List all the factors of the following numbers: (a) 90 (b) 105 (c) 132 (d) 360 (this number has 24 factors) (e) 840 (this number has 32 factors)

Understand the Question To list all factors of a number, we first find its prime factorisation by repeatedly dividing by prime numbers (). A factor is formed by multiplying any combination of these prime factors, including taking none (which gives ) up to taking all of them (which gives the number itself). For a number , every factor is of the form where , , etc. Part (a) List all the factors of 90 Step 1 · Prime Factorisation of 90 Divide by prime numbers: Step 2 · List All Factors of 90 Taking all combinations where , , and in increasing order: Answer > (a) Part (b) List all the factors of 105 Step 1 · Prime Factorisation of 105 Divide by prime numbers: Step 2 · List All Factors of 105 Taking all combinations where in increasing order: Answer > (b) Part (c) List all the factors of 132 Step 1 · Prime Factorisation of 132 Divide by prime numbers: Step 2 · List All Factors of 132 Taking all combinations where , , and in increasing order: Answer > (c) Part (d) List all the factors of 360 (this number has 24 factors) Step 1 · Prime Factorisation of 360 Divide by prime numbers: Step 2 · List All Factors of 360 Taking all combinations where , , and in increasing order: Answer > (d) Part (e) List all the factors of 840 (this number has 32 factors) Step 1 · Prime Factorisation of 840 Divide by prime numbers: Step 2 · List All Factors of 840 Taking all combinations where and in increasing order: Answer > (e) Common Mistakes Omitting 1 and the Number Itself: and the number itself are always factors of that number. Confusing Prime Factors with All Factors: The prime factors are only the prime divisors (e.g. for ), whereas all factors include all composite products of those primes. Missing Intermediate Factors: Check the total count using the formula where are the exponents in prime factorisation to verify no factors were missed.

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Question 1

List all the factors of the following numbers:

(a) 90

(b) 105

(d) 360 (this number has 24 factors)

(e) 840 (this number has 32 factors)

Solution

Understand the Question

To list all factors of a number, we first find its prime factorisation by repeatedly dividing by prime numbers ( $2, 3, 5, \dots$ ).
A factor is formed by multiplying any combination of these prime factors, including taking none (which gives $1$ ) up to taking all of them (which gives the number itself).
For a number $N = p_1^a \times p_2^b \times \dots$ , every factor is of the form $p_1^x \times p_2^y \times \dots$ where $0 \le x \le a$ , $0 \le y \le b$ , etc.

(a) List all the factors of 90

Step 1 · Prime Factorisation of 90

Divide $90$ by prime numbers:

\begin{aligned} 90 \div 2 &= 45 \\ 45 \div 3 &= 15 \\ 15 \div 3 &= 5 \\ 5 \div 5 &= 1 \end{aligned}

90 = 2 \times 3 \times 3 \times 5 = 2^1 \times 3^2 \times 5^1

Step 2 · List All Factors of 90

Taking all combinations $2^a \times 3^b \times 5^c$ where $a \in \{0, 1\}$ , $b \in \{0, 1, 2\}$ , and $c \in \{0, 1\}$ in increasing order:

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

Answer

(a) $1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90$

(b) List all the factors of 105

Step 1 · Prime Factorisation of 105

Divide $105$ by prime numbers:

\begin{aligned} 105 \div 3 &= 35 \\ 35 \div 5 &= 7 \\ 7 \div 7 &= 1 \end{aligned}

105 = 3 \times 5 \times 7 = 3^1 \times 5^1 \times 7^1

Step 2 · List All Factors of 105

Taking all combinations $3^a \times 5^b \times 7^c$ where $a, b, c \in \{0, 1\}$ in increasing order:

1, 3, 5, 7, 15, 21, 35, 105

Answer

(b) $1, 3, 5, 7, 15, 21, 35, 105$

(c) List all the factors of 132

Step 1 · Prime Factorisation of 132

Divide $132$ by prime numbers:

\begin{aligned} 132 \div 2 &= 66 \\ 66 \div 2 &= 33 \\ 33 \div 3 &= 11 \\ 11 \div 11 &= 1 \end{aligned}

132 = 2 \times 2 \times 3 \times 11 = 2^2 \times 3^1 \times 11^1

Step 2 · List All Factors of 132

Taking all combinations $2^a \times 3^b \times 11^c$ where $a \in \{0, 1, 2\}$ , $b \in \{0, 1\}$ , and $c \in \{0, 1\}$ in increasing order:

1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132

Answer

(c) $1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132$

(d) List all the factors of 360 (this number has 24 factors)

Step 1 · Prime Factorisation of 360

Divide $360$ by prime numbers:

\begin{aligned} 360 \div 2 &= 180 \\ 180 \div 2 &= 90 \\ 90 \div 2 &= 45 \\ 45 \div 3 &= 15 \\ 15 \div 3 &= 5 \\ 5 \div 5 &= 1 \end{aligned}

360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^3 \times 3^2 \times 5^1

Step 2 · List All Factors of 360

Taking all combinations $2^a \times 3^b \times 5^c$ where $a \in \{0, 1, 2, 3\}$ , $b \in \{0, 1, 2\}$ , and $c \in \{0, 1\}$ in increasing order:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360

Answer

(d) $1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360$

(e) List all the factors of 840 (this number has 32 factors)

Step 1 · Prime Factorisation of 840

Divide $840$ by prime numbers:

\begin{aligned} 840 \div 2 &= 420 \\ 420 \div 2 &= 210 \\ 210 \div 2 &= 105 \\ 105 \div 3 &= 35 \\ 35 \div 5 &= 7 \\ 7 \div 7 &= 1 \end{aligned}

840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7 = 2^3 \times 3^1 \times 5^1 \times 7^1

Step 2 · List All Factors of 840

Taking all combinations $2^a \times 3^b \times 5^c \times 7^d$ where $a \in \{0, 1, 2, 3\}$ and $b, c, d \in \{0, 1\}$ in increasing order:

1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840

Answer

(e) $1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840$

Common Mistakes

Omitting 1 and the Number Itself: $1$ and the number itself are always factors of that number.
Confusing Prime Factors with All Factors: The prime factors are only the prime divisors (e.g. $2, 3, 5$ for $90$ ), whereas all factors include all composite products of those primes.
Missing Intermediate Factors: Check the total count using the formula $(a+1)(b+1)\dots$ where $a, b$ are the exponents in prime factorisation to verify no factors were missed.

Question 1

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