How many 4-digit positive integers are there in which all 4 digits are even?

1. Translate the problem requirements: We need to find how many 4-digit numbers exist where every single digit (thousands, hundreds, tens, units) is even. The even digits are 0, 2, 4, 6, 8. A 4-digit number cannot start with 0.
2. Identify constraints for each digit position: Determine what digits are allowed in each position, considering that the first digit has special restrictions while other positions follow the general even digit rule.
3. Count choices for each position independently: Apply the multiplication principle by counting valid options for thousands place, then hundreds place, then tens place, then units place.
4. Apply multiplication principle to get total combinations: Multiply the number of choices for each position since each digit selection is independent of the others.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in simple terms. We want to count 4-digit numbers where every single digit is even.

First, what are the even digits? They are: 0, 2, 4, 6, 8. That's 5 even digits total.

Now, what makes a number a 4-digit number? It has to be between 1000 and 9999. This means the first digit (the thousands place) cannot be 0, because then it would only be a 3-digit number like 0234, which we actually write as 234.

So our requirements are:
• All 4 digits must be even (chosen from 0, 2, 4, 6, 8)
• The first digit cannot be 0 (to ensure it's truly a 4-digit number)

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

2. Identify constraints for each digit position

Let's think about each position in our 4-digit number: _ _ _ _

For the thousands place (first digit):
• Must be even
• Cannot be 0
• Available choices: 2, 4, 6, 8
• That's 4 choices

For the hundreds place (second digit):
• Must be even
• Can be 0 (no restriction here)
• Available choices: 0, 2, 4, 6, 8
• That's 5 choices

For the tens place (third digit):
• Must be even
• Can be 0 (no restriction here)
• Available choices: 0, 2, 4, 6, 8
• That's 5 choices

For the units place (fourth digit):
• Must be even
• Can be 0 (no restriction here)
• Available choices: 0, 2, 4, 6, 8
• That's 5 choices

Process Skill: APPLY CONSTRAINTS - Recognizing that only the first digit has the additional restriction

3. Count choices for each position independently

Now we can see our counting situation clearly:

Position 1 (thousands): 4 choices (2, 4, 6, 8)
Position 2 (hundreds): 5 choices (0, 2, 4, 6, 8)
Position 3 (tens): 5 choices (0, 2, 4, 6, 8)
Position 4 (units): 5 choices (0, 2, 4, 6, 8)

Let's think about this with a concrete example. If I choose 2 for the thousands place, I can still choose any of the 5 even digits for the hundreds place. If I choose 2 for thousands and 0 for hundreds, I can still choose any of the 5 even digits for the tens place. And so on.

Each choice is completely independent of the others, which means we can use the multiplication principle.

4. Apply multiplication principle to get total combinations

Since each digit choice is independent, we multiply the number of choices for each position:

Total number of 4-digit numbers with all even digits = \(4 \times 5 \times 5 \times 5\)

Let's calculate step by step:
\(4 \times 5 = 20\)
\(20 \times 5 = 100\)
\(100 \times 5 = 500\)

Therefore, there are 500 such 4-digit numbers.

Final Answer

The answer is 500, which corresponds to choice C.

Let's verify this makes sense: We have fewer choices for the first digit due to the restriction that it cannot be 0, but full freedom for the remaining three digits among all even digits. This gives us \(4 \times 5^3 = 4 \times 125 = 500\).

Common Faltering Points

Errors while devising the approach

1. Misidentifying even digits: Students might incorrectly include 1, 3, 5, 7, 9 as even digits or forget that 0 is an even digit. This fundamental error would lead to wrong counting from the start.

2. Missing the 4-digit constraint: Students often overlook that for a number to be truly 4-digit, the first digit cannot be 0. They might think all positions can have any even digit (including 0), leading to calculation of \(5 \times 5 \times 5 \times 5 = 625\) instead of the correct approach.

3. Confusing with permutation instead of counting: Some students might think this is about arranging specific digits rather than simply counting valid combinations, potentially leading them toward unnecessarily complex permutation formulas.

Errors while executing the approach

1. Arithmetic calculation errors: Students might make simple multiplication mistakes when computing \(4 \times 5 \times 5 \times 5\), especially in the step-by-step calculation (\(4 \times 5 = 20\), then \(20 \times 5 = 100\), then \(100 \times 5 = 500\)).

2. Incorrect application of multiplication principle: Students might add instead of multiply the choices for each position (\(4 + 5 + 5 + 5 = 19\)), misunderstanding that independent choices should be multiplied, not added.

3. Forgetting the constraint mid-calculation: Students might start correctly with 4 choices for the first digit but then mistakenly switch to 5 choices partway through their calculation, leading to inconsistent results.

Errors while selecting the answer

1. Selecting the trap answer 625: If students missed the constraint that the first digit cannot be 0, they would calculate \(5 \times 5 \times 5 \times 5 = 625\) and select choice A, which is likely included as a trap answer for this common error.

2. Second-guessing the calculation: Students might correctly calculate 500 but then doubt themselves because other answer choices like 600 or 400 seem 'close' and wonder if they made an error, leading them to change their correct answer.