1)(b+1)(c+1). Here: (3+1)(1+1)(2+1) = 4 · 2 · 3 = 24. Check: the number is 8 · 3 · 25 = 600, and yes, 600 has 24 divisors.
Q30. Divisibility puzzle
What is the smallest positive whole number that is divisible by every number from 1 to 6?
60.
This is LCM(1, 2, 3, 4, 5, 6). Prime factor each: 1 = 1, 2 = 2, 3 = 3, 4 = 2², 5 = 5, 6 = 2 · 3. LCM = highest power of each prime = 2² · 3 · 5 = 4 · 3 · 5 = 60. Check: 60 ÷ 1, 2, 3, 4, 5, 6 all give whole numbers.
Answer key summary
| # | Answer | # | Answer |
|---|---|---|---|
| 1 | Yes (7×12=84) | 16 | GCF = 6 |
| 2 | Yes | 17 | GCF = 12 |
| 3 | 9, 18, 27, 36, 45 | 18 | LCM = 12 |
| 4 | 1, 2, 3, 4, 6, 9, 12, 18, 36 | 19 | LCM = 36 |
| 5 | Yes | 20 | 12 bags (2 pencils, 3 erasers) |
| 6 | No | 21 | 9:00 AM |
| 7 | Yes | 22 | 3/4 |
| 8 | 522, 1050, 8124 | 23 | 7/24 |
| 9 | 0, 3, 6, or 9 | 24 | 24 |
| 10 | 29 prime; rest composite | 25 | 20, 40, 60 |
| 11 | 2, 3, 5, 7, 11, 13, 17, 19 | 26 | 24 bags (3/4/5) |
| 12 | 2² · 3 · 7 | 27 | 120 seconds |
| 13 | 2³ · 3 · 5 | 28 | 12 cm tile |
| 14 | 9 divisors | 29 | 24 divisors |
| 15 | 1, 2, 3, 4, 6, 12 | 30 | 60 |
About SOMATH & Young Fermats
SOMATH — School of Math is a math-focused school on the Upper West Side of Manhattan. Young Fermats — Pre-Algebra is our program for grades 5–6 (typically ages 10–12). Classes are small groups taught by cofounder Marcelo Ambrozio (Northwestern) — the same teacher who wrote the lesson above.
Multiples, divisors, divisibility, primes, and GCF/LCM are the first big vocabulary the class installs. The next several classes then use these ideas to attack fractions, ratios, and equivalent forms — the topics that decide whether a 6th grader glides into Algebra 1 or gets stuck in middle school arithmetic.
Free evaluation. Every new SOMATH family gets a free 30-minute placement session — we look at the arithmetic, the vocabulary, and the working memory, and place the student in the right level. Book a free evaluation → or call (646) 668-6151.
School of Math (SOMATH) · 226 W 79th St, 1st Floor · New York, NY 10024 · (646) 668-6151 · hello@schoolofmath.us