GCF and LCM WORD PROBLEMS - Free Printable

Problem: GCF and LCM Word Problems

We are tasked with solving a series of word problems involving the Greatest Common Factor (GCF) and Least Common Multiple (LCM). Let's solve each problem step by step.

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1. Joanne is campaigning for class president and plans to distribute some campaign materials: 20 flyers and 16 buttons. She wants each classroom to receive an identical set of campaign materials, without having any materials left over. What is the greatest number of classrooms Joanne can distribute materials to?

#### Solution:
- We need to find the greatest number of classrooms such that each classroom receives an identical set of flyers and buttons, with no leftovers.
- This means we need to find the Greatest Common Factor (GCF) of 20 and 16.
- Prime factorization:
- \( 20 = 2^2 \times 5 \)
- \( 16 = 2^4 \)
- The common prime factor is \( 2 \), and the smallest power of 2 in both factorizations is \( 2^2 = 4 \).
- Therefore, the GCF of 20 and 16 is \( 4 \).

#### Answer:
The greatest number of classrooms Joanne can distribute materials to is 4.

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2. Serena wants to create snack bags for a trip she is going on. She has 6 granola bars and 10 pieces of dried fruit. If the snack bags should be identical without any food left over, what is the greatest number of snack bags Serena can make?

#### Solution:
- We need to find the greatest number of identical snack bags such that each bag contains the same number of granola bars and dried fruit, with no leftovers.
- This means we need to find the Greatest Common Factor (GCF) of 6 and 10.
- Prime factorization:
- \( 6 = 2 \times 3 \)
- \( 10 = 2 \times 5 \)
- The common prime factor is \( 2 \).
- Therefore, the GCF of 6 and 10 is \( 2 \).

#### Answer:
The greatest number of snack bags Serena can make is 2.

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3. Matthew goes hiking every 12 days and swimming every 6 days. He did both kinds of exercise today. How many days from now will he go both hiking and swimming again?

#### Solution:
- We need to find the next time Matthew will do both hiking and swimming on the same day.
- This means we need to find the Least Common Multiple (LCM) of 12 and 6.
- Prime factorization:
- \( 12 = 2^2 \times 3 \)
- \( 6 = 2 \times 3 \)
- The LCM is found by taking the highest power of each prime factor:
- \( 2^2 \) (from 12)
- \( 3 \) (common in both)
- Therefore, the LCM of 12 and 6 is \( 2^2 \times 3 = 4 \times 3 = 12 \).

#### Answer:
Matthew will go both hiking and swimming again in 12 days.

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4. Mandy is making emergency-preparedness kits to share with friends. She has 12 bottles of water and 16 cans of food, which she would like to distribute equally among the kits, with nothing left over. What is the greatest number of kits Mandy can make?

#### Solution:
- We need to find the greatest number of kits such that each kit contains the same number of bottles of water and cans of food, with no leftovers.
- This means we need to find the Greatest Common Factor (GCF) of 12 and 16.
- Prime factorization:
- \( 12 = 2^2 \times 3 \)
- \( 16 = 2^4 \)
- The common prime factor is \( 2 \), and the smallest power of 2 in both factorizations is \( 2^2 = 4 \).
- Therefore, the GCF of 12 and 16 is \( 4 \).

#### Answer:
The greatest number of kits Mandy can make is 4.

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5. Edeena is packing equal numbers of apple slices and grapes for snacks. She bags the apple slices in groups of 18 and the grapes in groups of 9. What is the smallest number of grapes that she can pack?

#### Solution:
- We need to find the smallest number of grapes that can be packed such that the number of grapes is a multiple of both 18 (apple slices) and 9 (grapes).
- This means we need to find the Least Common Multiple (LCM) of 18 and 9.
- Prime factorization:
- \( 18 = 2 \times 3^2 \)
- \( 9 = 3^2 \)
- The LCM is found by taking the highest power of each prime factor:
- \( 2 \) (from 18)
- \( 3^2 \) (common in both)
- Therefore, the LCM of 18 and 9 is \( 2 \times 3^2 = 2 \times 9 = 18 \).

#### Answer:
The smallest number of grapes that Edeena can pack is 18.

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6. A club has 16 girls and 8 boys as members. The president wants to break the club into groups, with each group containing the same combination of girls and boys. The president also wants to make sure that no one is left out. What is the greatest number of groups the president can make?

#### Solution:
- We need to find the greatest number of groups such that each group has the same number of girls and boys, with no one left out.
- This means we need to find the Greatest Common Factor (GCF) of 16 and 8.
- Prime factorization:
- \( 16 = 2^4 \)
- \( 8 = 2^3 \)
- The common prime factor is \( 2 \), and the smallest power of 2 in both factorizations is \( 2^3 = 8 \).
- Therefore, the GCF of 16 and 8 is \( 8 \).

#### Answer:
The greatest number of groups the president can make is 8.

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7. Ariel is making flower arrangements. He has 7 roses and 14 daisies. If Ariel wants to make all the arrangements identical and have no flowers left over, what is the greatest number of flower arrangements that he can make?

#### Solution:
- We need to find the greatest number of identical flower arrangements such that each arrangement contains the same number of roses and daisies, with no leftovers.
- This means we need to find the Greatest Common Factor (GCF) of 7 and 14.
- Prime factorization:
- \( 7 = 7 \)
- \( 14 = 2 \times 7 \)
- The common prime factor is \( 7 \).
- Therefore, the GCF of 7 and 14 is \( 7 \).

#### Answer:
The greatest number of flower arrangements Ariel can make is 7.

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8. Wilma is thinking of a number that is divisible by both 17 and 8. What is the smallest possible number that Wilma could be thinking of?

#### Solution:
- We need to find the smallest number that is divisible by both 17 and 8.
- This means we need to find the Least Common Multiple (LCM) of 17 and 8.
- Prime factorization:
- \( 17 = 17 \) (prime)
- \( 8 = 2^3 \)
- Since 17 and 8 have no common factors other than 1, the LCM is simply their product:
- \( \text{LCM} = 17 \times 8 = 136 \).

#### Answer:
The smallest possible number Wilma could be thinking of is 136.

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Final Answers:

1. 4
2. 2
3. 12
4. 4
5. 18
6. 8
7. 7
8. 136

\[
\boxed{4, 2, 12, 4, 18, 8, 7, 136}
\]