Suppose there is a pile of​ quarters, dimes, and pennies with a total value of ​$1.07 How much of each coin can be present without being able to make change for a​ dollar? If there are multiple selections of coins that will​ work, choose the selection with the largest total number of coins

To solve this problem, we need to find combinations of quarters (25 cents), dimes (10 cents), and pennies (1 cent) that sum to 107 cents, but we must not be able to make change for exactly 100 cents with any subset of these coins.

### Step-by-Step Solution:

1. **Identify the total value constraint:**
   \[
   25q + 10d + 1p = 107
   \]
   where $ q $, $ d $, and $ p $ represent the number of quarters, dimes, and pennies respectively.

2. **Ensure no subset sums to 100 cents:**
   We will consider different scenarios, ensuring that at no point can we form exactly 100 cents.

3. **Find combinations:**
   We aim to find $ q $, $ d $, and $ p $ that fit the equation and the condition. Let's explore the possibilities systematically.

### Exploring the Combinations:
We will start by iterating through possible values of $ q $ and check if the remaining amount can be split into dimes and pennies such that no subset can make 100 cents.

#### Case Analysis:
- **Case 1: No quarters ($ q = 0 $):**
  \[
  10d + 1p = 107
  \]
  - For $ d = 0 $: $ p = 107 $ (impossible, no subset sums to 100)
  - For $ d = 1 $: $ p = 97 $ (impossible, no subset sums to 100)
  - For $ d = 2 $: $ p = 87 $ (impossible, no subset sums to 100)
  - Continue until $ d = 10 $: $ p = 7 $

  $ d = 10 $ and $ p = 7 $ gives a valid combination as:
  \[
  10 \times 10 + 7 = 107
  \]
  and no subset can make 100 cents.

- **Case 2: One quarter ($ q = 1 $):**
  \[
  25 + 10d + 1p = 107 \implies 10d + 1p = 82
  \]
  - For $ d = 0 $: $ p = 82 $ (impossible, no subset sums to 100)
  - For $ d = 1 $: $ p = 72 $ (impossible, no subset sums to 100)
  - For $ d = 2 $: $ p = 62 $ (impossible, no subset sums to 100)
  - Continue until $ d = 8 $: $ p = 2 $

  $ d = 8 $ and $ p = 2 $ gives:
  \[
  25 + 8 \times 10 + 2 = 107
  \]
  and no subset can make 100 cents.

- **Case 3: Two quarters ($ q = 2 $):**
  \[
  50 + 10d + 1p = 107 \implies 10d + 1p = 57
  \]
  - For $ d = 0 $: $ p = 57 $ (impossible, no subset sums to 100)
  - For $ d = 1 $: $ p = 47 $ (impossible, no subset sums to 100)
  - For $ d = 5 $: $ p = 7 $

  $ d = 5 $ and $ p = 7 $ gives:
  \[
  50 + 5 \times 10 + 7 = 107
  \]
  and no subset can make 100 cents.

- **Case 4: Three quarters ($ q = 3 $):**
  \[
  75 + 10d + 1p = 107 \implies 10d + 1p = 32
  \]
  - For $ d = 0 $: $ p = 32 $ (impossible, no subset sums to 100)
  - For $ d = 3 $: $ p = 2 $

  $ d = 3 $ and $ p = 2 $ gives:
  \[
  75 + 3 \times 10 + 2 = 107
  \]
  and no subset can make 100 cents.

### Optimal Solution:
Among all valid cases, the one with the largest total number of coins is:
- **$ q = 2, d = 5, p = 7 $**:
  \[
  2 \text{ quarters}, 5 \text{ dimes}, 7 \text{ pennies}
  \]
  Total number of coins = 2 + 5 + 7 = 14 coins.

### Conclusion:
The optimal selection of coins is:
- 2 quarters
- 5 dimes
- 7 pennies

This combination totals $1.07 and does not allow making change for $1.00.

Would you like more details or have any questions?

### Suggested Questions:
1. How would the solution change if we added nickels (5 cents)?
2. What is the method to solve this problem using linear algebra?
3. Can you explain how to use dynamic programming for similar problems?
4. How can we apply this method to solve problems with other denominations?
5. What are the practical applications of this type of problem-solving?

### Tip:
When dealing with coin problems, always check for subsets that might meet specific constraints to avoid errors in calculations.

Explore how to solve a coin problem involving quarters, dimes, and pennies where the total value is $1.07 and no subset can make exactly $1.00. Detailed step-by-step solution provided.

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