1,234 1,243 1,324 ..... .... +4,321 The addition problem above shows four of the 24 different integers that can be...

1. Translate the problem requirements: We need to find the sum of all 24 different 4-digit integers formed using digits 1, 2, 3, 4 exactly once in each integer (these are all permutations of the four digits).
2. Count digit frequency by position: Determine how many times each digit appears in each position (thousands, hundreds, tens, units) across all 24 integers.
3. Calculate positional contributions: For each position, calculate the total contribution by multiplying the sum of digits appearing there by the position's place value.
4. Sum all positional contributions: Add up the contributions from all four positions to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked to find. We have the digits 1, 2, 3, and 4, and we need to use each digit exactly once to create 4-digit numbers. Think of this like arranging 4 different colored balls in a row - there are many different ways to do it.

For example, we can make: 1234, 1243, 1324, 1342, 1423, 1432... and so on.

The question is asking: if we add up ALL possible 4-digit numbers we can make this way, what's the total sum?

How many such numbers exist? Well, for the first position we have 4 choices, for the second position we have 3 remaining choices, for the third we have 2 choices, and for the last position we have 1 choice left. So we have \(4 \times 3 \times 2 \times 1 = 24\) different numbers total.

Process Skill: TRANSLATE - Converting the arrangement problem into a clear counting and summation task

2. Count digit frequency by position

Here's the key insight: because we're dealing with all possible arrangements, there's a beautiful symmetry to this problem.

Let's think about the thousands place (the leftmost digit). Since we're making ALL possible arrangements, each of our 4 digits (1, 2, 3, 4) will appear in the thousands place the same number of times.

If each digit appears equally often in the thousands place, and we have 24 total numbers, then each digit appears \(24 \div 4 = 6\) times in the thousands place.

The same logic applies to every position! Each digit (1, 2, 3, 4) appears exactly 6 times in:
• The thousands place
• The hundreds place
• The tens place
• The units place

This symmetry occurs because when we create all possible permutations, each digit gets an equal 'share' of each position.

3. Calculate positional contributions

Now we can calculate how much each position contributes to our total sum.

Thousands place contribution:
Each of the digits 1, 2, 3, 4 appears 6 times in the thousands place.
When 1 is in thousands place, it contributes \(1000 \times 6 = 6,000\)
When 2 is in thousands place, it contributes \(2000 \times 6 = 12,000\)
When 3 is in thousands place, it contributes \(3000 \times 6 = 18,000\)
When 4 is in thousands place, it contributes \(4000 \times 6 = 24,000\)
Total from thousands place = \(6,000 + 12,000 + 18,000 + 24,000 = 60,000\)

Hundreds place contribution:
Each digit appears 6 times in the hundreds place.
Total from hundreds place = \((1 + 2 + 3 + 4) \times 6 \times 100 = 10 \times 6 \times 100 = 6,000\)

Tens place contribution:
Each digit appears 6 times in the tens place.
Total from tens place = \((1 + 2 + 3 + 4) \times 6 \times 10 = 10 \times 6 \times 10 = 600\)

Units place contribution:
Each digit appears 6 times in the units place.
Total from units place = \((1 + 2 + 3 + 4) \times 6 \times 1 = 10 \times 6 \times 1 = 60\)

4. Sum all positional contributions

Now we simply add up all the contributions from each position:

Total sum = Thousands + Hundreds + Tens + Units
Total sum = \(60,000 + 6,000 + 600 + 60\)
\(\text{Total sum} = 66,660\)

Let's verify this makes sense: we're adding 24 four-digit numbers, each using digits 1-4, so our answer should be substantial but not enormous. 66,660 seems reasonable.

Final Answer: E. 66,660

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the counting requirement: Students may think they need to find just a few specific numbers (like those shown in the problem) rather than understanding they need to sum ALL 24 possible permutations of the digits 1, 2, 3, 4.

2. Missing the symmetry insight: Students might attempt to list out all 24 numbers individually instead of recognizing that due to symmetry, each digit appears the same number of times in each position. This leads to a much more complex and error-prone approach.

3. Incorrect frequency calculation: Students may incorrectly calculate how many times each digit appears in each position, perhaps thinking it's not equal distribution or miscalculating \(24 \div 4 = 6\).

Errors while executing the approach

1. Place value errors: Students may confuse the place values when calculating contributions, such as using 100 instead of 1000 for the thousands place, or mixing up which coefficient goes with which position.

2. Arithmetic mistakes in positional calculations: Common errors include miscalculating \((1 + 2 + 3 + 4) = 10\), or making multiplication errors like \(6 \times 1000 = 6000\) for each digit's contribution.

3. Incomplete contribution calculation: Students might calculate the contribution for only one or two positions and forget to include all four place values (thousands, hundreds, tens, units) in their final calculation.

Errors while selecting the answer

1. Addition errors in final step: Students might make arithmetic mistakes when adding \(60,000 + 6,000 + 600 + 60\), potentially getting 66,600 instead of 66,660 or other similar mistakes.

2. Order of magnitude confusion: Students might select an answer that's off by a factor of 10 due to place value errors carried through from the execution phase, choosing answers like 6,666 (too small) or 666,600 (too large).