Factors divide into a number; multiples are built from it. Learn to find every factor, break any number into primes, count factors, and spot common factors and multiples.
Why can 24 chairs be set out in rows of 2, 3, 4, 6, 8 or 12, but 23 chairs only in one long row? Why do two buses that leave together come back together only every so often? The answers are factors and multiples. A factor divides into a number. A multiple is built from a number by multiplying it up. These two ideas look like opposites, but they are one relationship seen from two ends. This guide teaches both in full, including prime numbers, prime factorisation, how to count the factors of any number, and the ideas of common factors and common multiples that lead to the HCF and LCM.
A factor of a number is a whole number that divides into it exactly, with no remainder. For example, 4 is a factor of 12 because 12 ÷ 4 = 3 exactly. 5 is not a factor of 12, because 12 ÷ 5 leaves a remainder of 2.
The statement “4 is a factor of 12” can be said in several ways, all of which mean the same thing: 4 divides 12 exactly; 12 is divisible by 4; 12 = 4 × 3, so 4 is one of the numbers you multiply to make 12. Our Divisibility Rules article gives quick tests for whether one number divides another. Factors are the complete list of every number that passes the test.
Factors come in pairs, because every multiplication fact has two numbers. 12 = 1 × 12 = 2 × 6 = 3 × 4, so the factors of 12 are 1, 2, 3, 4, 6 and 12.
The most reliable way is the pair method. Work through 1, 2, 3, … in order. Whenever a number divides exactly, write down both it and its partner. Stop when the numbers you test meet the partners you have found.
1 × 24, 2 × 12, 3 × 8, 4 × 6. Next test 5: 24 ÷ 5 leaves a remainder, so 5 is not a factor. The next test would be 6, but 6 has already appeared as a partner of 4, so we are finished. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24 (8 factors).
If n = a × b with a ≤ b, then a cannot be bigger than √n. (If both were bigger than √n, their product would be bigger than n.) So you only need to test whole numbers up to √n. Every factor larger than √n is the partner of one you have already found.
(a) Factors of 36. √36 = 6, so test 1 to 6: 1 × 36, 2 × 18, 3 × 12, 4 × 9, 6 × 6. Factors: 1, 2, 3, 4, 6, 9, 12, 18, 36 (9 factors). Note that 6 × 6 gives the single factor 6.
(b) Is 97 prime? √97 is about 9.8, so test 2, 3, 5 and 7 (testing the primes up to 9 is enough). 97 is odd, its digit sum 16 is not a multiple of 3, it does not end in 0 or 5, and 7 × 13 = 91 and 7 × 14 = 98 so 7 does not divide it. No test works, so 97 is prime.
(c) Is 221 prime? √221 is about 14.9. Test the primes up to 14: 2, 3, 5, 7 and 11 all fail, but 13 divides it: 13 × 17 = 221. So 221 is not prime. Numbers like 91 (7 × 13) and 221 look prime but are not.
The list is missing two members.
The fix: 1 and the number itself are always factors. 18 = 1 × 18, 2 × 9, 3 × 6, so the full list is 1, 2, 3, 6, 9, 18. A good habit is to write the pairs first. If the pairs are written down, you cannot forget 1 and 18.
1 is a factor of every number, and every number is a factor of itself. These two facts are always true. So the smallest number of factors a number can have is one (the number 1 itself) and every other number has at least two.
These quick tests (from Divisibility Rules) save time when you search for factors:
| Divisible by | Test | Example |
|---|---|---|
| 2 | Last digit is even | 7 346 ✓ |
| 3 | Digit sum is divisible by 3 | 1 248: 1 + 2 + 4 + 8 = 15 ✓ |
| 4 | Last two digits divisible by 4 | 3 512: 12 ÷ 4 = 3 ✓ |
| 5 | Last digit is 0 or 5 | 8 735 ✓ |
| 6 | Divisible by both 2 and 3 | 138: even, 1 + 3 + 8 = 12 ✓ |
| 8 | Last three digits divisible by 8 | 5 128: 128 ÷ 8 = 16 ✓ |
| 9 | Digit sum is divisible by 9 | 4 761: 4 + 7 + 6 + 1 = 18 ✓ |
| 10 | Last digit is 0 | 4 570 ✓ |
| 11 | Alternating digit sum is 0 or a multiple of 11 | 2 728: 2 − 7 + 2 − 8 = −11 ✓ |
A prime number has exactly two factors: 1 and itself. A composite number has more than two factors. The number 1 is neither prime nor composite, because it has only one factor. The number 2 is prime, and it is the only even prime, because every other even number is also divisible by 2.
The prime numbers up to 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47. There are 15 of them. An easy way to list primes is the Sieve of Eratosthenes, named after an ancient Greek mathematician: write the numbers, cross out 1, then cross out the multiples of 2, 3, 5 and 7 (but not 2, 3, 5 and 7 themselves). You only need primes up to 7, because 7² = 49 is the last square under 50 and 11² = 121 is already too big. What remains is prime.
This mistake is very common, because 1 has no “other” factors.
