Greatest Common Factor (GCD) Exercises

O greatest common divisor (MDC), between two or more numbers, is a number that divides all of them and is also the largest possible number.

We can determine the GCD by finding all the divisors of each number and then finding the greatest common divisor between them.

However, a practical way to calculate MDC is from the decomposition into prime factors. In this case, the GCD is given by the product of the lowest exponent common factors.

To learn more about this subject, check out a list of greatest common divisor (GCD) exercises with resolution.

Greatest common factor (GCD) exercise list

---

Question 1. Find all divisors of 8 and 12 and determine the GCD between them.

---

Question 2. Find all divisors of 6 and 9 and 15 and determine the GCD between them.

---

Question 3. Decompose the numbers 18 and 21 into prime factors and calculate the GCD between them.

---

Question 4.

Decompose the numbers 72, 81 and 126 into prime factors and calculate the GCD between them.

---

Question 5. What is the largest number by which we can simultaneously divide the numbers 48 and 98?

---

Question 6. A teacher has 16 meters of blue ribbon and 24 meters of red ribbon. She wants to cut them into pieces that are the same size but as long as possible.

How big will each ribbon be and how many blue and red ribbons will she get?

---

Question 7. A merchant wants to place 5200 tomatoes and 3400 potatoes in boxes such that each box has the same quantity and is as large as possible.

Determine the number of tomatoes and potatoes in each box and the number of boxes needed.

---

Question 8. A producer of whole juice has three branches and wants to transport the bottles produced, per day, in each of them, in trucks that carry the same amount and that is the largest possible.

If daily productions are 240, 300, and 360 bottles, how many bottles must each truck carry? How many trucks per branch?

---

Resolution of question 1

Divisors of each number:

D(8) = {1, 2, 4, 8}
D(12) = {1, 2, 3, 4, 6, 12}

Common divisors: 1, 2 and 4
Greatest common divisor: 4

GCD(8,12) = 4

Resolution of question 2

Divisors of each number:

D(6) = {1, 2, 3, 6}
D(9) = {1, 3, 9}
D(15) = {1, 3, 5, 15}

Common divisors: 1, 2, 3
Greatest common divisor: 3

GCD(6, 9, 15) = 3

Resolution of question 3

Decomposition into prime factors of 18:

18 | 2
9 | 3
3 | 3
1 ⇒ 18 = 2. 3. 3

Decomposition into prime factors of 21:

21 | 3
7 | 7
1 ⇒ 21 = 3. 7

So 18 and 21 only have one factor in common: 3

So GCD(18, 21) = 3.

Resolution of question 4

Decomposition into prime factors of 72:

72 | 2
36 | 2
18 | 2
9 | 3
3 | 3
1 ⇒ 72 = 2. 2. 2. 3. 3

Decomposition into prime factors of 81:

81 | 3
27 | 3
9 | 3
3 | 3
1 ⇒ 81 = 3. 3. 3. 3

Decomposition into prime factors of 126:

126 | 2
63 | 3
21 | 3
7 | 7
1 ⇒ 126 = 2. 3. 3. 7

MDC(72, 81, 126) = 3. 3 = 9

Resolution of question 5

The largest number by which we can divide 48 and 98 simultaneously is the GCD between them.

Decomposition into prime factors of 48:

48 | 2
24 | 2
12 | 2
6 | 2
3 | 3
1 ⇒ 48 = 2. 2. 2. 2. 3

Decomposition into prime factors of 98:

98 | 2
49 | 7
7 | 7
1 ⇒ 98 = 2. 7. 7

GCD(48, 98) = 2

So the largest number that we can divide both the numbers 48 and 98 by is the number 2.

Resolution of question 6

The longest possible length, equal between the blue and red ribbons, is the MDC between 16 and 24.

Decomposition into prime factors of 16:

16 | 2
8 | 2
4 | 2
2 | 2
1 ⇒ 16 = 2. 2. 2. 2

Decomposition into prime factors of 24:

24 | 2
12 | 2
6 | 2
3 | 3
1 ⇒ 24 = 2. 2. 2. 3

GCD(16, 24) = 2. 2. 2 = 8

Therefore, each piece of tape should be 8 meters long.

16: 8 = 2 ⇒ will be 2 blue ribbons.
24: 8 = 3 ⇒ will be 3 red ribbons.

Resolution of question 7

The largest amount in each box, the same for tomatoes and potatoes, is the MDC between 5200 and 3400.

Decomposition into prime factors of 5200:

5200 | 2
2600 | 2
1300 | 2
650 | 2
325 | 5
65 | 5
13 | 13
1 ⇒ 5200 = 2. 2. 2. 2. 5. 5. 13

Decomposition into prime factors of 3400:

3400 | 2
1700 | 2
850 | 2
425 | 5
85 | 5
17 |17
1 ⇒ 5200 = 2. 2. 2. 5. 5. 17

MDC(5200, 3400) = 2. 2. 2. 5. 5 = 200

Therefore, each box should have 200 tomatoes or potatoes.

5200: 200 = 26 ⇒ that's 26 boxes of tomatoes.
3400: 200 = 17 ⇒ that's 17 crates of potatoes.

In all, you will need 26 + 17 = 43 boxes.

Resolution of question 8

The largest number of bottles transported in each truck, the same for the three branches, is the MDC between 240, 300 and 360.

Decomposition into prime factors of 240:

240 | 2
120 | 2
60 | 2
30 | 2
15 | 3
5 | 5
1 ⇒ 240 = 2. 2. 2. 2. 3. 5

Decomposition into prime factors of 300:

300 | 2
150 | 2
75 | 3
25 | 5
5 | 5
1 ⇒ 300 = 2. 2. 3. 5. 5

Decomposition into prime factors of 360:

360 | 2
180 | 2
90 | 2
45 | 3
15 | 3
5 | 5
1 ⇒ 360 = 2. 2. 2. 3. 3. 5

MDC(240, 300, 360) = 2. 2. 3. 5 = 60

Therefore, each truck must transport 60 bottles of juice.

240: 60 = 4 ⇒ there will be 4 trucks for the branch that produces 240 bottles.
300: 60 = 5 ⇒ there will be 5 trucks for the branch that produces 300 bottles.
360: 60 = 6 ⇒ there will be 6 trucks for the branch that produces 360 bottles.

You may also be interested:

• List of Least Common Multiple Exercises – MMC
• List of exercises on multiples and divisors
• List of Prime and Composite Number Exercises