How many different numbers are possible when 2 or more different numbers from the following set are added together? {0.1,...

1. Translate the problem requirements: We need to find how many distinct sum values are possible when we select 2 or more different numbers from the set \(\{0.1, 0.01, 0.001, 0.0001, 0.00001\}\) and add them together.
2. Recognize the structure of decimal numbers: Each number in our set represents a different decimal place (tenths, hundredths, thousandths, etc.), meaning when we add any combination, we get a unique decimal representation.
3. Count valid subsets systematically: Calculate the total number of ways to choose 2 or more numbers from our 5-number set, which equals total possible subsets minus the cases we exclude (empty set and single-number subsets).
4. Verify uniqueness of sums: Confirm that each different combination of numbers produces a unique sum due to the decimal place structure.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is really asking for. We have a set of five decimal numbers: \(\{0.1, 0.01, 0.001, 0.0001, 0.00001\}\). We need to find how many different sum values we can create by adding together 2 or more of these numbers.

The key word here is "different" - meaning each unique combination of numbers will give us a different sum, and we want to count how many such distinct sums are possible.

For example, if we pick 0.1 and 0.01, we get 0.11. If we pick 0.1, 0.01, and 0.001, we get 0.111. Each selection gives us a different result.

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

2. Recognize the structure of decimal numbers

Now here's the beautiful insight that makes this problem much simpler: each number in our set represents a different decimal place position.

• 0.1 affects the tenths place
• 0.01 affects the hundredths place
• 0.001 affects the thousandths place
• 0.0001 affects the ten-thousandths place
• 0.00001 affects the hundred-thousandths place

When we add any combination of these numbers together, each decimal place is either "on" (if we include that number) or "off" (if we don't include it). Since each number affects a completely different decimal place, every different combination of numbers will produce a unique sum.

For instance:
• Just \(0.1 + 0.01 = 0.11000\)
• Just \(0.1 + 0.001 = 0.10100\)
• All three: \(0.1 + 0.01 + 0.001 = 0.11100\)

Each combination creates a unique decimal "fingerprint."

3. Count valid subsets systematically

Since each combination of our 5 numbers creates a unique sum, we need to count how many ways we can choose 2 or more numbers from our set of 5.

Let's think about this step by step:

First, let's count ALL possible combinations (including those we don't want yet):
• With 5 numbers, each number can either be "chosen" or "not chosen"
• This gives us \(2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32\) total possibilities

But we need to exclude the cases that don't meet our "2 or more" requirement:
• Empty set (choosing no numbers): 1 way
• Choosing exactly 1 number: 5 ways (we can pick any one of the 5 numbers)

So our answer is: \(32 - 1 - 5 = 26\) different sums

Process Skill: APPLY CONSTRAINTS - Systematically excluding cases that don't meet the problem requirements

4. Verify uniqueness of sums

Let's double-check our reasoning by confirming that each combination truly produces a unique sum.

Because our numbers are 0.1, 0.01, 0.001, 0.0001, and 0.00001, we can think of any sum as a 5-digit binary pattern where each position represents whether we include that decimal place.

For example:
• Pattern 11000: Include 0.1 and 0.01 → Sum = 0.11000
• Pattern 10100: Include 0.1 and 0.001 → Sum = 0.10100
• Pattern 01110: Include 0.01, 0.001, and 0.0001 → Sum = 0.01110

Since each binary pattern corresponds to exactly one sum, and we've counted 26 valid patterns (those with at least two 1's), we indeed have 26 different possible sums.

4. Final Answer

We have determined that there are 26 different numbers possible when 2 or more different numbers from the given set are added together.

This matches answer choice D. 26.

Our systematic approach of counting all possible subset combinations (32 total) and subtracting the invalid cases (1 empty set + 5 single-number cases) gives us \(32 - 6 = 26\) valid combinations, each producing a unique sum due to the distinct decimal place structure of our numbers.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "different numbers are possible"

Students often confuse this with counting the number of ways to select numbers rather than counting the number of distinct sum values. They might try to calculate combinations like \(\mathrm{C}(5,2) + \mathrm{C}(5,3) + \mathrm{C}(5,4) + \mathrm{C}(5,5) = 10 + 10 + 5 + 1 = 26\), which coincidentally gives the right answer but for the wrong reason. The key insight they miss is that we need to count unique sums, not selection methods.

2. Missing the constraint "2 or more different numbers"

Students frequently overlook the "2 or more" requirement and include single numbers in their count. This would lead them to calculate \(2^5 - 1 = 31\) (excluding only the empty set) instead of \(2^5 - 1 - 5 = 26\). The constraint is crucial but easy to miss when rushing through the problem.

3. Not recognizing the decimal place structure

Some students might worry about whether different combinations could produce the same sum and attempt complex verification. They miss the elegant insight that each number affects a different decimal place (tenths, hundredths, etc.), guaranteeing that every combination produces a unique sum. Without this recognition, they might get bogged down in unnecessary calculations.

Errors while executing the approach

1. Arithmetic errors in subset counting

When calculating \(2^5 = 32\), students sometimes make basic arithmetic mistakes, especially if they expand it as \(2 \times 2 \times 2 \times 2 \times 2\) step by step. Additionally, when subtracting the invalid cases \((32 - 1 - 5 = 26)\), simple arithmetic errors can lead to wrong final answers like 25 or 27.

2. Incorrect identification of invalid cases

Students might incorrectly count the number of single-element subsets. Instead of recognizing there are exactly 5 ways to choose one number from the set of 5, they might miscalculate this as 4 or 6, leading to final answers of 27 or 25 respectively.

Errors while selecting the answer

No likely faltering points