The fix: a prime needs exactly two factors, 1 and itself. The number 1 has only one factor, itself. So it fails the test. There is also a deeper reason: if 1 were prime, then 6 = 2 × 3 = 1 × 2 × 3 = 1 × 1 × 2 × 3 and so on, and prime factorisation would no longer be unique (section 6).
Every factor of a number is exactly one of three kinds: the number 1, a prime factor, or a composite factor. A prime factor is a factor that is itself a prime number. A composite factor is a factor that is itself composite.
The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. The prime factors are 2, 3 and 5. The composite factors are 6, 10, 15 and 30. The number 1 is in neither group.
This mixes up “prime factors” (a small subset) with “all factors”.
The fix: only some factors are prime. 6 is a factor of 30, but 6 = 2 × 3, so it is composite. A number can have many factors but only a few prime ones: 30 has eight factors and only three of them (2, 3, 5) are prime.
To write a number as a product of prime factors means to break it down until only primes remain, all multiplied together. There are two usual methods.
Method 1: the factor tree. Split the number into any two factors. Keep splitting every composite number until every branch ends in a prime.
Method 2: repeated division. Divide by the smallest prime that goes in, then again, until you reach 1.
360 ÷ 2 = 180, 180 ÷ 2 = 90, 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, 5 ÷ 5 = 1. The primes used are 2, 2, 2, 3, 3, 5, so 360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5. The small raised numbers are indices (powers); see Exponents & Powers.
(a) 36 = 4 × 9 = (2 × 2) × (3 × 3) = 2² × 3².
(b) 100 = 2 × 50 = 2 × 2 × 25 = 2 × 2 × 5 × 5 = 2² × 5².
(c) 90 = 2 × 45 = 2 × 3 × 15 = 2 × 3 × 3 × 5 = 2 × 3² × 5.
(d) 1 001 = 7 × 143 = 7 × 11 × 13 = 7 × 11 × 13. A number that looks hard can break into three small primes.
No matter which method or which starting split you choose, you always end with the same primes (the order may differ). For example, 36 = 4 × 9 and 36 = 2 × 18 = 2 × 2 × 9 = 2 × 2 × 3 × 3 both give 2² × 3². This is the Fundamental Theorem of Arithmetic: every whole number greater than 1 is either prime or can be written as a product of primes in exactly one way (ignoring the order). A version of the key ideas appears in Euclid’s Elements, written about 2,300 years ago. The first complete, clear statement and proof is usually credited to the mathematician Carl Friedrich Gauss in 1801.
4 and 9 are factors, but neither is a prime number.
The fix: “prime factorisation” means every number in the product must be prime. 4 = 2 × 2 and 9 = 3 × 3, so 36 = 2 × 2 × 3 × 3 = 2² × 3². After you finish, check each number in your answer: is it prime? Then multiply back to confirm: 2 × 2 × 3 × 3 = 36 ✓.
You could list all the factors and count them, but the prime factorisation gives a shortcut. If n = pa × qb × rc (with p, q, r different primes), then the number of factors of n is (a + 1)(b + 1)(c + 1). The reason: a factor of n can use the prime p zero, one, … or a times (a + 1 choices), the prime q anywhere from zero to b times (b + 1 choices), and so on. Every combination of choices gives a different factor.
| n | Prime factorisation | Number of factors |
|---|---|---|
| 24 | 2³ × 3 | (3 + 1)(1 + 1) = 8 |
| 36 | 2² × 3² | (2 + 1)(2 + 1) = 9 |
| 100 | 2² × 5² | (2 + 1)(2 + 1) = 9 |
| 360 | 2³ × 3² × 5 | (3 + 1)(2 + 1)(1 + 1) = 24 |
| 720 | 2⁴ × 3² × 5 | (4 + 1)(2 + 1)(1 + 1) = 30 |
Perfect squares have an odd number of factors. Most numbers have an even number of factors because they pair up (4 and 6 for 24). In a perfect square one pair is a number with itself, such as 6 × 6 = 36, and that pair contributes only one factor. That is why 36 has 9 factors. In the prime factorisation of a perfect square, every index is even, so every (index + 1) is odd, and a product of odd numbers is odd. Another consequence: a number with exactly three factors is always the square of a prime (4, 9, 25, 49, …).
A school has 100 lockers, all closed. Student 1 opens every locker. Student 2 changes every 2nd locker (closes it if open, opens it if closed). Student 3 changes every 3rd locker, and so on up to student 100. Which lockers are open at the end?
Locker n is changed once by each student whose number is a factor of n. So it is changed as many times as n has factors. A locker ends up open only if it is changed an odd number of times, so it needs an odd number of factors. These are the perfect squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, which is 10 lockers.
Type any whole number from 1 to 2000, or tap one of the examples. You get its factor pairs, its prime and composite factors, its prime factorisation and a check on the number of factors.
