One side of a parking stall is defined by a straight stripe that consists of n painted sections of equal...

1. Translate the problem requirements: We have n painted sections of equal length separated by \(\mathrm{(n-1)}\) unpainted sections, where each unpainted section is half the length of a painted section. The total distance from start of first painted section to end of last painted section is 203 inches. We need to find n where the unpainted section length is an integer greater than 2.
2. Set up the length relationship: Express the total length in terms of the painted section length, recognizing that we have n painted sections and \(\mathrm{(n-1)}\) gaps between them.
3. Apply the constraint conditions: Use the fact that unpainted sections are integers greater than 2 to determine which values of the painted section length are valid.
4. Test feasible values: Check which combination of n and section lengths satisfies both the total length requirement and the integer constraint.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking at. Imagine a parking stripe that looks like this:

[PAINTED]—gap—[PAINTED]—gap—[PAINTED]—gap—[PAINTED]

We have n painted sections, and between each pair of consecutive painted sections, there's an unpainted gap. So if we have n painted sections, we have \(\mathrm{(n-1)}\) gaps between them.

Each painted section has the same length - let's call this length 'p' inches. Each unpainted gap is half as long as a painted section, so each gap is \(\mathrm{p/2}\) inches long.

The total distance from the start of the first painted section to the end of the last painted section is 203 inches. Also, we're told that the length of each unpainted section (which is \(\mathrm{p/2}\)) must be an integer greater than 2.

Process Skill: TRANSLATE - Converting the parking stripe description into clear mathematical relationships

2. Set up the length relationship

Now let's think about the total length. We need to add up:

• All the painted sections: n sections × p inches each = np inches
• All the unpainted gaps: \(\mathrm{(n-1)}\) gaps × \(\mathrm{p/2}\) inches each = \(\mathrm{(n-1)p/2}\) inches

The total length is: \(\mathrm{np + (n-1)p/2 = 203}\)

Let's simplify this equation. We can factor out p:

\(\mathrm{p[n + (n-1)/2] = 203}\)

\(\mathrm{p[n + n/2 - 1/2] = 203}\)

\(\mathrm{p[3n/2 - 1/2] = 203}\)

\(\mathrm{p[(3n-1)/2] = 203}\)

Therefore: \(\mathrm{p = 406/(3n-1)}\)

3. Apply the constraint conditions

We know that each unpainted section has length \(\mathrm{p/2}\), and this must be an integer greater than 2.

So: \(\mathrm{p/2 > 2}\), which means \(\mathrm{p > 4}\)

Also: \(\mathrm{p/2}\) must be an integer, which means p must be even

Substituting our expression for p:

\(\mathrm{p = 406/(3n-1)}\) must be even and greater than 4

Since \(\mathrm{406 = 2 \times 203}\), for p to be even, \(\mathrm{(3n-1)}\) must be odd. Since 3n is always odd when n is even and even when n is odd, \(\mathrm{(3n-1)}\) is even when n is even and odd when n is odd. So n must be odd.

Process Skill: APPLY CONSTRAINTS - Using the integer condition to limit our possible values

4. Test feasible values

We need \(\mathrm{(3n-1)}\) to divide 406, and n to be odd, and \(\mathrm{p = 406/(3n-1) > 4}\).

From \(\mathrm{p > 4}\): \(\mathrm{406/(3n-1) > 4}\), so \(\mathrm{406 > 4(3n-1)}\), so \(\mathrm{406 > 12n-4}\), so \(\mathrm{410 > 12n}\), so \(\mathrm{n < 34.17}\)

Let's find the factors of 406: \(\mathrm{406 = 2 \times 7 \times 29}\)

So the factors of 406 are: 1, 2, 7, 14, 29, 58, 203, 406

We need \(\mathrm{(3n-1)}\) to be one of these factors, with n being odd and \(\mathrm{n < 34.17}\):

• If \(\mathrm{3n-1 = 1}\), then \(\mathrm{n = 2/3}\) (not integer)
• If \(\mathrm{3n-1 = 2}\), then \(\mathrm{n = 1}\), \(\mathrm{p = 203}\) (\(\mathrm{p/2 = 101.5}\), not integer)
• If \(\mathrm{3n-1 = 7}\), then \(\mathrm{n = 8/3}\) (not integer)
• If \(\mathrm{3n-1 = 14}\), then \(\mathrm{n = 5}\), \(\mathrm{p = 29}\) (\(\mathrm{p/2 = 14.5}\), not integer)
• If \(\mathrm{3n-1 = 29}\), then \(\mathrm{n = 10}\), \(\mathrm{p = 14}\) (\(\mathrm{p/2 = 7}\), integer > 2) ✓
• If \(\mathrm{3n-1 = 58}\), then \(\mathrm{n = 59/3}\) (not integer)
• If \(\mathrm{3n-1 = 203}\), then \(\mathrm{n = 68}\) (too large)

Let's verify \(\mathrm{n = 10}\): We have 10 painted sections of 14 inches each, and 9 unpainted sections of 7 inches each.

Total length = \(\mathrm{10(14) + 9(7) = 140 + 63 = 203}\) inches ✓

4. Final Answer

The value of n is 10.

This matches answer choice C.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding the stripe structure

Students often incorrectly assume there are n gaps instead of \(\mathrm{(n-1)}\) gaps between n painted sections. They visualize the stripe as having gaps at both ends or fail to realize that n painted sections only create \(\mathrm{(n-1)}\) spaces between them. This leads to setting up the wrong equation from the start.

Faltering Point 2: Misinterpreting the constraint about unpainted sections

Students may overlook that "the length of each unpainted section is an integer greater than 2" refers to the actual length \(\mathrm{p/2}\), not just that it needs to be greater than 2. They might miss that \(\mathrm{p/2}\) must be an integer, which means p must be even - a crucial constraint for solving the problem.

Faltering Point 3: Incorrect setup of the total length equation

Students might incorrectly include gaps at the beginning or end of the stripe when the problem clearly states the total length is "from the beginning of the first painted section to the end of the last painted section." This leads to adding extra gap lengths that shouldn't be included.

Errors while executing the approach

Faltering Point 1: Algebraic manipulation errors when simplifying the length equation

When simplifying \(\mathrm{p[n + (n-1)/2] = 203}\), students often make errors in combining fractions or factoring. Common mistakes include incorrectly simplifying \(\mathrm{(n-1)/2}\) or making arithmetic errors when getting to \(\mathrm{p[(3n-1)/2] = 203}\).

Faltering Point 2: Incomplete factor analysis of 406

Students may not systematically find all factors of 406 or make errors in the factorization (\(\mathrm{406 = 2 \times 7 \times 29}\)). Without all factors, they cannot check all possible values of \(\mathrm{(3n-1)}\), leading to missing the correct answer or checking incorrect values.

Faltering Point 3: Errors in applying multiple constraints simultaneously

Students struggle to keep track of all conditions: n must be an integer, \(\mathrm{(3n-1)}\) must divide 406, p must be even, \(\mathrm{p > 4}\), and \(\mathrm{p/2}\) must be an integer. They often forget one constraint while checking others, or make computational errors when testing specific values.

Errors while selecting the answer

Faltering Point 1: Failing to verify the solution

Students may find \(\mathrm{n = 10}\) but fail to verify by calculating the actual total length: 10 painted sections of 14 inches each plus 9 unpainted sections of 7 inches each. Without this verification (\(\mathrm{10\times14 + 9\times7 = 203}\)), they might doubt their answer or select a different option.