There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all...

1. Translate the problem requirements: We need to find the sum of all possible three-digit numbers formed using only digits 1, 2, and 3 (with repetition allowed). Since we can use any of the 3 digits in each of the 3 positions, we have \(3 \times 3 \times 3 = 27\) total numbers.
2. Recognize the symmetry pattern: Since we're systematically using each digit in each position, every digit (1, 2, 3) will appear exactly the same number of times in the hundreds place, tens place, and units place across all 27 numbers.
3. Calculate frequency of each digit per position: Determine how many times each digit appears in each position (hundreds, tens, units) by dividing the total count evenly.
4. Apply positional values and sum: Multiply each digit by its positional value (100, 10, or 1), account for how many times it appears, then sum all contributions to get the final answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding exactly what we're being asked to do. We need to find the sum of ALL possible three-digit numbers that can be made using only the digits 1, 2, and 3.

Since we can use any of these three digits in any position (and we can repeat digits), let's think about how many numbers we can make:

• For the hundreds place: we have 3 choices (1, 2, or 3)
• For the tens place: we have 3 choices (1, 2, or 3)
• For the units place: we have 3 choices (1, 2, or 3)

So the total number of different three-digit integers = \(3 \times 3 \times 3 = 27\) numbers.

This confirms what the problem told us - there are indeed 27 different numbers.

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

2. Recognize the symmetry pattern

Here's the key insight: because we're systematically creating ALL possible combinations, there's a beautiful balance or symmetry to how the digits get distributed.

Think about it this way - we're not randomly picking numbers. We're methodically creating every single possible combination. This means that each digit (1, 2, and 3) will appear exactly the same number of times in each position.

For example, the digit '1' will appear in the hundreds place the same number of times as the digit '2' appears in the hundreds place, and the same number of times as the digit '3' appears in the hundreds place.

This same balance happens for the tens place and the units place too.

This symmetry is what makes this problem manageable - we don't need to list out all 27 numbers!

3. Calculate frequency of each digit per position

Now let's figure out exactly how many times each digit appears in each position.

Since we have 27 total numbers, and each digit (1, 2, 3) appears equally often in each position, each digit appears:
\(27 \div 3 = 9\) times in each position

Let's verify this makes sense:

• In the hundreds place: digit 1 appears 9 times, digit 2 appears 9 times, digit 3 appears 9 times
• In the tens place: digit 1 appears 9 times, digit 2 appears 9 times, digit 3 appears 9 times
• In the units place: digit 1 appears 9 times, digit 2 appears 9 times, digit 3 appears 9 times

Total check: \(9 + 9 + 9 = 27\) numbers ✓

4. Apply positional values and sum

Now we can calculate the total sum by considering each digit's contribution in each position.

Hundreds place contribution:

• Digit 1 appears 9 times: \(9 \times 1 \times 100 = 900\)
• Digit 2 appears 9 times: \(9 \times 2 \times 100 = 1,800\)
• Digit 3 appears 9 times: \(9 \times 3 \times 100 = 2,700\)
• Subtotal from hundreds place: \(900 + 1,800 + 2,700 = 5,400\)

Tens place contribution:

• Digit 1 appears 9 times: \(9 \times 1 \times 10 = 90\)
• Digit 2 appears 9 times: \(9 \times 2 \times 10 = 180\)
• Digit 3 appears 9 times: \(9 \times 3 \times 10 = 270\)
• Subtotal from tens place: \(90 + 180 + 270 = 540\)

Units place contribution:

• Digit 1 appears 9 times: \(9 \times 1 \times 1 = 9\)
• Digit 2 appears 9 times: \(9 \times 2 \times 1 = 18\)
• Digit 3 appears 9 times: \(9 \times 3 \times 1 = 27\)
• Subtotal from units place: \(9 + 18 + 27 = 54\)

Final Answer

Total sum = \(5,400 + 540 + 54 = 5,994\)

Looking at our answer choices, this matches choice E: 5,994.

We can also verify our logic with a quick pattern check: Notice that 5,400 + 540 + 54 follows the same digit pattern (5,4), just shifted by place value, which makes sense given the symmetric nature of our problem.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the counting principle
Students may incorrectly think that since we're looking for "different" three-digit integers, repetition of digits is not allowed. This would lead them to calculate \(3 \times 2 \times 1 = 6\) numbers instead of \(3 \times 3 \times 3 = 27\) numbers. The key insight they miss is that "different integers" means different combinations, not that each digit within a number must be unique.

2. Attempting to list all 27 numbers individually
Many students will start writing out all possible numbers (111, 112, 113, 121, 122, etc.) instead of recognizing the symmetry pattern. While this brute force approach can work, it's time-consuming and error-prone on the GMAT. Students may miss numbers or include duplicates, leading to calculation errors.

3. Missing the symmetry insight
Students may fail to recognize that each digit (1, 2, 3) appears equally often in each position due to the systematic nature of creating all combinations. Without this crucial insight, they cannot use the elegant frequency-based calculation method and must rely on more cumbersome approaches.

Errors while executing the approach

1. Incorrect frequency calculation
When calculating how many times each digit appears in each position, students may incorrectly divide or use wrong logic. For example, they might think each digit appears \(27 \div 9 = 3\) times instead of \(27 \div 3 = 9\) times in each position, confusing the total numbers with the number of different digits.

2. Place value errors
Students often make mistakes when applying positional values, such as forgetting to multiply by 100 for hundreds place, 10 for tens place, etc. They might calculate contributions like \(9 \times 3 \times 1 = 27\) for the hundreds place instead of \(9 \times 3 \times 100 = 2,700\).

3. Arithmetic mistakes in final summation
Even with correct individual calculations, students may make addition errors when combining the contributions from different positions. For instance, they might incorrectly add 5,400 + 540 + 54, especially under time pressure.

Errors while selecting the answer

1. Forgetting to include all positional contributions
Students might calculate the contribution from only one or two positions (like just hundreds and tens) and select an incomplete answer. For example, they might get \(5,400 + 540 = 5,940\) and try to match it to the closest answer choice instead of including the units place contribution of 54.