Marc goes to the store with exactly $1.00 in change. He has at least one of each coin less than half a dollar coin but he does not have half a dollar coin. a. What is the least number of coins he could have, b. What is the greatest number of coins he could have?

Let's break down Marc's situation. He has **exactly $1.00** in change, and the coins available to him are pennies (1¢), nickels (5¢), and dimes (10¢). He has at least one of each coin, but **no half-dollar coins** (50¢ coins). Our goal is to determine:

**a.** The least number of coins he could have.  
**b.** The greatest number of coins he could have.

### **Part a: Least Number of Coins**
To minimize the number of coins, we should use the highest denomination possible while ensuring that Marc has at least one of each coin. Let's follow a systematic approach:

1. **Dimes (10¢)** are the highest denomination. Let's try to use the maximum number of dimes to cover as much of the dollar as possible.  
   - If Marc has 9 dimes, this gives 90¢, which is the closest we can get to 1.00 without exceeding it.
   
2. Next, Marc needs to make up the remaining 10¢ using **nickels (5¢)** and **pennies (1¢)**.
   - To complete the remaining 10¢, Marc could use **2 nickels**.

Now, Marc has:
- 9 dimes = 90¢  
- 2 nickels = 10¢

So, the **least number of coins** Marc could have is:
- **9 dimes + 2 nickels = 11 coins.**

### **Part b: Greatest Number of Coins**
To maximize the number of coins, Marc should use the smallest denominations as much as possible. This would involve using **pennies (1¢)**, **nickels (5¢)**, and **dimes (10¢)** while still having at least one of each.

1. **Pennies (1¢)**: The maximum number of pennies he can use is 99 (because 100 pennies would be a dollar, but he has to include nickels and dimes as well).
   - We start by using **99 pennies** for 99¢.
   
2. To complete the remaining 1¢ and ensure he has at least one nickel and one dime, we can use:
   - **1 nickel** (5¢) to cover the remainder of the dollar.

Thus, Marc can use:
- 99 pennies = 99¢  
- 1 nickel = 5¢  
- 1 dime = 10¢ (which can be used to balance the value)

Now, Marc has:
- 99 pennies  
- 1 nickel  
- 1 dime  

So, the **greatest number of coins** Marc could have is:
- **99 pennies + 1 nickel + 1 dime = 101 coins**.

### **Summary**:
- **a. The least number of coins Marc could have is 11.**
- **b. The greatest number of coins Marc could have is 101.**

Would you like further details or explanations on any part of this solution? Here are some related questions you might find interesting:

1. How would the solution change if Marc could also have half-dollar coins?
2. How does the number of coins change if Marc's total amount was $2.00 instead of $1.00?
3. What if Marc had no pennies and could only use nickels and dimes—how would that affect the number of coins?
4. How can this problem be modeled with an equation to represent the total value of the coins?
5. What is the largest number of pennies Marc could have if he could only use pennies and nickels?

**Tip:** When solving coin problems, it's helpful to start with the highest denomination to minimize the number of coins or the lowest denomination to maximize it.

Explore how Marc can use pennies, nickels, and dimes to make exactly $1.00, and discover the least and greatest number of coins he could have. This problem involves optimization and coin counting strategies. Suitable for Grades 4-6.

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