Jessica is making stationery sets from 10 sheets of paper and 15 envelopes. If she wants all the sets to be identical without any paper or envelopes left over, what is the greatest number of sets Jessica can make?


To determine the greatest number of identical stationery sets Jessica can make, we need to find the greatest common divisor (GCD) of 10 (sheets of paper) and 15 (envelopes). The GCD will tell us how many sets she can create without leaving any paper or envelopes unused.

### Step 1: Find the prime factorizations
- \( 10 = 2 \times 5 \)
- \( 15 = 3 \times 5 \)

### Step 2: Identify the common factors
The only common factor between 10 and 15 is 5.

### Step 3: Conclusion
The greatest common divisor (GCD) is 5, meaning Jessica can make 5 identical sets with no leftovers. Each set will contain:
- \( \frac{10}{5} = 2 \) sheets of paper
- \( \frac{15}{5} = 3 \) envelopes

Thus, **the greatest number of sets Jessica can make is 5**.

Would you like further clarification or additional details?

### Related Questions:
1. What is the least common multiple (LCM) of 10 and 15, and what does it represent in this context?
2. If Jessica had 20 sheets of paper and 30 envelopes, how many sets could she make?
3. How would the number of sets change if Jessica had an odd number of paper sheets, such as 13?
4. Can you explain how the Euclidean algorithm can be used to find the GCD of 10 and 15?
5. If Jessica uses 3 sheets of paper and 4 envelopes per set, how many sets can she make from 30 sheets and 40 envelopes?

### Tip:
When solving problems involving divisibility, always look for the GCD to maximize the number of identical groups you can form!

Jessica is making stationery sets from 10 sheets of paper and 15 envelopes. If she wants all the sets to be identical without any paper or envelopes left over, what is the greatest number of sets Jessica can make?

Discover how to find the greatest number of identical stationery sets Jessica can make from 10 sheets of paper and 15 envelopes. This problem explores the concept of the greatest common divisor (GCD) and provides a step-by-step solution. Perfect for students in Grades 5-7.

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