Number Systems • Topic 2 of 4

Fundamental Theorem of Arithmetic

What is the Fundamental Theorem of Arithmetic? Every composite number greater than 1 can be expressed as a product of prime numbers, and this factorisation is unique apart from the order of the factors. This unique product is called the prime factorisation of the number.

Example: $84 = 2 \times 2 \times 3 \times 7 = 2^{2} \times 3 \times 7$ (written with exponents). No other set of primes multiplies to 84.

Why "unique"? Think of primes (2, 3, 5, 7, 11, …) as the "atoms" of arithmetic. Just as a molecule is built from a fixed set of atoms, every composite number is built from exactly one multiset of primes. The order in which we write them does not matter: $2 \times 3 \times 7 = 7 \times 3 \times 2$.

Key facts:

  • Prime numbers have exactly two factors, 1 and themselves; they cannot be split further
  • Composite numbers are products of two or more primes
  • The number $1$ is neither prime nor composite
  • Prime factorisation is unique — this is what makes HCF and LCM well-defined

Finding HCF and LCM by prime factorisation. Write each number as a product of primes. Then:

  • HCF = product of each common prime raised to the smallest power that appears
  • LCM = product of every prime that appears, each raised to the greatest power

For example, $12 = 2^{2}\times 3$ and $18 = 2\times 3^{2}$. The HCF uses the smaller power of each common prime: $2^{1}\times 3^{1} = 6$. The LCM uses the larger power of every prime: $2^{2}\times 3^{2} = 36$.

The HCF–LCM relationship. For any two positive integers $a$ and $b$:

$\mathrm{HCF}(a,b) \times \mathrm{LCM}(a,b) = a \times b$

This is a quick way to find one if the other is known. It works for two numbers only — it is not valid for three or more numbers, where you must use full factorisation.

An important application. The theorem lets us decide things like "can $6^{n}$ end in 0?" A number ends in 0 only if it has both 2 and 5 as factors. Since $6^{n} = 2^{n}\times 3^{n}$ has no factor of 5, it can never end in 0. Similarly, an expression such as $7\times 11\times 13 + 13 = 13(7\times 11 + 1) = 13 \times 78$ is composite because we have exhibited a factor other than 1 and itself.

Common mistakes to avoid:

  • Leaving a composite factor in the answer, e.g. writing $60 = 4 \times 15$ instead of $2^{2}\times 3 \times 5$. Factorisation must go all the way down to primes.
  • Using $\mathrm{HCF}\times\mathrm{LCM} = a\times b$ for three numbers — it only holds for two.
  • Mixing up the rules: HCF takes the smallest powers of common primes; LCM takes the largest powers of all primes.
  • Calling 1 a prime number — it is neither prime nor composite.
Prime factorization tree of 140 splitting into 2, 2, 5 and 7Factor Tree — every composite is a unique product of primes14027023557green = prime140 = 2 × 2 × 5 × 7 = 2^2 × 5 × 7HCF and LCM from prime factorizations of 12 and 18HCF & LCM from prime factors12 = 2^2 × 318 = 2 × 3^2HCF = smallest powers2^1 × 3^1 = 6LCM = greatest powers2^2 × 3^2 = 36Check: HCF × LCM = 6 × 36 = 216 = 12 × 18.
Solved Examples
1
Worked Example
Find the prime factorisation of 240.
Solution
  1. Step 1: $240 \div 2 = 120$, $120 \div 2 = 60$, $60 \div 2 = 30$, $30 \div 2 = 15$.
  2. Step 2: $15 \div 3 = 5$, and $5 \div 5 = 1$ (stop at 1).
  3. Step 3: Collect the prime divisors: $2 \times 2 \times 2 \times 2 \times 3 \times 5$.
  4. Step 4: Write with exponents: $2^{4} \times 3 \times 5$.

Answer: $240 = 2^{4} \times 3 \times 5$

2
Worked Example
Find the prime factorisation of 5005.
Solution
  1. Step 1: $5005 \div 5 = 1001$.
  2. Step 2: $1001 \div 7 = 143$.
  3. Step 3: $143 \div 11 = 13$, and 13 is prime.
  4. Step 4: So $5005 = 5 \times 7 \times 11 \times 13$.

Answer: $5005 = 5 \times 7 \times 11 \times 13$

3
Worked Example
Find $\mathrm{HCF}(12, 18)$ and $\mathrm{LCM}(12, 18)$ using prime factorisation.
Solution
  1. Step 1: $12 = 2^{2} \times 3$ and $18 = 2 \times 3^{2}$.
  2. Step 2: HCF = product of common primes to the smallest power: $2^{1} \times 3^{1} = 6$.
  3. Step 3: LCM = product of all primes to the greatest power: $2^{2} \times 3^{2} = 36$.

Answer: $\mathrm{HCF} = 6,\ \mathrm{LCM} = 36$

4
Worked Example
Find the HCF and LCM of 96 and 404 by prime factorisation, and verify the product relation.
Solution
  1. Step 1: $96 = 2^{5} \times 3$ and $404 = 2^{2} \times 101$.
  2. Step 2: HCF = smallest power of the common prime 2: $2^{2} = 4$.
  3. Step 3: LCM = $2^{5} \times 3 \times 101 = 9696$.
  4. Step 4: Check: $\mathrm{HCF}\times\mathrm{LCM} = 4 \times 9696 = 38784 = 96 \times 404$.

