Lesson 4 of 15, 10 min read
Divisibility rules, prime numbers and prime factorization
Factors and multiples, quick divisibility rules for 2, 3, 4, 5, 6, 9 and 10, prime and composite numbers, and how to find the prime factorization of a number with a factor tree.
Factors and multiples
\(6 \cdot 7 = 42\), so \(6\) and \(7\) are factors of \(42\), and \(42\) is a multiple of both \(6\) and \(7\). Another way to say it: \(42\) is divisible by \(6\), because \(42 \div 6\) leaves no remainder.
- The multiples of a number go on forever: the multiples of \(8\) are \(8, 16, 24, 32, \dots\)
- The factors of a number are limited. To find them all, look for factor pairs, starting from \(1\) and stopping when the pairs start repeating.
Divisibility rules
You can often tell whether a number is divisible by another without doing the division.
| A number is divisible by | if | Example |
|---|---|---|
| \(2\) | its last digit is even | \(5{,}318\) |
| \(3\) | the sum of its digits is divisible by \(3\) | \(822\): \(8 + 2 + 2 = 12\) |
| \(4\) | its last two digits form a number divisible by \(4\) | \(7{,}324\): \(24\) |
| \(5\) | its last digit is \(0\) or \(5\) | \(6{,}095\) |
| \(6\) | it's divisible by both \(2\) and \(3\) | \(822\) |
| \(9\) | the sum of its digits is divisible by \(9\) | \(4{,}617\): \(4 + 6 + 1 + 7 = 18\) |
| \(10\) | its last digit is \(0\) | \(3{,}450\) |
Prime and composite numbers
The primes less than \(40\) are
Two facts trip people up:
- \(1\) is neither prime nor composite. It has only one factor.
- \(2\) is the only even prime. Every other even number has \(2\) as a factor.
To test whether a number is prime, try dividing by \(2, 3, 5, 7, 11, \dots\) You can stop once the prime you're testing, squared, is bigger than the number. For example, \(89\) isn't divisible by \(2\), \(3\), \(5\) or \(7\), and \(11^2 = 121 > 89\), so \(89\) is prime.
Prime factorization
Every composite number can be written as a product of primes in exactly one way, apart from the order. That product is its prime factorization.
A factor tree is a neat way to find it. Split the number into any two factors, then keep splitting until every branch ends in a prime.
Prime factorizations make the next lesson, on GCF and LCM, much faster.
Practice
Quick check
Which number is prime?
\(51 = 3 \cdot 17\), \(57 = 3 \cdot 19\) and \(91 = 7 \cdot 13\). Only \(59\) has no factors other than \(1\) and itself.
Quick check
What is the prime factorization of \(60\)?
\(60 = 4 \cdot 15 = 2^2 \cdot 3 \cdot 5\). The options \(6 \cdot 10\) and \(2^2 \cdot 15\) are correct products but use composite factors.
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