If n is a negative integer and q is a positive integer, then there is only one negative integer m...

1. Translate the problem requirements: We need to find a negative integer m and non-negative remainder r such that \(-100 = \mathrm{m} \times 30 + \mathrm{r}\), where \(0 \leq \mathrm{r} \leq 30\). The key constraint is that r must be non-negative, which is different from typical remainder calculations.
2. Set up the division framework: Start with regular division of \(|-100|\) by 30 to understand the basic relationship, then adjust for the negative dividend while maintaining the non-negative remainder constraint.
3. Apply the non-negative remainder constraint: Since we need \(\mathrm{r} \geq 0\) and we're dealing with a negative dividend, we must choose the negative integer m that ensures the remainder falls within the required range [0, 30].
4. Verify the solution: Check that our values of m and r satisfy both the equation \(-100 = \mathrm{m} \times 30 + \mathrm{r}\) and the constraint \(0 \leq \mathrm{r} \leq 30\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this problem is really asking in plain English. We need to find a way to write -100 as:

\(-100 = \text{(some negative integer)} \times 30 + \text{(some remainder)}\)

The catch is that the remainder must be between 0 and 30 (including 0, but not exceeding 30). This is different from how we normally think about division with negative numbers.

In everyday division, we might say \(-100 \div 30 = -3.33...\), but here we need to be more careful. We need to find a negative integer m and a non-negative remainder r such that:
\(-100 = \mathrm{m} \times 30 + \mathrm{r}\), where \(0 \leq \mathrm{r} \leq 30\)

Process Skill: TRANSLATE - Converting the formal mathematical language into a clear requirement

2. Set up the division framework

Let's think about this step by step. If we were dividing 100 by 30 (ignoring the negative sign for a moment), we'd get:
\(100 = 3 \times 30 + 10\)

This tells us that \(100 = 90 + 10\), which checks out.

But our number is -100, not 100. If we try to use the same approach:
\(-100 = -3 \times 30 + (-10)\)

This gives us \(-100 = -90 + (-10) = -100\) ✓

However, there's a problem! Our remainder is -10, but the problem requires the remainder to be non-negative (between 0 and 30).

3. Apply the non-negative remainder constraint

Since our remainder of -10 doesn't satisfy the constraint \(0 \leq \mathrm{r} \leq 30\), we need to adjust our approach.

Here's the key insight: we can "borrow" from the quotient to make the remainder positive. Think of it like this:

Instead of: \(-100 = -3 \times 30 + (-10)\)
We can write: \(-100 = -4 \times 30 + 20\)

Let's check this: \(-4 \times 30 + 20 = -120 + 20 = -100\) ✓

Now our remainder is 20, which satisfies our constraint because \(0 \leq 20 \leq 30\).

This works because we're essentially saying: "Instead of being 3 groups of 30 short (with 10 extra shortage), we're 4 groups of 30 short (with 20 left over)."

Process Skill: APPLY CONSTRAINTS - Ensuring our solution meets all given requirements

4. Verify the solution

Let's double-check our answer:

• We found m = -4 and r = 20
• Does \(-100 = (-4) \times 30 + 20\)? Let's see: \((-4) \times 30 + 20 = -120 + 20 = -100\) ✓
• Is m a negative integer? Yes, -4 is negative ✓
• Is \(0 \leq \mathrm{r} \leq 30\)? Yes, \(0 \leq 20 \leq 30\) ✓

Therefore, the remainder when -100 is divided by 30 is 20.

Final Answer

The remainder when -100 is divided by 30 is 20.

This corresponds to answer choice E.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the remainder constraint

Students often miss that the remainder r must satisfy \(0 \leq \mathrm{r} \leq \mathrm{q}\), meaning it must be non-negative. They might think they can use standard division rules where remainders can be negative, leading them to accept r = -10 as a valid remainder.

2. Confusion about the quotient being negative

The problem specifically states that m must be a negative integer, but students might overlook this constraint. They could attempt to use a positive quotient, which would violate the given conditions.

3. Misinterpreting the division format

Students might not recognize that this is asking for the division algorithm format \(\mathrm{n} = \mathrm{mq} + \mathrm{r}\) and instead try to use regular decimal division \((-100 \div 30 = -3.33...)\), missing the requirement for integer quotient and specific remainder constraints.

Errors while executing the approach

1. Stopping at the first division attempt

After finding \(-100 = -3 \times 30 + (-10)\), students might accept r = -10 as the final answer without checking if it satisfies the non-negative remainder constraint \((0 \leq \mathrm{r} \leq 30)\).

2. Incorrect adjustment calculation

When students realize they need to make the remainder positive, they might incorrectly calculate the adjustment. For example, they might try \(-100 = -2 \times 30 + (-40)\) instead of properly "borrowing" one more group of 30 to get \(-100 = -4 \times 30 + 20\).

3. Arithmetic errors in verification

Students might make calculation mistakes when checking their work, such as computing \((-4) \times 30 + 20\) incorrectly, leading them to doubt their correct answer or accept an incorrect one.

Errors while selecting the answer

1. Choosing the negative remainder

Even after working through the problem, students might select answer choice B (-10) because they remember getting -10 during their initial calculation and don't recognize that this doesn't satisfy the remainder constraints.

2. Confusing quotient with remainder

Students might accidentally select the quotient value (-4) if it were among the choices, or get confused about which value represents the remainder r versus the quotient m in their final equation.