Jack is making snack bags. He has 18 baby carrots and 42 pretzels. He wants to divide them equally among the bags. What is the greatest number of snack bags he can make?




To determine the greatest number of snack bags Jack can make with 18 baby carrots and 42 pretzels, we need to find the greatest common divisor (GCD) of the two numbers.

1. **Factors of 18:** 
   - 1, 2, 3, 6, 9, 18

2. **Factors of 42:** 
   - 1, 2, 3, 6, 7, 14, 21, 42

3. **Common factors of 18 and 42:** 
   - 1, 2, 3, 6

4. **Greatest Common Divisor:** 
   - The GCD is 6.

Thus, the greatest number of snack bags Jack can make is **6 bags**. 

Each bag would contain:
- \( \frac{18}{6} = 3 \) baby carrots 
- \( \frac{42}{6} = 7 \) pretzels 

Would you like more details or have any questions? Here are some related questions you might find interesting:

1. How many baby carrots and pretzels would be left over after making the bags?
2. If Jack had 24 baby carrots instead, how many bags could he make then?
3. What would be the total number of snacks in each bag if he combined both snacks?
4. How would the situation change if Jack wanted to make only 4 bags instead?
5. Can you explain how to find the GCD using the prime factorization method?

**Tip:** Always look for the GCD when dividing items into equal parts to maximize the number of groups.

Jack is making snack bags. He has 18 baby carrots and 42 pretzels. He wants to divide them equally among the bags. What is the greatest number of snack bags he can make?

Discover how to calculate the greatest number of snack bags Jack can make using 18 baby carrots and 42 pretzels. This guide explains finding the greatest common divisor (GCD) and includes step-by-step solutions. Ideal for Grades 5-7.

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