When each of the children in a room was given the same number of crayons from a box containing 45...

1. Translate the problem requirements: We have 45 crayons total. Each child gets the same number of crayons. After distribution, the crayons remaining equals (number of children - 3). We need to find which option cannot be the number of children.
2. Set up the division relationship: If there are n children and each gets k crayons, then \(\mathrm{nk + remainder = 45}\), where \(\mathrm{remainder = n - 3}\).
3. Create the constraint equation: Substitute to get \(\mathrm{nk + (n - 3) = 45}\), which simplifies to \(\mathrm{n(k + 1) = 48}\).
4. Test divisibility for each answer choice: For n to be valid, it must be a divisor of 48, and the remainder \(\mathrm{(n - 3)}\) must be non-negative and less than n.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in plain English: We have a box with 45 crayons. These crayons are being shared equally among some children in a room. After giving each child the same number of crayons, there are some crayons left over in the box. The problem tells us that the number of leftover crayons is exactly 3 less than the number of children.

For example, if there were 8 children, then there would be \(\mathrm{8 - 3 = 5}\) crayons remaining in the box after distribution.

We need to figure out which of the given answer choices could NOT represent the number of children in the room.

Process Skill: TRANSLATE - Converting the word problem into mathematical relationships

2. Set up the division relationship

When we divide things equally, we use the basic division relationship: Total = (Number of groups) × (Amount per group) + Remainder

In our crayon situation:
• Total crayons = 45
• Number of groups = number of children (let's call this n)
• Amount per group = crayons each child gets (let's call this k)
• Remainder = crayons left in box = \(\mathrm{n - 3}\)

So our relationship becomes: \(\mathrm{45 = n \times k + (n - 3)}\)

This means: \(\mathrm{n \times k + n - 3 = 45}\)

3. Create the constraint equation

Let's rearrange our equation to make it easier to work with:
\(\mathrm{n \times k + n - 3 = 45}\)
\(\mathrm{n \times k + n = 45 + 3}\)
\(\mathrm{n \times k + n = 48}\)
\(\mathrm{n \times (k + 1) = 48}\)

This tells us that \(\mathrm{n \times (k + 1)}\) must equal 48. In other words, the number of children times (crayons per child plus 1) must equal 48.

For this to work, the number of children (n) must be a factor of 48.

4. Test divisibility for each answer choice

Now we need to check each answer choice to see if it could work. For a number to be valid:
1. It must divide 48 evenly (be a factor of 48)
2. The remainder \(\mathrm{(n - 3)}\) must make sense (be non-negative and less than n)

Process Skill: APPLY CONSTRAINTS - Checking both mathematical and logical constraints

Let's find the factors of 48 first: \(\mathrm{48 = 2^4 \times 3^1}\), so the factors are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

Checking each answer choice:

1. \(\mathrm{n = 3}\): 3 is a factor of 48. If \(\mathrm{k + 1 = 48 ÷ 3 = 16}\), then \(\mathrm{k = 15}\). Remainder = \(\mathrm{3 - 3 = 0}\). ✓ This works!
2. \(\mathrm{n = 8}\): 8 is a factor of 48. If \(\mathrm{k + 1 = 48 ÷ 8 = 6}\), then \(\mathrm{k = 5}\). Remainder = \(\mathrm{8 - 3 = 5}\). ✓ This works!
3. \(\mathrm{n = 12}\): 12 is a factor of 48. If \(\mathrm{k + 1 = 48 ÷ 12 = 4}\), then \(\mathrm{k = 3}\). Remainder = \(\mathrm{12 - 3 = 9}\). ✓ This works!
4. \(\mathrm{n = 14}\): Let's check if 14 divides 48: \(\mathrm{48 ÷ 14 = 3.43...}\) This is not a whole number! Since 14 is not a factor of 48, there's no way to have 14 children and satisfy our equation. ✗ This doesn't work!
5. \(\mathrm{n = 16}\): 16 is a factor of 48. If \(\mathrm{k + 1 = 48 ÷ 16 = 3}\), then \(\mathrm{k = 2}\). Remainder = \(\mathrm{16 - 3 = 13}\). ✓ This works!

5. Final Answer

The answer is D. 14.

We found that 14 cannot be the number of children because 14 is not a factor of 48. For the crayon distribution to work as described in the problem, the number of children must be a factor of 48, and 14 is not.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the remainder condition: Students often misread "the number of crayons left in the box was 3 less than the number of children" as "3 crayons were left" or "the number of children was 3 less than the crayons left." This fundamental misunderstanding leads to setting up the wrong equation entirely.

2. Confusing the division relationship: Students may struggle to properly set up the division equation, especially understanding that remainder = \(\mathrm{n - 3}\), where n is the number of children. They might write \(\mathrm{45 = n \times k + 3}\) instead of \(\mathrm{45 = n \times k + (n - 3)}\).

3. Missing the constraint that remainder must be valid: Students often forget that the remainder \(\mathrm{(n - 3)}\) must be non-negative and less than the number of children (n). This means \(\mathrm{n ≥ 3}\), and the remainder must be less than n for the division to make sense.

Errors while executing the approach

1. Arithmetic errors in factorization: When finding factors of 48, students commonly make errors like missing factors (forgetting 6 or 24) or incorrectly calculating \(\mathrm{48 ÷ 14}\), potentially rounding 3.43 to 3 and thinking it works.

2. Incorrect algebraic manipulation: Students may make errors when rearranging \(\mathrm{n \times k + n - 3 = 45}\) to get \(\mathrm{n(k + 1) = 48}\). Common mistakes include sign errors or forgetting to factor out n correctly.

3. Testing wrong values: When checking if each answer choice works, students might substitute incorrectly or make computational errors when calculating k values or remainders for each case.

Errors while selecting the answer

1. Selecting a working option instead of the non-working one: This is a critical error where students find that options A, B, C, and E all work mathematically, but forget that the question asks for what could NOT be the number of children, leading them to select one of the working options instead of D.

2. Misunderstanding the question type: Students might get confused by the negative phrasing "could NOT have been" and select the first option they verify as correct, rather than continuing to check all options to find the one that doesn't work.