Finding Common Ground · Lesson 2 of 7
Prime Factorisation and All the Factors
“Build a number from primes and use its prime pieces to reveal its factors.”
• Identify prime and composite numbers and use division to factorise. • Explain why different factorisation routes reach the same prime factors. • Construct factors by taking subproducts of prime factors. • Test a general claim with an appropriate counterexample.
Break a number into prime pieces
Listing every factor becomes cumbersome when numbers grow. Prime factors give a compact description from which we can rebuild those factors. A prime is greater than 1 and has exactly two positive factors, 1 and itself; a composite number has more. The Sieve of Eratosthenes finds primes in a range by repeatedly crossing out multiples of each uncrossed prime, starting with 2. The number 1 is not prime.
Writing a positive integer greater than 1 as a product containing only prime numbers. A prime number has itself as its complete prime factorisation.
For example, 90 can begin as 3 × 30 or 2 × 45. Continue breaking composite factors: either route ends at 2 × 3 × 3 × 5, possibly in a different order. In a division layout, divide repeatedly by a prime, write each quotient on the next row, and stop when the final quotient is prime or 1. This is especially convenient for numbers such as 1200.
Problem
Find the prime factorisation of 105.
- 1.Divide 105 by 3 to obtain 35.
- 2.Divide 35 by 5 to obtain 7, which is prime.
- 3.Read the prime divisors and final prime: 105 = 3 × 5 × 7.
Problem
Show that 90 = 3 × 30 and 90 = 2 × 45 lead to the same prime pieces.
- 1.First route: 3 × 30 = 3 × 2 × 15 = 3 × 2 × 3 × 5.
- 2.Second route: 2 × 45 = 2 × 5 × 9 = 2 × 5 × 3 × 3.
- 3.Both contain one 2, two 3s and one 5; only their order differs.
Make factors from subproducts
If 840 = 2 × 2 × 2 × 3 × 5 × 7, the subproduct 2 × 2 × 7 = 28 is a factor: the unused primes make 2 × 3 × 5 = 30, so 840 = 28 × 30. A proposed factor needing three 3s cannot be assembled because 840 has only one 3. The empty selection corresponds to the factor 1.
Problem
Use 225 = 3 × 3 × 5 × 5 to list its positive factors.
- 1.Choose zero, one or two copies of 3 and zero, one or two copies of 5.
- 2.Form each distinct product: 1, 3, 9; 5, 15, 45; 25, 75, 225.
- 3.In increasing order the factors are 1, 3, 5, 9, 15, 25, 45, 75, 225.
A conjecture is a claim suggested by a pattern but not yet established for every case. Anshu suggests that a larger number must have a longer prime factorisation. Compare 96 = 2 × 2 × 2 × 2 × 2 × 3 with 121 = 11 × 11: 121 is larger yet has only two prime factors when repetitions are counted. This one counterexample disproves the claim.
In 225 = 3 × 3 × 5 × 5, each repeated copy matters. A factor such as 75 needs two 5s, while 125 needs three and is not a factor of 225.
Quiz
Which is a prime number?
Which is the complete prime factorisation of 90?
Is 3 × 3 × 3 a factor of 840 = 2³ × 3 × 5 × 7?
What does one counterexample do to an “always” claim?
How many distinct positive factors does 225 = 3² × 5² have?
Practice Problems
- Use the division method to factorise 1200 completely.
- List all factors of 90 using its prime factors; check that none are repeated.
- Determine whether 14, 28 and 27 are factors of 840 from its prime pieces.
- Find all factors of 105 from 3 × 5 × 7.
- Produce a different counterexample to “the larger the number, the longer its prime factorisation.” Explain what “longer” means.
- Factorise 132 and construct at least six distinct factors.
- List all factors of 360 systematically and verify that there are 24.
- List all factors of 840 systematically and verify that there are 32.
Key Takeaways
• A prime number is greater than 1 and has only 1 and itself as positive factors. • Continue a factorisation until every factor is prime; routes may change the order, not the prime pieces. • A factor is a subproduct of the prime factorisation, with 1 included. • Repeated primes cannot be used more often than they occur. • One valid counterexample disproves a universal conjecture.