A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37...

1. Translate the problem requirements: We have \(32 \text{ schools} \times 2 \text{ classes} = 64\) total classes, with 37 teachers. Each teacher teaches 1-3 classes. We need to find the minimum and maximum possible values of n (teachers who teach exactly 3 classes).
2. Set up the constraint equation: Express the total classes (64) in terms of teachers teaching 1, 2, and 3 classes respectively, creating a mathematical relationship between these variables.
3. Find the minimum value of n: Determine if it's possible for \(\mathrm{n} = 0\) by checking if 37 teachers can cover 64 classes using only 1 and 2 classes per teacher, then verify feasibility.
4. Find the maximum value of n: Maximize n by minimizing teachers who teach 1 or 2 classes, using the constraint that we need exactly 37 teachers total and exactly 64 classes total.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we know in plain English:

• We have 32 elementary schools, each trying out a program in exactly 2 classes
• This gives us a total of \(32 \times 2 = 64\) classes that need teachers
• We have exactly 37 teachers available to teach these classes
• Each class has exactly 1 teacher (no team teaching)
• Each teacher must teach at least 1 class but no more than 3 classes

We need to find the smallest and largest possible number of teachers who teach exactly 3 classes. Let's call this number 'n'.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical constraints

2. Set up the constraint equation

Let's think about this systematically. If we have 37 teachers total, we can group them by how many classes they teach:

• Let \(\mathrm{a}\) = number of teachers who teach exactly 1 class
• Let \(\mathrm{b}\) = number of teachers who teach exactly 2 classes
• Let \(\mathrm{n}\) = number of teachers who teach exactly 3 classes

Now we can write two equations in plain English:

1) All teachers must be accounted for: \(\mathrm{a} + \mathrm{b} + \mathrm{n} = 37\)
2) All classes must be covered: \((1 \times \mathrm{a}) + (2 \times \mathrm{b}) + (3 \times \mathrm{n}) = 64\)

This gives us: \(\mathrm{a} + 2\mathrm{b} + 3\mathrm{n} = 64\)

3. Find the minimum value of n

To minimize n, let's see if n can equal 0. This means no teachers teach 3 classes.

If \(\mathrm{n} = 0\), our equations become:
• \(\mathrm{a} + \mathrm{b} = 37\) (we need 37 teachers)
• \(\mathrm{a} + 2\mathrm{b} = 64\) (we need to cover 64 classes)

Let's solve this step by step:
From the first equation: \(\mathrm{a} = 37 - \mathrm{b}\)
Substituting into the second equation: \((37 - \mathrm{b}) + 2\mathrm{b} = 64\)
Simplifying: \(37 + \mathrm{b} = 64\)
Therefore: \(\mathrm{b} = 27\)

This gives us \(\mathrm{a} = 37 - 27 = 10\)

Let's check: 10 teachers teaching 1 class + 27 teachers teaching 2 classes = \(10 + 54 = 64\) classes ✓
And we used exactly \(10 + 27 = 37\) teachers ✓

Since both values are positive integers, \(\mathrm{n} = 0\) is possible.

Process Skill: APPLY CONSTRAINTS - Checking that our solution satisfies all given conditions

4. Find the maximum value of n

To maximize n, we want to minimize the number of teachers teaching 1 or 2 classes.

The most efficient way is to have as many teachers as possible teaching 3 classes, and fill the remaining classes with teachers teaching 2 classes (since 2 > 1).

Let's work backwards. If we have n teachers teaching 3 classes, they cover \(3\mathrm{n}\) classes.
Remaining classes to cover: \(64 - 3\mathrm{n}\)
Remaining teachers available: \(37 - \mathrm{n}\)

For maximum efficiency, let's try to use only teachers teaching 2 classes for the remainder:
If all remaining teachers teach 2 classes: \(2(37 - \mathrm{n})\) classes would be covered

We need: \(3\mathrm{n} + 2(37 - \mathrm{n}) \geq 64\)
Expanding: \(3\mathrm{n} + 74 - 2\mathrm{n} \geq 64\)
Simplifying: \(\mathrm{n} + 74 \geq 64\)
Therefore: \(\mathrm{n} \geq -10\)

But we also need: \(3\mathrm{n} + 2(37 - \mathrm{n}) \leq 64\) (we can't exceed 64 classes)
This gives us: \(\mathrm{n} + 74 \leq 64\), so \(\mathrm{n} \leq -10\)

Wait, this suggests we need exactly \(\mathrm{n} = -10\), which is impossible. Let me reconsider.