A multiple of a number is what you get by multiplying it by a whole number: 1, 2, 3, 4, … The first multiple of a number is the number itself. (Zero is a multiple of every number too, but we normally start counting from the number itself.)
| Number | First 10 multiples |
|---|---|
| 4 | 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 |
| 6 | 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 |
| 7 | 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 |
Multiples are found by skip-counting, as in the Times Tables Trainer. Unlike factors, which form a short, complete list, multiples never end: you can always multiply by a bigger number. To test whether a number is a multiple of another, divide: 91 ÷ 7 = 13 exactly, so 91 is a multiple of 7 (7 × 13 = 91).
The two ideas are linked. “4 is a factor of 12” and “12 is a multiple of 4” say the same thing. Every factor of a number is less than or equal to it. Every (positive) multiple of a number is greater than or equal to it.
| Factors | Multiples | |
|---|---|---|
| Direction | Divide into the number | Built by multiplying the number up |
| How many? | A finite list | Infinitely many |
| Size compared to the number | Less than or equal to it | Greater than or equal to it |
| Smallest | 1 | The number itself |
| Largest | The number itself | None (they never end) |
This confuses a multiple (8 is built from 4) with a factor (which must divide into 4).
The fix: 8 is a multiple of 4 because 4 × 2 = 8. A factor of 4 must divide into 4 exactly, so the factors of 4 are only 1, 2 and 4. A quick test: factors are small (never bigger than the number); multiples are big (never smaller than the number).
A common multiple of two numbers is a number that is a multiple of both. The smallest one is the lowest common multiple (LCM).
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, …
Multiples of 6: 6, 12, 18, 24, 30, 36, …
Common multiples: 12, 24, 36, 48, 60, … The lowest is 12. Every common multiple is a multiple of the LCM.
Three buses leave a station together. After that, bus A leaves every 4 minutes, bus B every 6 minutes and bus C every 10 minutes. When do all three next leave together?
We need the LCM of 4, 6 and 10. The multiples of 10 are 10, 20, 30, 40, 50, 60. The first one that 4 and 6 also divide is 60 (60 ÷ 4 = 15 and 60 ÷ 6 = 10). All three leave together again after 60 minutes.
Another everyday case is gears: a gear with 12 teeth meshing with a gear with 18 teeth returns to its starting position after LCM(12, 18) = 36 teeth have passed, which is 3 turns of the small gear and 2 turns of the large gear. The methods for larger numbers, and word problems, are in HCF & LCM and HCF & LCM Word Problems.
A common factor of two numbers divides both of them. The biggest one is the highest common factor (HCF). List the factors of each number and look for the overlap.
A sheet of card 24 cm by 36 cm is cut into identical squares with no waste. What is the largest possible square, and how many squares are there?
The side must divide both 24 and 36, so it is a common factor. The largest is the HCF, 12 cm. Then there are (24 ÷ 12) × (36 ÷ 12) = 2 × 3 = 6 squares.
Type two whole numbers from 1 to 200. See their multiples, their factors, the lowest common multiple and the highest common factor.
Tap the correct answer for each question. Your score is tracked as you go, and your best score is saved on this device.
1. 1 × 20, 2 × 10, 4 × 5: the factors are 1, 2, 4, 5, 10, 20.
2. 84 = 4 × 21 = 2 × 2 × 3 × 7 = 2² × 3 × 7.
3. Multiples of 10 are also multiples of 5, so the common multiples are 10, 20, 30, 40.
1. 1 × 28, 2 × 14, 4 × 7: the factors are 1, 2, 4, 7, 14, 28.
2. 90 = 2 × 45 = 2 × 3 × 15 = 2 × 3 × 3 × 5 = 2 × 3² × 5.
3. 72 = 2³ × 3². The number of factors is (3 + 1)(2 + 1) = 4 × 3 = 12.
4. (a) 23, 29, 31, 37. (b) 91 = 7 × 13, so it has factors other than 1 and 91 and is not prime.
5. Multiples of 12: 12, 24, 36. Multiples of 18: 18, 36. LCM = 36. Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 18: 1, 2, 3, 6, 9, 18. HCF = 6.
6. Multiples of 9 between 50 and 80: 54, 63, 72. 54 = 2 × 3³ has 2 × 4 = 8 factors. 63 = 3² × 7 has 3 × 2 = 6 factors. 72 has 12 factors. The number is 63.
7. 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 (15 factors). The number is odd because 144 is a perfect square (12 × 12): the pair 12 × 12 gives only one factor, so the other 14 factors pair up and 12 is left over.
Factors become meaningful when learners build them: give each learner 24 counters or squares and ask them to make every possible rectangle (Figure 1). They discover the pairs for themselves, and discover that a prime number of counters allows only one rectangle. A hundred square on which learners cross out multiples is a good activity for the Sieve of Eratosthenes. Factor trees are most useful when learners see that different starting splits give the same primes. Encourage them to check an answer by multiplying back. The locker puzzle in section 7 is a good enrichment problem that rewards understanding of why squares have an odd number of factors. Keep the language of factors and multiples apart by pairing the words with direction: factors go into a number, multiples go up from it.
This article is original EDUSAMBAM writing for students up to O-Level and similar courses. Exact examination requirements differ between boards, so compare your practice with the specification and past papers of your own board.
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