Answer: $\mathrm{HCF} = 4,\ \mathrm{LCM} = 9696$ (verified)

5
Worked Example
Find the HCF and LCM of 6, 72 and 120 by prime factorisation.
Solution
  1. Step 1: $6 = 2 \times 3$, $72 = 2^{3} \times 3^{2}$, $120 = 2^{3} \times 3 \times 5$.
  2. Step 2: HCF = smallest power of each prime common to all three: $2^{1} \times 3^{1} = 6$.
  3. Step 3: LCM = greatest power of every prime appearing: $2^{3} \times 3^{2} \times 5 = 360$.

Answer: $\mathrm{HCF} = 6,\ \mathrm{LCM} = 360$

6
Worked Example
Given $\mathrm{HCF}(a, b) = 9$, $\mathrm{LCM}(a, b) = 90$ and $a = 18$, find $b$.
Solution
  1. Step 1: Use $\mathrm{HCF} \times \mathrm{LCM} = a \times b$.
  2. Step 2: $9 \times 90 = 18 \times b$, so $810 = 18b$.
  3. Step 3: $b = 810 \div 18 = 45$.

Answer: $b = 45$

7
Worked Example
The HCF of two numbers is 23 and their LCM is 1449. If one number is 161, find the other.
Solution
  1. Step 1: Apply $\mathrm{HCF} \times \mathrm{LCM} = $ product of the numbers.
  2. Step 2: $23 \times 1449 = 161 \times \text{other}$, so $33327 = 161 \times \text{other}$.
  3. Step 3: $\text{other} = 33327 \div 161 = 207$.

Answer: The other number is 207

8
Worked Example
Find the HCF and LCM of 510 and 92 and verify $\mathrm{HCF}\times\mathrm{LCM} = $ product.
Solution
  1. Step 1: $510 = 2 \times 3 \times 5 \times 17$ and $92 = 2^{2} \times 23$.
  2. Step 2: HCF = smallest power of the common prime 2: $2^{1} = 2$.
  3. Step 3: LCM = $2^{2} \times 3 \times 5 \times 17 \times 23 = 23460$.
  4. Step 4: Check: $2 \times 23460 = 46920 = 510 \times 92$.

Answer: $\mathrm{HCF} = 2,\ \mathrm{LCM} = 23460$ (verified)

9
Worked Example
Explain why $7 \times 11 \times 13 + 13$ is a composite number.
Solution
  1. Step 1: Take 13 common: $7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1)$.
  2. Step 2: $7 \times 11 + 1 = 77 + 1 = 78$, so the expression equals $13 \times 78$.
  3. Step 3: It has factors other than 1 and itself, so it is composite.

Answer: It equals $13 \times 78$, hence composite

10
Worked Example
Check whether $6^{n}$ can end with the digit 0 for any natural number $n$.
Solution
  1. Step 1: A number ends in 0 only if it is divisible by 10, i.e. by both 2 and 5.
  2. Step 2: Prime-factorise: $6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}$.
  3. Step 3: There is no factor of 5 in $6^{n}$, so it is never divisible by 10.
  4. Step 4: By uniqueness of prime factorisation, no power of 6 can supply a factor 5.

Answer: No, $6^{n}$ can never end in 0

11
Worked Example
Three bells toll at intervals of 9, 12 and 15 minutes. If they toll together at 8 a.m., when do they next toll together?
Solution
  1. Step 1: They next toll together after $\mathrm{LCM}(9, 12, 15)$ minutes.
  2. Step 2: $9 = 3^{2}$, $12 = 2^{2} \times 3$, $15 = 3 \times 5$.
  3. Step 3: LCM = $2^{2} \times 3^{2} \times 5 = 180$ minutes $= 3$ hours.
  4. Step 4: So they toll together again 3 hours after 8 a.m.

Answer: At 11 a.m. (after 180 minutes)

12
Worked Example
A number $N = 2^{3} \times 3^{2} \times 5$ is multiplied by $2 \times 3 \times 5$. Find the prime factorisation of the result.
Solution
  1. Step 1: Write the multiplier with exponents: $2^{1} \times 3^{1} \times 5^{1}$.
  2. Step 2: Add exponents of equal bases: for 2, $3 + 1 = 4$; for 3, $2 + 1 = 3$; for 5, $1 + 1 = 2$.
  3. Step 3: So the product is $2^{4} \times 3^{3} \times 5^{2}$.

Answer: $2^{4} \times 3^{3} \times 5^{2}$

Key Points

  • Every composite number > 1 has a unique prime factorisation (order ignored)
  • Primes are the building blocks; the number 1 is neither prime nor composite
  • To factorise, divide repeatedly by the smallest prime until you reach 1
  • HCF = product of common primes to the smallest power; LCM = product of all primes to the greatest power
  • For two numbers, $\mathrm{HCF}\times\mathrm{LCM} = a\times b$ (not valid for three or more)
  • Use the theorem to test divisibility, e.g. $6^{n}$ never ends in 0 since it has no factor 5
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The Fundamental Theorem of Arithmetic says every composite number is a product of:
Explanation: Unique prime factorisation.
Q2.The prime factorisation of $12$ is:
Explanation: $12=2^2\cdot3$.
Q3.HCF$(a,b)\times$LCM$(a,b)=$
Explanation: Product of the two numbers.
Q4.LCM$(4,6)=$
Explanation: $12$ is the least common multiple.
Practice more MCQs → 📝 Assignment · 35 marks