Let's be more systematic. We need exactly 64 classes covered:
\(3\mathrm{n} + 2\mathrm{b} + \mathrm{a} = 64\), where \(\mathrm{a} + \mathrm{b} + \mathrm{n} = 37\)

To maximize n, minimize a (set \(\mathrm{a} = 0\) if possible):
If \(\mathrm{a} = 0\): \(\mathrm{b} + \mathrm{n} = 37\) and \(3\mathrm{n} + 2\mathrm{b} = 64\)
Substituting \(\mathrm{b} = 37 - \mathrm{n}\): \(3\mathrm{n} + 2(37 - \mathrm{n}) = 64\)
\(3\mathrm{n} + 74 - 2\mathrm{n} = 64\)
\(\mathrm{n} = 64 - 74 = -10\)

Since n can't be negative, we need some teachers to teach only 1 class.

Let's try small values systematically:
If \(\mathrm{n} = 13\): We need \(3(13) + 2\mathrm{b} + \mathrm{a} = 64\), so \(39 + 2\mathrm{b} + \mathrm{a} = 64\), giving us \(2\mathrm{b} + \mathrm{a} = 25\)
Also: \(\mathrm{a} + \mathrm{b} + 13 = 37\), so \(\mathrm{a} + \mathrm{b} = 24\)

From these: \(\mathrm{b} = 25 - 24 = 1\) and \(\mathrm{a} = 24 - 1 = 23\)
Check: \(23 + 1 + 13 = 37\) ✓ and \(23 + 2 + 39 = 64\) ✓

Trying \(\mathrm{n} = 14\): \(2\mathrm{b} + \mathrm{a} = 22\) and \(\mathrm{a} + \mathrm{b} = 23\)
This gives \(\mathrm{b} = 22 - 23 = -1\), which is impossible.

Therefore, the maximum value of n is 13.

Final Answer

The minimum value of n is 0 and the maximum value of n is 13.

Therefore, the answer is A) 0 and 13.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint "at least 1, but not more than 3 classes"

Students often misread this as "exactly 1, 2, or 3 classes" without recognizing that ALL teachers must teach within this range. They might try to use teachers who teach 0 classes or more than 3 classes, leading to invalid solutions.

2. Not setting up the dual constraint system properly

Many students focus only on the teacher count (37 teachers) but forget that the classes must also be exactly covered (64 classes). They might set up only one equation instead of the two required constraints: \(\mathrm{a} + \mathrm{b} + \mathrm{n} = 37\) and \(\mathrm{a} + 2\mathrm{b} + 3\mathrm{n} = 64\).

3. Confusing the optimization objective

Students sometimes try to find the total range of possible values rather than specifically finding the minimum and maximum values of n (teachers who teach exactly 3 classes). They might attempt to optimize the wrong variable or look for average values instead.

Errors while executing the approach

1. Arithmetic errors in substitution and solving

When solving the system of equations, students frequently make calculation mistakes. For example, when finding the minimum \(\mathrm{n}=0\) case, they might incorrectly calculate that \(\mathrm{b}=27\) and \(\mathrm{a}=10\), or make errors when substituting \(\mathrm{a} = 37-\mathrm{b}\) into the second equation.

2. Not checking feasibility of solutions

Students often find values for a, b, and n but forget to verify that all values are non-negative integers. For instance, when testing maximum values of n, they might accept negative values for a or b without recognizing these are impossible.

3. Stopping at the first working solution

When finding the maximum value of n, students might test \(\mathrm{n}=13\), see that it works, and assume it's the maximum without testing \(\mathrm{n}=14\) to confirm that higher values are impossible. They fail to systematically verify the boundary conditions.

Errors while selecting the answer

1. Reversing the order of minimum and maximum

Even after correctly calculating that the minimum n is 0 and maximum n is 13, students sometimes select answer choices that list these values in reverse order, choosing "13 and 0" instead of "0 and 13".

No likely faltering points for other aspects of answer selection in this particular question.