Last Year a company gave bonuses to a number of employees, but only in the three amounts of $750, $1{,}500,...

1. Translate the problem requirements: We need to find the minimum number of bonuses given that the total is \(\$64,800\), using only three amounts (\(\$750, \$1,500, \$7,350\)), with at least one bonus of each amount given.
2. Set up the constraint equation: Express the total bonus amount as a combination of the three bonus types with unknown quantities.
3. Apply the minimization strategy: To minimize total bonuses, maximize the use of the largest bonus amount while respecting the constraint that each amount appears at least once.
4. Test the optimal allocation: Calculate if using maximum \(\$7,350\) bonuses (with minimum required \(\$750\) and \(\$1,500\) bonuses) satisfies our total exactly.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what the problem is asking us to find. We have a company that gave bonuses in exactly three amounts: \(\$750\), \(\$1,500\), and \(\$7,350\). The total amount spent on bonuses was \(\$64,800\), and we know that at least one person received each of the three bonus amounts.

Our goal is to find the smallest possible number of employees who could have received bonuses. Think of it like this: if we want to minimize the number of bonuses, we should try to use as many large bonuses as possible, since fewer large bonuses can add up to the same total as many small bonuses.

Let's call the number of \(\$750\) bonuses 'a', the number of \(\$1,500\) bonuses 'b', and the number of \(\$7,350\) bonuses 'c'. We know that \(\mathrm{a} \geq 1\), \(\mathrm{b} \geq 1\), and \(\mathrm{c} \geq 1\) (each amount must be given at least once).

2. Set up the constraint equation

Now we need to express our total bonus amount mathematically. Since we have 'a' bonuses of \(\$750\), 'b' bonuses of \(\$1,500\), and 'c' bonuses of \(\$7,350\), our total is:

\(750\mathrm{a} + 1500\mathrm{b} + 7350\mathrm{c} = 64800\)

We can simplify this by dividing everything by 150 (the greatest common factor of all the bonus amounts):

\(5\mathrm{a} + 10\mathrm{b} + 49\mathrm{c} = 432\)

This is much easier to work with! Now we want to minimize the total number of bonuses, which is \(\mathrm{a} + \mathrm{b} + \mathrm{c}\), subject to the constraints that \(\mathrm{a} \geq 1\), \(\mathrm{b} \geq 1\), \(\mathrm{c} \geq 1\).

3. Apply the minimization strategy

To minimize the total number of bonuses, we should use as many of the largest bonuses (\(\$7,350\)) as possible. Since we need at least one of each type, let's start with the minimum: \(\mathrm{a} = 1\), \(\mathrm{b} = 1\), and see how many \(\$7,350\) bonuses we can give.

Substituting \(\mathrm{a} = 1\) and \(\mathrm{b} = 1\) into our equation:

\(5(1) + 10(1) + 49\mathrm{c} = 432\)

\(5 + 10 + 49\mathrm{c} = 432\)

\(15 + 49\mathrm{c} = 432\)

\(49\mathrm{c} = 417\)

\(\mathrm{c} = 417 \div 49 = 8.51...\)

Since c must be a whole number, we can't use exactly 8.51 bonuses of \(\$7,350\). Let's try \(\mathrm{c} = 8\):

\(5\mathrm{a} + 10\mathrm{b} + 49(8) = 5\mathrm{a} + 10\mathrm{b} + 392 = 432\)

\(5\mathrm{a} + 10\mathrm{b} = 40\)

\(\mathrm{a} + 2\mathrm{b} = 8\)

4. Test the optimal allocation

Now we need to find values of a and b (both \(\geq 1\)) that satisfy \(\mathrm{a} + 2\mathrm{b} = 8\).

Let's try different values:

• If \(\mathrm{b} = 1\): \(\mathrm{a} + 2(1) = 8\), so \(\mathrm{a} = 6\)
• If \(\mathrm{b} = 2\): \(\mathrm{a} + 2(2) = 8\), so \(\mathrm{a} = 4\)
• If \(\mathrm{b} = 3\): \(\mathrm{a} + 2(3) = 8\), so \(\mathrm{a} = 2\)
• If \(\mathrm{b} = 4\): \(\mathrm{a} + 2(4) = 8\), so \(\mathrm{a} = 0\) (not allowed since \(\mathrm{a} \geq 1\))

All valid combinations give us the same total: \(\mathrm{a} + \mathrm{b} + \mathrm{c} = 6 + 1 + 8 = 15\), or \(4 + 2 + 8 = 14\), or \(2 + 3 + 8 = 13\).

Wait, let's try \(\mathrm{c} = 9\) to see if we can do better:

\(5\mathrm{a} + 10\mathrm{b} + 49(9) = 5\mathrm{a} + 10\mathrm{b} + 441 = 432\)

\(5\mathrm{a} + 10\mathrm{b} = -9\)

This gives us negative values, which isn't possible.

So our best solution uses \(\mathrm{c} = 8\), and the minimum total is 13 bonuses (with \(\mathrm{a} = 2\), \(\mathrm{b} = 3\), \(\mathrm{c} = 8\)).

Let's verify: \(2(\$750) + 3(\$1,500) + 8(\$7,350) = \$1,500 + \$4,500 + \$58,800 = \$64,800\) ✓

Final Answer

The fewest number of bonuses the company could have given is 13.

This corresponds to answer choice D.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the optimization goal: Students might think they need to maximize the number of bonuses instead of minimizing it. This confusion stems from not carefully reading "fewest number of bonuses" and leads to using as many small bonuses as possible rather than large ones.

2. Overlooking the constraint requirements: Students may miss that "each of the three amounts was given to at least one employee" means \(\mathrm{a} \geq 1\), \(\mathrm{b} \geq 1\), \(\mathrm{c} \geq 1\). They might try to solve by setting some variables to zero, which violates the problem conditions.

3. Not simplifying the equation: Students might work directly with \(750\mathrm{a} + 1500\mathrm{b} + 7350\mathrm{c} = 64800\) instead of dividing by the GCD (150) to get \(5\mathrm{a} + 10\mathrm{b} + 49\mathrm{c} = 432\). This makes calculations much more complex and error-prone.

Errors while executing the approach

1. Arithmetic errors in division: When calculating \(\mathrm{c} = 417 \div 49\), students might incorrectly compute this as exactly 8 or 9 instead of recognizing it's 8.51... Since c must be an integer, they need to test both \(\mathrm{c} = 8\) and potentially other values systematically.

2. Forgetting to check all valid combinations: After finding \(\mathrm{a} + 2\mathrm{b} = 8\) with \(\mathrm{c} = 8\), students might only test one combination (like \(\mathrm{a} = 6, \mathrm{b} = 1\)) and conclude the answer is 15, missing the optimal solution of \(\mathrm{a} = 2, \mathrm{b} = 3, \mathrm{c} = 8\) that gives 13 total bonuses.

3. Not verifying the final calculation: Students might make errors when checking their solution by substituting back into the original equation \(750\mathrm{a} + 1500\mathrm{b} + 7350\mathrm{c} = 64800\), leading to incorrect validation of their answer.

Errors while selecting the answer

1. Selecting the first valid solution found: Students might find one valid combination (like \(\mathrm{a} = 6, \mathrm{b} = 1, \mathrm{c} = 8\) giving 15 bonuses) and immediately select answer choice E (14) as the closest option, without systematically checking all possible combinations to find the true minimum of 13.