Two street lights blink at different speeds, a teacher splits a pile of supplies into identical kits, a band lines up in equal rows. Behind each of these situations hide factors and multiples. In this chapter you will learn to spot prime and composite numbers, test divisibility in seconds, and find the greatest common factor (GCF) and the least common multiple (LCM) to solve real word problems.
1. Factors and multiples
A factor of a whole number \(n\) is a whole number that divides \(n\) with no remainder. A multiple of \(n\) is the product of \(n\) and any whole number \(1, 2, 3, \dots\)
Since \(4 \times 6 = 24\), the numbers \(4\) and \(6\) are factors of \(24\), and \(24\) is a multiple of both \(4\) and \(6\). A number always has a finite list of factors, but its list of multiples never ends.
The picture shows why the factors of \(12\) come in pairs: \(12\) squares can be arranged as \(1 \times 12\), \(2 \times 6\) or \(3 \times 4\). So the factors of \(12\) are \(1, 2, 3, 4, 6\) and \(12\).
- Start with \(1 \times n\).
- Test \(2, 3, 4, \dots\) in order. When a number divides \(n\), write the pair.
- Stop when the second number of a pair is smaller than the first one (or equal to it).
- Write all the numbers of the pairs from smallest to largest.
Pairs: \(1 \times 36\), \(2 \times 18\), \(3 \times 12\), \(4 \times 9\), \(6 \times 6\). We stop at \(6 \times 6\). The factors of \(36\) are \(1, 2, 3, 4, 6, 9, 12, 18, 36\).
A factor is never larger than the number itself, and a multiple is never smaller than the number. Do not mix up the two words: the factors of \(10\) are \(1, 2, 5, 10\), but its multiples are \(10, 20, 30, \dots\)
2. Prime and composite numbers
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.
The prime numbers below \(50\) are \(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\). Notice that \(2\) is the only even prime number: every other even number has \(2\) as an extra factor.
To test whether a number is prime, divide it by the primes \(2, 3, 5, 7, \dots\) in order. You can stop as soon as the square of the prime you are testing is bigger than your number. For \(97\), testing \(2, 3, 5, 7\) is enough because \(11 \times 11 = 121\) is already too big.
Some composite numbers look prime. \(51\) seems like it should be prime, but \(51 = 3 \times 17\). Always test the small primes before you decide.
3. Divisibility rules
A number is divisible by another when the division leaves no remainder. These rules let you check it without dividing.
| Divisible by | Rule |
|---|---|
| \(2\) | The last digit is \(0, 2, 4, 6\) or \(8\). |
| \(3\) | The sum of the digits is divisible by \(3\). |
| \(4\) | The number formed by the last two digits is divisible by \(4\). |
| \(5\) | The last digit is \(0\) or \(5\). |
| \(6\) | The number is divisible by both \(2\) and \(3\). |
| \(9\) | The sum of the digits is divisible by \(9\). |
| \(10\) | The last digit is \(0\). |
Last digit \(2\): even, so divisible by \(2\). Digit sum: \(4 + 5 + 7 + 2 = 18\), which is divisible by \(3\) and by \(9\). Last two digits: \(72 = 4 \times 18\), so divisible by \(4\). Being divisible by \(2\) and \(3\), it is divisible by \(6\) too. It does not end in \(0\) or \(5\), so it is not divisible by \(5\) or \(10\). Indeed, \(4{,}572 = 36 \times 127\).
4. Prime factorization
Every composite number can be written as a product of prime numbers, and this product is unique (apart from the order). A factor tree helps you find it: split the number into two factors, then keep splitting until every branch ends on a prime.
Here \(60 = 6 \times 10 = (2 \times 3) \times (2 \times 5)\), so \(60 = 2 \times 2 \times 3 \times 5\). With exponents, we write \(60 = 2^2 \times 3 \times 5\). Whatever first split you choose (\(4 \times 15\), \(3 \times 20\), …), you always end with the same primes.
