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Secondary 2 · Mathematics
Numbers · Natural numbers

Greatest common divisor

Before you start, make sure you know: Prime numbers and prime factorization (Secondary Year 1 (first cycle))
Before you start, make sure you know: Multiples, divisors, and divisibility rules (Secondary Year 1 (first cycle))

At a camp in Jarabacoa they have two climbing ropes, $20$ and $28$ meters long. They want to cut them into equal pieces, as long as possible, with nothing left over. The greatest common divisor answers that question.

What the GCD is

To find it, we write the divisors of each number and look for the largest one they share.

Venn diagram showing the divisors of 24 and 36 and highlighting the common divisors, marking 12 as the greatest common divisor. 8, 24 9, 18, 36 1, 2, 3, 4, 6, 12 GCD = 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 an 8th-grade 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 an 8th-grade 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

Calculating the GCD by prime factorization

For large numbers, listing divisors takes too long. The quick method: factor each number and take only the common factors, raised to the smallest exponent.

Calculation of the GCD of 12 and 18 from their prime factorization by selecting the common factors with the smallest exponent. 12 = 2² · 3 18 = 2 · 3² M.C.D. = 2 · 3 = 6 only factors common, with the smallest exponent

With the factor tree

Each number is split into two factors until all of them are prime. This gives the prime factorization.

Factor trees of 72 and 108 and calculation of their GCD equal to 36. 72236 218 29 33 108254 227 39 33 72 = 2³ · 3² 108 = 2² · 3³ M.C.D. = 2² · 3² = 36

With the vertical bar

Write the numbers to the left of a bar and divide only by primes that divide all of them at once. Stop when there is no common prime left.

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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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Two numbers are called co-prime if their only common factor is $1$. Check whether the following pairs of numbers are co-prime by calculating their highest common factor:…

Two numbers are called co-prime if their only common factor is $1$. Check whether the following pairs of numbers are co-prime by calculating their highest common factor:

  1. a) $8$ and $15$
  2. b) $14$ and $21$
  3. c) $9$ and $20$
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

The Euclidean algorithm

There is another method, a very old one, that doesn't need factoring: successive divisions.

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The Euclidean algorithm is an ancient and effective method for finding the $m.c.d.$ of two numbers using successive division. It consists of dividing the larger number…

The Euclidean algorithm is an ancient and effective method for finding the $m.c.d.$ of two numbers using successive division. It consists of dividing the larger number by the smaller one; if the remainder is not zero, the divisor is divided by the remainder, and so on until the remainder is zero. The last divisor used is the $m.c.d.$

Use this method to find the $m.c.d.(525, 225)$. Check the result by doing the usual prime factorisation afterwards.

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

Relationship between the GCD and the LCM

For two numbers, the GCD and the LCM are connected by a very useful formula.

Diagram that checks the equality between the product of 12 and 18 and the product of their GCD and their LCM. 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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Find two natural numbers, which we will call $x$ and $y$, knowing that their product is $x \times y = 11\,250$ and that their lowest common multiple is…

Find two natural numbers, which we will call $x$ and $y$, knowing that their product is $x \times y = 11\,250$ and that their lowest common multiple is $m.c.m.(x, y) = 750$.

Solve
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We have two natural numbers $A$ and $B$ such that their $m.c.d.(A, B) = 15$. If we multiply the number $A$ by $12$ and the number $B$ by $12$, what will be the new…

We have two natural numbers $A$ and $B$ such that their $m.c.d.(A, B) = 15$. If we multiply the number $A$ by $12$ and the number $B$ by $12$, what will be the new highest common factor of the resulting numbers? Justify your answer by applying the properties of factors.

Solve

Applications of the GCD

The GCD shows up whenever you want to share out or cut into equal parts as large as possible, with nothing left over: ribbons, teams, tiles, aid boxes.

Diagram showing two ropes of 20 and 28 meters cut into equal pieces of 4 meters, the greatest common divisor. 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 Jarabacoa, the instructors have two climbing ropes: one measures $20$ meters and the other $28$ meters. They want to cut them into equal pieces…

At an adventure camp in Jarabacoa, the instructors have two climbing ropes: one measures $20$ meters and the other $28$ meters. 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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A craft shop has two silk ribbons, a crimson one that is $160$ cm long and a turquoise one that is $120$ cm long. They want to cut both ribbons into pieces of the same…

A craft shop has two silk ribbons, a crimson one that is $160$ cm long and a turquoise one that is $120$ cm long. They want to cut both ribbons into pieces of the same length, so that the pieces are as long as possible and no fabric is left over.

Two uncut ribbons drawn to the same scale: the crimson one of 160 cm and the turquoise one of 120 cmCrimson: 160 cmTurquoise: 120 cm
  1. a) How long must each piece be?
  1. b) How many crimson pieces will you get? And how many turquoise pieces?
Solve
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A food bank has received $210$ tins of tuna, $126$ packets of rice and $84$ jars of pulses. They want to prepare the maximum possible number of identical aid boxes…

A food bank has received $210$ tins of tuna, $126$ packets of rice and $84$ jars of pulses. They want to prepare the maximum possible number of identical aid boxes containing all three types of food, with nothing left over.

  1. a) How many aid boxes will they be able to prepare?
  2. b) How many items of each type will there be in each box?
Solve
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A logistics company wants to transport three types of goods in identical cubic containers, as large as possible, to make the most of the space in a large cargo…

A logistics company wants to transport three types of goods in identical cubic containers, as large as possible, to make the most of the space in a large cargo warehouse. The quantities of goods are measured by volume:

  • Technology boxes: $1\,512$ $m^3$

  • Food boxes: $1\,848$ $m^3$

  • Textile boxes: $1\,260$ $m^3$

If each container must be filled with a single type of good and no space can be left over in any of them, what must the volume of each cubic container be? How many textile containers will there be in the warehouse?

Technology V₁ = 1512 m³ Food V₂ = 1848 m³ Textiles V₃ = 1260 m³ HCF V = l³ l l l
Solve
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A marine wildlife rescue center has received a donation of three heavy-duty hoses to clean the tanks. The lengths of the hoses are $156$ meters, $144$ meters and $108$…

A marine wildlife rescue center has received a donation of three heavy-duty hoses to clean the tanks. The lengths of the hoses are $156$ meters, $144$ meters and $108$ meters.

They want to cut them into pieces of equal length, with nothing left over, to distribute them among different volunteer teams. The head of the center sets the condition that the pieces must be as long as possible, but for safety reasons, no piece can measure more than $15$ meters.

What length should each piece be to meet all conditions? How many volunteers will receive a hose?

Solve

Back to the beginning: $20 = 2^2 \times 5$ and $28 = 2^2 \times 7$. Their GCD is $2^2 = 4$ m, so each piece is $4$ m long: you get $5$ pieces from the first rope and $7$ from the second.

Summary

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