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ESO Y2 · Mathematics
Numbers · Natural numbers

Highest common factor

Two ropes, $20$ and $28$ metres long. You want to cut them into equal pieces, as long as possible, with nothing left over. That is the question the highest common factor answers.

Before you start, make sure you know: Prime numbers and prime factorisation (1st ESO)
Before you start, make sure you know: Multiples, factors and divisibility tests (1st ESO)

What the h.c.f. is

Venn diagram showing the factors of 24 and 36 and highlighting the common factors, marking 12 as the highest common factor. 8, 24 9, 18, 36 1, 2, 3, 4, 6, 12 h.c.f. = 12 (the largest common one)
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Look at the factors of the numbers $24$ and $36$: $Div(24) = \{1, 2, 3, 4, 6, 8, 12, 24\}$ $Div(36) = \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$ From these lists: a) Write down…

Look at the factors of the numbers $24$ and $36$:

  • $Div(24) = \{1, 2, 3, 4, 6, 8, 12, 24\}$

  • $Div(36) = \{1, 2, 3, 4, 6, 9, 12, 18, 36\}$

From these lists:

  1. a) Write down the common factors of both numbers.
  2. b) Identify the highest of these common factors. What do we call this number?
Solve
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In a Year 8 class, there are $24$ male students and $16$ female students. The PE teacher wants to form equal-sized teams, as large as possible, so that each team…

In a Year 8 class, there are $24$ male students and $16$ female students. The PE teacher wants to form equal-sized teams, as large as possible, so that each team contains only boys or only girls.

  1. a) How many students will there be in each team?
  2. b) How many boys' teams and how many girls' teams will be formed?
Solve

Finding the h.c.f. by prime factorisation

The quick method for large numbers: break each number down into primes and take only the common factors, raised to the lowest power.

Calculation of the h.c.f. of 12 and 18 from their prime factorisation, selecting the common factors with the lowest power. 12 = 2² · 3 18 = 2 · 3² m.c.d. = 2 · 3 = 6 only factors common to both, with the lowest power
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Calculate the greatest common divisor (g.c.d.) of the following numbers using the prime factorization method (remember: you take the common factors raised to the lowest…

Calculate the greatest common divisor (g.c.d.) of the following numbers using the prime factorization method (remember: you take the common factors raised to the lowest exponent):

  1. a) $12$ and $18$
  2. b) $20$ and $30$
  3. c) $15$ and $25$
Solve
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Find the Highest Common Factor of the following groups of numbers using prime factorisation: a) $m.c.d.(120, 150)$ b) $m.c.d.(72, 108, 144)$ c) $m.c.d.(45, 60, 105)$…

Find the Highest Common Factor of the following groups of numbers using prime factorisation:

  1. a) $m.c.d.(120, 150)$
  2. b) $m.c.d.(72, 108, 144)$
  3. c) $m.c.d.(45, 60, 105)$

Remember to first express each number as a product of its prime factors and then select the common factors raised to the lowest power.

Solve

Relationship between the h.c.f. and the l.c.m.

For two numbers, the h.c.f. and the l.c.m. are linked by a very useful formula.

Diagram checking the equality between the product of 12 and 18 and the product of their h.c.f. and l.c.m. 12 · 18 = 216 m.c.d. = 6 m.c.m. = 36 6 · 36 = 216 ✓ = 12 · 18
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It is known that the product of two natural numbers is $2160$ and that their lowest common multiple ($m.c.m.$) is $180$. Using the relationship between the product of…

It is known that the product of two natural numbers is $2160$ and that their lowest common multiple ($m.c.m.$) is $180$. Using the relationship between the product of two numbers, their $m.c.d.$ and their $m.c.m.$, find out what the Highest Common Factor of those numbers is.

Solve
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Use the Euclidean algorithm to find the $m.c.d.(2\,310, 1\,485)$. Once found, state whether the numbers are coprime and justify your answer based on the divisors…

Use the Euclidean algorithm to find the $m.c.d.(2\,310, 1\,485)$. Once found, state whether the numbers are coprime and justify your answer based on the divisors obtained during the algorithm process.

Solve

Applications of the h.c.f.

The h.c.f. turns up whenever you want to share out or cut into equal parts that are as large as possible, with nothing left over.

Diagram showing the division of two lengths into equal pieces of maximum size corresponding to their highest common factor. Rope A: 20 m Rope B: 28 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m 4 m m.c.d.(20, 28) = 4 m (5 pieces of A and 7 pieces of B)
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At an adventure camp in the Pyrenees, the instructors have two climbing ropes: one measures $20$ metres and the other $28$ metres. They want to cut them into equal…

At an adventure camp in the Pyrenees, the instructors have two climbing ropes: one measures $20$ metres and the other $28$ metres. They want to cut them into equal pieces of the maximum possible length, with no rope left over. How long should each piece be?

20 m 28 m ? m Max piece
Solve
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We want to decorate a wall in the music room measuring $120$ cm high by $180$ cm wide using square pieces of card. If we want the pieces of card to be as large as…

We want to decorate a wall in the music room measuring $120$ cm high by $180$ cm wide using square pieces of card. If we want the pieces of card to be as large as possible and cover the wall exactly without having to cut them:

  1. a) What length should each side of the pieces of card be?
  2. b) How many pieces of card will we need in total?
180 cm 120 cm x x
Solve
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An architect is designing the floor of an e-sports club trophy room. The room is rectangular, measuring $12,6$ metres long by $7,2$ metres wide. The intention is to…

An architect is designing the floor of an e-sports club trophy room. The room is rectangular, measuring $12,6$ metres long by $7,2$ metres wide. The intention is to cover the floor with square tiles of the largest possible size, so that none need to be cut and the design is perfectly symmetrical.

However, the architect wants to leave a small ventilation channel $60$ cm wide along the entire perimeter of the room, so the actual area to be tiled will be a smaller inner rectangle.

Calculate the side length of the largest tile that can be used to cover that inner rectangle and how many tiles will be needed in total.

Square tile Side = HCF 12,6 m (1 260 cm) 7,2 m (720 cm) 1260 - 2(60) = 1140 cm 720 - 2(60) = 600 cm Ventilation channel 60 cm 60
Solve

Back to the start: You factorise $20 = 2^2 \cdot 5$ and $28 = 2^2 \cdot 7$; their h.c.f. is $2^2 = 4\text{ m}$, so you will get pieces of $4\text{ m}$ ($5$ pieces from the first rope and $7$ from the second).

Summary

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