5. Greatest common factor (GCF)
The GCF of two numbers is the largest number that is a factor of both.
The factors of \(12\) are \(1, 2, 3, 4, 6, 12\) and the factors of \(18\) are \(1, 2, 3, 6, 9, 18\). The overlap of the diagram holds the common factors, and the largest is \(6\). So \(\text{GCF}(12, 18) = 6\).
- Write the prime factorization of each number.
- Keep only the primes that appear in both, each with its smaller exponent.
- Multiply them.
\(48 = 2^4 \times 3\) and \(72 = 2^3 \times 3^2\). The smaller exponents are \(2^3\) and \(3^1\), so \(\text{GCF}(48, 72) = 2^3 \times 3 = 24\).
6. Least common multiple (LCM)
The LCM of two numbers is the smallest positive number that is a multiple of both.
The multiples of \(4\) are \(4, 8, 12, 16, \dots\) and the multiples of \(6\) are \(6, 12, 18, \dots\) The first number on both lines is \(12\), so \(\text{LCM}(4, 6) = 12\). The next common multiples, \(24\) and \(36\), are multiples of the LCM.
- Write the prime factorization of each number.
- Keep every prime that appears in either number, each with its larger exponent.
- Multiply them.
\(12 = 2^2 \times 3\) and \(18 = 2 \times 3^2\). The larger exponents are \(2^2\) and \(3^2\), so \(\text{LCM}(12, 18) = 4 \times 9 = 36\).
For two numbers \(a\) and \(b\): \(\text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b\). For \(12\) and \(18\): \(6 \times 36 = 216 = 12 \times 18\).
7. The distributive property with the GCF
The distributive property says \(a(b + c) = ab + ac\). Read from right to left, it lets you factor out the GCF of a sum.
\(\text{GCF}(36, 48) = 12\). Since \(36 = 12 \times 3\) and \(48 = 12 \times 4\), we get \(36 + 48 = 12(3 + 4) = 12 \times 7 = 84\).
The numbers left inside the parentheses, \(3\) and \(4\), have no common factor other than \(1\). If they did, the number you factored out was not the greatest one.
Do not forget the \(1\) when a term equals the common factor. For example, \(15 + 45 = 15(1 + 3)\), not \(15(3)\).
8. Word problems with the GCF and the LCM
Ask yourself one question to choose the right tool.
- Are you splitting things into equal groups, equal pieces or equal rows, as large as possible? Use the GCF.
- Are two events repeating, and you want to know when they happen together again, or the smallest amount that fits both patterns? Use the LCM.
A coach has \(24\) apples and \(36\) oranges and wants to make identical fruit bags with nothing left over. The most bags is \(\text{GCF}(24, 36) = 12\). Each bag holds \(24 \div 12 = 2\) apples and \(36 \div 12 = 3\) oranges.
Two buses leave the station together at 9:00 a.m. One leaves every \(8\) minutes, the other every \(12\) minutes. They leave together again after \(\text{LCM}(8, 12) = 24\) minutes, at 9:24 a.m.
Key takeaways
- A factor divides a number exactly; a multiple is the number times \(1, 2, 3, \dots\)
- A prime number has exactly two factors; \(1\) is neither prime nor composite; \(2\) is the only even prime.
- Use the divisibility rules for \(2, 3, 4, 5, 6, 9\) and \(10\) before dividing.
- Prime factorization is unique: \(60 = 2^2 \times 3 \times 5\).
- GCF: common primes, smaller exponents. LCM: all primes, larger exponents.
- \(\text{GCF}(a, b) \times \text{LCM}(a, b) = a \times b\).
- Factor out the GCF to rewrite a sum: \(36 + 48 = 12(3 + 4)\).
- Splitting into equal groups means GCF; repeating events means LCM.
Test yourself: quick challenge for Grade 6
Speed drill for Grade 6: how many in 60 seconds?
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