Greatest Common Factor (GCF) Worksheets - Free Printable

Problem Analysis:

The image contains a series of word problems related to GCF (Greatest Common Factor) and LCM (Least Common Multiple). Each problem requires the application of these concepts to solve real-world scenarios. Below, I will solve each problem step by step.

---

Problem 1:

A high-flying event manager conducts an event every 2 days. A local artist performs at his events every 3 days. If both are scheduled for today, how long will it be until they meet again at the same time?

#### Solution:
- The event manager conducts events every 2 days.
- The artist performs every 3 days.
- To find when they will meet again at the same time, we need to calculate the LCM (Least Common Multiple) of 2 and 3.

#### Steps:
1. Prime factorization:
- \( 2 = 2 \)
- \( 3 = 3 \)

2. LCM Calculation:
- The LCM is the product of the highest powers of all prime factors involved.
- Here, the prime factors are 2 and 3.
- \( \text{LCM}(2, 3) = 2 \times 3 = 6 \).

#### Answer:
They will meet again at the same time in 6 days.

---

Problem 2:

Brian built a doghouse for his pet. He cut a piece of wood into 24 pieces of equal length and another piece of wood into 30 pieces of equal length. What is the greatest possible length of each piece?

#### Solution:
- Brian cuts one piece of wood into 24 pieces of equal length.
- He cuts another piece of wood into 30 pieces of equal length.
- To find the greatest possible length of each piece, we need to determine the GCF (Greatest Common Factor) of 24 and 30.

#### Steps:
1. Prime factorization:
- \( 24 = 2^3 \times 3 \)
- \( 30 = 2 \times 3 \times 5 \)

2. GCF Calculation:
- The GCF is the product of the lowest powers of all common prime factors.
- The common prime factors are 2 and 3.
- \( \text{GCF}(24, 30) = 2^1 \times 3^1 = 6 \).

#### Answer:
The greatest possible length of each piece is 6 units.

---

Problem 3:

Scott wants to arrange identical flower pots on either side of a table so that there are no gaps between the flower pots. He has two types of flower pots: square ones with sides measuring 18 cm and round ones with diameters of 12 cm. What is the least size table he can buy so that there are no gaps between the flower pots and all of them fit along the table's edge?

#### Solution:
- The square flower pots have sides of 18 cm.
- The round flower pots have diameters of 12 cm.
- To ensure there are no gaps between the flower pots, the table's width must be a multiple of both 18 cm and 12 cm.
- We need to find the LCM (Least Common Multiple) of 18 and 12.

#### Steps:
1. Prime factorization:
- \( 18 = 2 \times 3^2 \)
- \( 12 = 2^2 \times 3 \)

2. LCM Calculation:
- The LCM is the product of the highest powers of all prime factors involved.
- The prime factors are 2 and 3.
- \( \text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36 \).

#### Answer:
The least size table Scott can buy is 36 cm wide.

---

Problem 4:

Itsuo’s house has 2 living rooms, 4 bedrooms, and 2 bathrooms. He wants to install carpet tiles in all the rooms. Each room should have the same number of tiles, and there should be no leftover tiles. What is the greatest number of tiles Itsuo can use for each room if he has 100 tiles available?

#### Solution:
- Itsuo has 100 tiles in total.
- He needs to divide these tiles equally among 2 living rooms, 4 bedrooms, and 2 bathrooms, which totals:
\[
2 + 4 + 2 = 8 \text{ rooms}
\]
- To find the greatest number of tiles per room without any leftovers, we need to determine the GCF (Greatest Common Factor) of 100 and 8.

#### Steps:
1. Prime factorization:
- \( 100 = 2^2 \times 5^2 \)
- \( 8 = 2^3 \)

2. GCF Calculation:
- The GCF is the product of the lowest powers of all common prime factors.
- The only common prime factor is 2.
- \( \text{GCF}(100, 8) = 2^2 = 4 \).

#### Verification:
- If each room gets 4 tiles, the total number of tiles used is:
\[
4 \times 8 = 32 \text{ tiles}
\]
- Since 100 tiles are available, this division is valid.

#### Answer:
The greatest number of tiles Itsuo can use for each room is 4 tiles.

---

Problem 5:

The factory sold an automobile part every 3 minutes and a rubber ball every 5 minutes. If the factory starts selling both parts at the same time, after how many minutes will the factory sell both parts at the same time again?

#### Solution:
- The factory sells automobile parts every 3 minutes.
- The factory sells rubber balls every 5 minutes.
- To find when both parts will be sold at the same time again, we need to calculate the LCM (Least Common Multiple) of 3 and 5.

#### Steps:
1. Prime factorization:
- \( 3 = 3 \)
- \( 5 = 5 \)

2. LCM Calculation:
- The LCM is the product of the highest powers of all prime factors involved.
- The prime factors are 3 and 5.
- \( \text{LCM}(3, 5) = 3 \times 5 = 15 \).

#### Answer:
The factory will sell both parts at the same time again after 15 minutes.

---

Final Answers:

1. \(\boxed{6}\)
2. \(\boxed{6}\)
3. \(\boxed{36}\)
4. \(\boxed{4}\)
5. \(\boxed{15}\)