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The problem is asking to find the greatest possible length Samuel could cut both the yellow ribbon (99 m) and the purple ribbon (165 m) into smaller pieces, ensuring no ribbon is left over. This requires finding the **greatest common divisor (GCD)** of 99 and 165.

Let's work through the steps to find the GCD:

1. **Prime Factorization of 99**:  
   \( 99 = 3^2 \times 11 \)

2. **Prime Factorization of 165**:  
   \( 165 = 3 \times 5 \times 11 \)

3. **Common Factors**:  
   Both numbers share the factors \( 3 \) and \( 11 \).

4. **Multiply the Common Factors**:  
   \( 3 \times 11 = 33 \)

Thus, the greatest possible length that Samuel could make the smaller pieces of ribbon is **33 meters**.

Would you like more details on the method or have any questions?

Here are some related questions:
1. How do you find the greatest common divisor (GCD) using the Euclidean algorithm?
2. What are the different methods to find the GCD of two numbers?
3. How does the concept of divisibility help in cutting materials like ribbons or wood?
4. Can prime factorization be used for very large numbers efficiently?
5. Why is finding the GCD useful in real-life situations like packaging or construction?

**Tip:** Whenever dealing with problems involving cutting or dividing into equal parts, look for the GCD to determine the largest possible segment size.

Samuel has 2 rolls of ribbon. He has 99 m of yellow ribbon and 165 m of purple ribbon. He wants to cut all of the ribbon into smaller pieces of equal length, with no ribbon left over. Work out the greatest possible length that Samuel could make the smaller pieces of ribbon.

Learn how to find the greatest possible length Samuel can use to cut 99 m of yellow ribbon and 165 m of purple ribbon into equal smaller pieces. This problem covers key number theory concepts like prime factorization and finding the greatest common divisor. Suitable for Grades 6-8.

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