If all positive integers that have at least 1 digit equal to 2 are listed in increasing order, what is...

1. Translate the problem requirements: We need to find positive integers that contain at least one digit equal to 2, list them in increasing order \(\mathrm{2, 12, 20, 21, 22, 23, ...}\), and identify which number appears in the 100th position on this list.
2. Count systematically by number of digits: Start with 1-digit numbers, then 2-digit numbers, then 3-digit numbers, keeping a running count until we reach the 100th number.
3. Apply organized counting within each digit category: For each category (1-digit, 2-digit, etc.), systematically count how many numbers contain at least one digit 2.
4. Track cumulative count and identify the target: Keep a running total of numbers found and determine when we cross the 100th position to identify the specific number.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we're looking for. We need positive integers that contain at least one digit equal to 2. This means numbers like 2, 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42, etc.

We're arranging these in increasing order: \(\mathrm{2, 12, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 32, 42, 52, 62, 72, 82, 92, 102, 112, 120, 121, 122, ...}\)

Our goal is to find which number appears in the 100th position on this list.

Process Skill: TRANSLATE - Converting the problem statement into a clear counting task

2. Count systematically by number of digits

Let's organize our counting by looking at 1-digit numbers first, then 2-digit numbers, then 3-digit numbers.

1-digit numbers containing 2:
Only the number 2 itself.
Count so far: 1

2-digit numbers containing 2:
Let's think about this systematically. A 2-digit number can have 2 in the tens place or the units place (or both).

- Numbers with 2 in tens place: \(\mathrm{20, 21, 22, 23, 24, 25, 26, 27, 28, 29}\) (that's 10 numbers)
- Numbers with 2 in units place: \(\mathrm{12, 22, 32, 42, 52, 62, 72, 82, 92}\) (that's 9 numbers)
- But we counted 22 twice, so we subtract 1

Total 2-digit numbers: \(\mathrm{10 + 9 - 1 = 18}\) numbers
Cumulative count: \(\mathrm{1 + 18 = 19}\)

3. Apply organized counting within each digit category

Since we only have 19 numbers so far, we need to continue into 3-digit numbers.

3-digit numbers containing 2:
For 3-digit numbers from 100-199, we need at least one digit to be 2:

Numbers 100-119:
- 102, 112 (that's 2 numbers)
Cumulative count: \(\mathrm{19 + 2 = 21}\)

Numbers 120-129:
- All of these contain 2 in the tens place: 120, 121, 122, 123, 124, 125, 126, 127, 128, 129 (that's 10 numbers)
Cumulative count: \(\mathrm{21 + 10 = 31}\)

Numbers 130-199:
We need numbers with 2 in the units place: 132, 142, 152, 162, 172, 182, 192 (that's 7 numbers)
Cumulative count: \(\mathrm{31 + 7 = 38}\)

Numbers 200-299:
All numbers from 200-299 contain 2 in the hundreds place, so all 100 numbers qualify.
Cumulative count: \(\mathrm{38 + 100 = 138}\)

Since \(\mathrm{138 > 100}\), our 100th number is somewhere in the 200s.

4. Track cumulative count and identify the target

We had 38 numbers before reaching 200. So we need the \(\mathrm{(100 - 38) = 62}\)nd number in the range 200-299.

Since all numbers from 200-299 contain the digit 2, we simply count:
200 (1st), 201 (2nd), 202 (3rd), ...

The 62nd number in this sequence is: \(\mathrm{200 + (62 - 1) = 200 + 61 = 261}\)

Process Skill: APPLY CONSTRAINTS - Systematically tracking our count to ensure we don't miss or double-count any numbers

Final Answer

The 100th positive integer containing at least one digit equal to 2 is 261.

This matches answer choice A) 261.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding "at least one digit equal to 2"
Students might incorrectly interpret this as "exactly one digit equal to 2" and exclude numbers like 22, 122, 222, etc. This would lead to systematic undercounting throughout the solution.

2. Failing to recognize the need for systematic organization
Students may attempt to list numbers randomly (2, 12, 20, 21, 32, 42...) without organizing by digit count, making it nearly impossible to accurately track their position and avoid missing numbers.

3. Not planning for overlap cases
When counting 2-digit numbers, students might not anticipate that numbers like 22 contain the digit 2 in multiple positions, leading to double-counting errors if they don't plan their counting strategy carefully.

Errors while executing the approach

1. Double-counting numbers with multiple 2s
When counting 2-digit numbers, students might count 22 both in the "tens place = 2" category (20, 21, 22...) and the "units place = 2" category (12, 22, 32...), inflating their count by 1 and throwing off all subsequent calculations.

2. Arithmetic errors in cumulative counting
With multiple stages of counting (1-digit: 1, 2-digit: 18, 3-digit 100-199: 19, etc.), students may make simple addition errors when tracking their cumulative totals, leading to incorrect position tracking.

3. Miscounting within the 200-299 range
Students might forget that they need the 62nd number starting from 200, and incorrectly calculate \(\mathrm{200 + 62 = 262}\) instead of \(\mathrm{200 + (62-1) = 261}\), not accounting for the fact that 200 itself is the 1st number in that range.

Errors while selecting the answer

1. Off-by-one error in final calculation
Even with correct counting up to the 200s, students might calculate \(\mathrm{200 + 62 = 262}\) instead of \(\mathrm{200 + 61 = 261}\), forgetting that when looking for the 62nd number starting from 200, they need to add 61 to the starting point since 200 is the first number.

2. Selecting a nearby answer choice without verification
Students who get close but make small errors might see their incorrect result (like 262) in the answer choices and select it without double-checking their cumulative count or final calculation.