Factors and Multiples - SS1 Mathematics Lesson Note

A number \(a\) is a factor of \(a\) integer \(b\ \)if \(b\) can be divided by \(a\) without remainder. \(2,\ 3,\ 4,\ 6,\ 8,\ 12\ \)and\(\ 24\) are factors of \(24\) as each number can divide \(24\) without remainder. Amongst the factors of \(24\), \(2\ and\ 3\) are prime numbers, thus we call them prime factors of \(24\) whilst \(4,\ 6,\ 8,\ 12\) and \(24\) are composite numbers, thus composite factors of \(24\). Breaking 24 into its factors is called factorization while breaking 30 into its prime factors only is referred to as prime factorization.

Given a set of numbers with common factors, the highest among these factors is termed the highest common factor (HCF) or greatest common factor (GCF).

Example 4 Find the Highest common factor of \(28\), \(42\) and \(56\).

Solution Factors of \(28\) are \(2,\ 4,\ 7,\ 14\ \)and\(\ 28\)

Factors of \(42\) are \(2,\ 3,\ 6,\ 7,\ 14,\ 21\ \)and\(\ 42\)

Factors of \(56\) are \(2,\ 4,\ 7,\ 8,\ 14,\ 28\ \)and\(\ 56\)

Common factors of \(28\), \(42\) and \(56\) are \(2\), \(7\) and \(14\).

Therefore, the Highest Common Factor (HCF) = 14.

Example 5 Using prime factorization, find the HCF of \(28\), \(42\) and \(56\)

Solution \(Step\ I\): factorize each number using its prime factors

2	28
2	14
7	7
	1

2	42
3	21
7	7
	1

2	56
2	28
2	14
7	7
	1

Thus, \(28 = 2 \times 2 \times 7\),

\(42 = 2 \times 3 \times 7\),

\(56 = 2 \times 2 \times 2 \times 7\)

From the factorization above, the only common factors are one set of \(2\) and \(7\) each. Thus, the HCF = \(2 \times 7 = 14\)

Multiples of a number \(n\) can be obtained by multiplying \(n\) by the counting numbers \(1,\ 2,\ 3,\ 4,\ \ldots\) to infinity, that is the multiples of \(n\) will be \(n \times 1,\ n \times 2,\ n \times 3,\ n \times 4,\ \ldots = n,\ 2n,\ 3n,\ 4n,\ \ldots\) and the multiples of 6 are \(6,\ 12,\ 18,\ 24,\ 30,\ 36,\ 42,\ 48,\ \ldots\)

Given a set of numbers with common multiples, the lowest amongst these multiples is termed the lowest common multiple (LCM).

Example 6 Find the LCM of \(3,\ 4\ \)and\(\ 6\)

Solution Multiples of \(3\) = \(3,\ 6,\ 9,\ 12,\ 15,\ 18,\ 21,\ 24,\ 27,\ 30,\ \ldots\)

Multiples of \(4 = 4,\ 8,\ 12,\ 16,\ 20,\ 24,\ 28,\ 32,\ 36,\ \ldots\)

Multiples of 6 = \(6,\ 12,\ 18,\ 24,\ 30,\ 36,\ 42,\ 48,\ 54,\ \ldots\)

The common multiples observed are \(12,\ 24,\ \ldots\)

Hence, the Lowest Common Multiple is \(12\)

Alternatively, using the prime factorization method, divide each number repeatedly until we completely divide out.

2	3	4	6
2	3	2	3
3	3	1	3
	1	1	1

Thus, the LCM = \(2 \times 2 \times 3 = 12\)