Loading...
There are 27 different three-digit integers that can be formed using only the digits 1, 2 and 3. If all 27 of the integers were listed, what would their sum be?
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:
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
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!
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:
Total check: \(9 + 9 + 9 = 27\) numbers ✓
Now we can calculate the total sum by considering each digit's contribution in each position.
Hundreds place contribution:
Tens place contribution:
Units place contribution:
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.
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.
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.
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.