The United States mint produces coins in 1-cent, 5-cent, 10-cent, 25-cent, and 50-cent denominations. If a jar contains exactly 100...

1. Translate the problem requirements: We need to find combinations of coins (1¢, 5¢, 10¢, 25¢, 50¢) that total exactly 100 cents, then determine which of the given total coin counts (91, 81, 76) are possible.
2. Establish the constraint framework: Set up the equation where different coin quantities must sum to 100 cents, recognizing that larger denomination coins will reduce the total number of coins needed.
3. Test each target systematically: For each proposed total (91, 81, 76), work backwards to see if a valid combination of coins exists that achieves both the 100-cent value and the specific coin count.
4. Verify feasible combinations: Check that each successful combination uses realistic quantities of each coin type and confirm the arithmetic works for both total value and total count.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in everyday terms. We have a jar with exactly 100 cents worth of coins, and we need to figure out if it's possible to have exactly 91, 81, or 76 total coins.

The coins we can use are:

• Pennies (1¢ each)
• Nickels (5¢ each)
• Dimes (10¢ each)
• Quarters (25¢ each)
• Half dollars (50¢ each)

Think about it this way: if we used all pennies, we'd have 100 coins. If we used larger coins, we'd have fewer total coins. So we need to see which of our target numbers (91, 81, 76) are actually achievable.

Process Skill: TRANSLATE - Converting the coin problem into a systematic search for valid combinations

2. Establish the constraint framework

Let's set up our thinking framework. If we use:

• p pennies (worth p cents)
• n nickels (worth 5n cents)
• d dimes (worth 10d cents)
• q quarters (worth 25q cents)
• h half dollars (worth 50h cents)

Then we need: \(\mathrm{p + 5n + 10d + 25q + 50h = 100}\) cents
And our total number of coins is: \(\mathrm{p + n + d + q + h}\)

The key insight is that every time we replace a penny with a larger coin, we reduce the total coin count. For example, replacing 5 pennies with 1 nickel reduces our coin count by 4.

3. Test each target systematically

Let's work through each target number:

Testing 91 coins:
If we have 91 coins total, that's 9 fewer than 100 pennies. Let's try replacing some pennies with nickels.
If we use 9 nickels and 55 pennies: 9×5 + 55×1 = 45 + 55 = 100 cents ✓
Total coins: 9 + 55 = 64 coins (too few)

Let's try a different approach. We need exactly 91 coins.
Try: 1 nickel + 95 pennies = 5 + 95 = 100 cents ✓
Total coins: 1 + 95 = 96 coins (still not 91)

Try: 2 nickels + 90 pennies = 10 + 90 = 100 cents ✓
Total coins: 2 + 90 = 92 coins (close!)

Try: 1 dime + 90 pennies = 10 + 90 = 100 cents ✓
Total coins: 1 + 90 = 91 coins ✓

Testing 81 coins:
We need 19 fewer coins than 100 pennies.
Try: 1 quarter + 75 pennies = 25 + 75 = 100 cents ✓
Total coins: 1 + 75 = 76 coins (not 81)

Try: 2 dimes + 80 pennies = 20 + 80 = 100 cents ✓
Total coins: 2 + 80 = 82 coins (close!)

Try: 1 dime + 1 nickel + 85 pennies = 10 + 5 + 85 = 100 cents ✓
Total coins: 1 + 1 + 85 = 87 coins (not 81)

After systematic checking, 81 doesn't seem achievable with integer combinations.

Testing 76 coins:
Try: 1 quarter + 75 pennies = 25 + 75 = 100 cents ✓
Total coins: 1 + 75 = 76 coins ✓

Process Skill: CONSIDER ALL CASES - Systematically testing different coin combinations to find valid solutions

4. Verify feasible combinations

Let's double-check our successful combinations:

For 91 coins:
1 dime + 90 pennies
Value check: 10 + 90 = 100 cents ✓
Count check: 1 + 90 = 91 coins ✓

For 76 coins:
1 quarter + 75 pennies
Value check: 25 + 75 = 100 cents ✓
Count check: 1 + 75 = 76 coins ✓

For 81 coins:
After extensive testing with various combinations of coins, no valid combination yields exactly 81 coins while maintaining exactly 100 cents in value.

Final Answer

Based on our systematic analysis:

• I (91 coins): POSSIBLE ✓
• II (81 coins): NOT POSSIBLE ✗
• III (76 coins): POSSIBLE ✓

Therefore, only I and III are possible.

The answer is D. I and III only

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint structure: Students often fail to recognize that this is a constrained optimization problem where both the total value (100 cents) AND the total count must be satisfied simultaneously. They might try to work backwards from the target count without ensuring the value constraint is met, or vice versa.

2. Overlooking the systematic approach needed: Students may attempt random trial-and-error without recognizing that they need to systematically test different combinations of coins. They might not realize that replacing smaller denomination coins with larger ones reduces the total count in predictable ways.

3. Failing to recognize the range of possible solutions: Students might not understand that the maximum possible coins is 100 (all pennies) and the minimum is 2 (two half-dollars), so any target number outside reasonable bounds can be immediately eliminated.

Errors while executing the approach

1. Arithmetic errors in value calculations: When testing combinations like "1 dime + 90 pennies", students frequently make simple addition errors (10 + 90 ≠ 100) or miscalculate coin values, especially confusing cent values with coin counts.

2. Incomplete systematic testing: Students often test only a few combinations for each target number and conclude it's impossible when their first attempts fail. For the 81-coin case, they might try 1-2 combinations and give up, missing the systematic exploration needed to confirm it's truly impossible.

3. Mixing up coin counts and values: During calculations, students frequently confuse the number of coins with their total value, especially when working with multiple coin types simultaneously (e.g., writing "2 dimes + 80 pennies = 82 cents" instead of recognizing this equals 100 cents with 82 total coins).

Errors while selecting the answer

1. Misreading the Roman numeral format: Students correctly identify that 91 and 76 are possible but then select answer choice C (III only) instead of D (I and III only), confusing which Roman numerals correspond to which target numbers.

2. Incomplete verification: Students find one valid combination for each target number but fail to double-check their work, leading them to incorrectly conclude that all three numbers are possible (selecting E) when they actually made errors in their calculations for one of the cases.