Lowest Common Multiple
Theory
The lowest common multiple (LCM) of two numbers is the smallest number that is a multiple of both of them. The LCM of \(6\) and \(10\) is \(30\). This Year 7 maths page lists multiples, uses a shortcut and prime factors, links the LCM with the HCF, and solves problems such as buses leaving together again.
The multiples of a number are its times table: the multiples of \(6\) are \(6, 12, 18, 24, \ldots\) A common multiple of two numbers is in both lists. The lowest common multiple (LCM) is the smallest one.
Method 1: list the multiples. Multiples of \(5\): \(5, 10, 15, 20, 25, 30, 35, \ldots\) Multiples of \(7\): \(7, 14, 21, 28, 35, \ldots\) The first number in both lists is \(35\), so the LCM of \(5\) and \(7\) is \(35\).
Shortcut: go up the multiples of the larger number and stop at the first one that the smaller number divides into exactly.
Method 2: prime factors. Use each prime the most times it appears in either number, then multiply. \(18 = 2 \times 3 \times 3\) and \(30 = 2 \times 3 \times 5\), so the LCM is \(2 \times 3 \times 3 \times 5 = 90\). In index form, use the larger power of each prime.
Using the HCF. \(\text{HCF} \times \text{LCM}\) equals the two numbers multiplied. The HCF of \(20\) and \(45\) is \(5\), so the LCM is \(20 \times 45 \div 5 = 180\). Numbers with no common factor except \(1\) have LCM equal to their product: \(5 \times 7 = 35\).
The link between the HCF and the LCM of two numbers \(a\) and \(b\):
| HCF | LCM | |
|---|---|---|
| Uses | only the shared primes | every prime, the most times it appears |
| Size | never bigger than the smaller number | never smaller than the larger number |
| \(8\) and \(10\) | \(2\) | \(40\) |
How to find the LCM by listing
- Write the multiples of the larger number in order.
- Test each one: does the smaller number divide into it exactly?
- Stop at the first one that works. That is the LCM.
How to find the LCM with prime factors
- Write each number as a product of primes.
- Use each prime the most times it appears in either number (the larger power).
- Multiply, then check that both numbers divide into the answer.
How to spot an LCM problem
- Two things repeat (buses, lights, laps, gear teeth, packs) and you want them to happen together again or to match exactly.
- Find the LCM of the two repeat sizes.
- Answer the question: a time, a number of packs, or a number of turns.
(ii) Use the shortcut to find the LCM of \(8\) and \(14\).
(i) Multiples of \(6\): \(6, 12, 18, 24, 30, \ldots\) Multiples of \(10\): \(10, 20, 30, \ldots\)
(ii) Go up the multiples of \(14\): \(14\), \(28\) and \(42\) are not multiples of \(8\), but \(56\) is.
| \(30 \div 6\) | \(=\) | \(5\) |
| \(30 \div 10\) | \(=\) | \(3\) |
| \(56 \div 8\) | \(=\) | \(7\) |
(i) \(30\)
(ii) \(56\)
(i) \(5\) and \(7\)
(ii) \(20\) and \(45\)
(i) \(5\) and \(7\) have no common factor except \(1\), so multiply them.
(ii) Listing: \(20, 40, 60, 80, 100, 120, 140, 160, 180\) and \(45, 90, 135, 180\). Or use the HCF: \(20 = 4 \times 5\) and \(45 = 9 \times 5\), so the HCF is \(5\).
| \(5 \times 7\) | \(=\) | \(35\) |
| \(20 \times 9\) | \(=\) | \(180\) |
(i) \(35\)
(ii) \(180\) (\(20 \times 45 \div 5 = 20 \times 9\))
\(28 = 2 \times 2 \times 7\) and \(42 = 2 \times 3 \times 7\). The prime \(2\) appears at most two times, \(3\) one time and \(7\) one time.
| \(2 \times 2 \times 3 \times 7\) | \(=\) | \(84\) |
The LCM is \(84\). Check: \(84 \div 28 = 3\) and \(84 \div 42 = 2\).
The lowest common denominator is the LCM of \(8\) and \(12\). Go up the multiples of \(12\): \(12\) is not a multiple of \(8\), but \(24\) is.
| \(\dfrac{3}{8} + \dfrac{5}{12}\) | \(=\) | \(\dfrac{9}{24} + \dfrac{10}{24}\) |
| \(\) | \(=\) | \(\dfrac{19}{24}\) |
The lowest common denominator is \(24\) and the sum is \(\dfrac{19}{24}\).
(i) After how many seconds are they next at the start line together?
(ii) How many laps has each runner done?
\(72 = 2^3 \times 3^2\) and \(90 = 2 \times 3^2 \times 5\). Use the larger power of each prime.
| \(2^3 \times 3^2 \times 5\) | \(=\) | \(8 \times 9 \times 5\) |
| \(\) | \(=\) | \(360\) |
| \(360 \div 72\) | \(=\) | \(5\) |
| \(360 \div 90\) | \(=\) | \(4\) |
(i) After \(360\) seconds (\(6\) minutes)
(ii) Ava \(5\) laps, Ben \(4\) laps
Multiples of \(21\): \(21\) is not a multiple of \(14\), but \(42\) is.
| \(42 \div 14\) | \(=\) | \(3\) |
| \(42 \div 21\) | \(=\) | \(2\) |
\(42\) minutes after 6:30 am, at 7:12 am.
(ii) Gear A has \(16\) teeth and gear B has \(24\) teeth. Two marked teeth touch. How many turns does each gear make before they touch again?
(i) The LCM of \(10\) and \(15\) is \(30\).
(ii) The LCM of \(16\) and \(24\) is \(48\) teeth.
| \(30 \div 10\) | \(=\) | \(3\) |
| \(30 \div 15\) | \(=\) | \(2\) |
| \(48 \div 16\) | \(=\) | \(3\) |
| \(48 \div 24\) | \(=\) | \(2\) |
(i) \(3\) packs of cups and \(2\) packs of plates (\(30\) of each)
(ii) Gear A turns \(3\) times and gear B turns \(2\) times.
Common pitfalls
Frequently asked questions
What is the lowest common multiple?
The smallest number that is a multiple of both numbers.
What is a quick way to find the LCM?
Go up the multiples of the larger number and stop at the first one the smaller number divides. For \(8\) and \(14\): \(14, 28, 42, 56\), so the LCM is \(56\).
How do you find the LCM using prime factors?
Use each prime the most times it appears in either number and multiply. \(28 = 2 \times 2 \times 7\) and \(42 = 2 \times 3 \times 7\) give \(2 \times 2 \times 3 \times 7 = 84\).
How are the HCF and the LCM connected?
HCF times LCM equals the two numbers multiplied. For \(18\) and \(30\): \(6 \times 90 = 540 = 18 \times 30\).
When do I use the LCM in a problem?
When repeating things must happen together again, or amounts must match exactly, like buses, lights, laps or packs.