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ESO Y1 · Mathematics
Numbers · Divisibility

Highest common factor

Two wooden strips measure $36$ cm and $48$ cm. You want to cut them into pieces of the same length, as long as possible, with no wood left over. That length is the highest common factor.

What the H.C.F. is

Lists of the factors of 12 and 18 with their common factors highlighted and 6 marked as the highest common factor. factors of 121, 2, 3, 4, 6, 12 factors of 181, 2, 3, 6, 9, 18 common 1, 2, 3,6 M.C.D.(12,18) = 6

The most direct method (for small numbers): list all the factors of each number and pick the largest of the common ones.

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Find all the divisors of the numbers $12$ and $18$. Then, state which are the common divisors and point out which of them is the Greatest Common Divisor (GCD). You can…

Find all the divisors of the numbers $12$ and $18$. Then, state which are the common divisors and point out which of them is the Greatest Common Divisor (GCD).

You can use a diagram like the one in the figure: write the divisors of each number in each circle and, in the area where they overlap, the ones that are common.

Diagram of two overlapping circles, one for the divisors of 12 and another for the divisors of 18, empty to be filled inDivisors of 12Divisors of 18Common area
Solve
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Work out in your head the $MCD$ of the following pairs of numbers. These are small numbers where you can identify the highest common factor at a glance: a) $MCD(4, 8)$…

Work out in your head the $MCD$ of the following pairs of numbers. These are small numbers where you can identify the highest common factor at a glance:

  1. a) $MCD(4, 8)$
  2. b) $MCD(5, 10)$
  3. c) $MCD(7, 14)$
  4. d) $MCD(10, 20)$

What do all these results have in common?

Solve

Finding the H.C.F. with prime factors

With large numbers, it is quicker to break them down into prime factors.

Diagram of finding the H.C.F. by prime factorisation, selecting common factors with the lowest power. A = 2² · 3 · 5B = 2 · 3² 5 is not in B → it is left out of 2² and 2¹, the lowest is 2¹ M.C.D.(A,B) = 2 · 3 = 6
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Look at the prime factorisations of these two numbers and find their $MCD$: $A = 2^{2} \cdot 3 \cdot 5$ $B = 2 \cdot 3^{2}$ Remember the rule: "For the HCF, we take the…

Look at the prime factorisations of these two numbers and find their $MCD$:

  • $A = 2^{2} \cdot 3 \cdot 5$

  • $B = 2 \cdot 3^{2}$

Remember the rule: "For the HCF, we take the common factors raised to the lowest power".

Solve
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Calculate the Highest Common Factor ($M.C.D.$) of the following sets of numbers using prime factorisation: a) $48$ and $72$ b) $30, 45$ and $60$ c) $120$ and $150$

Calculate the Highest Common Factor ($M.C.D.$) of the following sets of numbers using prime factorisation:

  1. a) $48$ and $72$
  2. b) $30, 45$ and $60$
  3. c) $120$ and $150$
Solve
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Look at the following prime factorisations of three numbers and, without performing the multiplications, state their $\text{HCF}$: $A = 2^3 \cdot 3^2 \cdot 5$…

Look at the following prime factorisations of three numbers and, without performing the multiplications, state their $\text{HCF}$:

$A = 2^3 \cdot 3^2 \cdot 5$

$B = 2^2 \cdot 3^3$

$C = 2^2 \cdot 3^2 \cdot 7$

Solve

Coprime numbers

Prime factorisations of 14 and 15 showing that they share no prime factors and their H.C.F. is 1. 14 = 2 · 7 15 = 3 · 5 no prime in common→ M.C.D.(14,15) = 1 14 and 15 are coprime, even though neither of them is prime
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Two numbers are said to be coprime if their only common factor is $1$. Check whether the following pairs of numbers are coprime by calculating their $M.C.D.$: a) $14$…

Two numbers are said to be coprime if their only common factor is $1$. Check whether the following pairs of numbers are coprime by calculating their $M.C.D.$:

  1. a) $14$ and $15$
  2. b) $22$ and $33$
  3. c) $21$ and $25$
Solve

"Share into equal parts" problems

When you have to group or share several quantities into equal parts that are as large as possible (or into the greatest number of identical groups), with nothing left over, the H.C.F. is the tool to use.

Illustration of 45 and 60 books grouped on equal shelves of 15 books each. 45 books 15 15 15 60 books 15 15 15 15 each box: 15 books (H.C.F.)
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In the technology workshop, the students have two wooden battens measuring $36$ cm and $48$ cm in length. They need to cut them into pieces of the same length, with this…

In the technology workshop, the students have two wooden battens measuring $36$ cm and $48$ cm in length. They need to cut them into pieces of the same length, with this length being the maximum possible.

  1. a) What will be the length of each piece?
  2. b) How many pieces will they get in total from the two battens?
Solve
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At the school, a day of group activities has been organized. There are $54$ students in 1st ESO and $42$ students in 2nd ESO. The students are to be split into teams so…

At the school, a day of group activities has been organized. There are $54$ students in 1st ESO and $42$ students in 2nd ESO. The students are to be split into teams so that each team consists only of students from the same year (students from 1st and 2nd year are not mixed). All teams must have exactly the same number of people, and that number must be as large as possible without anyone being left over.

  1. a) How many students will there be in each team?
  1. b) How many teams of 1st ESO and how many teams of 2nd ESO will be formed?
Solve
Practice★★☆Not started
For a food bank, a supermarket donates $36$ tins of pulses, $48$ cartons of milk and $60$ bags of rice. They want to prepare the maximum possible number of identical…

For a food bank, a supermarket donates $36$ tins of pulses, $48$ cartons of milk and $60$ bags of rice. They want to prepare the maximum possible number of identical bags to distribute, such that each bag contains the same number of items of each type.

  1. a) What is the maximum number of bags they can prepare?
  2. b) What will each bag contain?
Solve

Back to the beginning: Breaking down into prime factors, $36 = 2^2 \cdot 3^2$ and $48 = 2^4 \cdot 3$. The $H.C.F.(36, 48) = 2^2 \cdot 3 = 12$ cm, so the maximum length of each piece is $12$ cm (you will get $3$ pieces from the first strip and $4$ from the second).

Summary

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