A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according...

1. Translate the problem requirements: Break down what each constraint means - first digit has 8 valid options (2-9), second digit has 2 valid options (0 or 1), third digit has 10 options (0-9), but we must exclude cases where both second and third digits are 0 simultaneously
2. Calculate total unrestricted combinations: Find how many codes would be possible if we ignore the restriction about second and third digits both being 0
3. Identify and subtract prohibited cases: Determine how many codes violate the constraint (second digit = 0 AND third digit = 0) and subtract these from the total
4. Verify the final count: Confirm our answer makes logical sense given the constraints and matches one of the provided options

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what each constraint means in plain English:

First digit constraint: Cannot be 0 or 1, so it can only be 2, 3, 4, 5, 6, 7, 8, or 9. That gives us 8 valid options for the first digit.

Second digit constraint: Must be 0 or 1. That gives us exactly 2 valid options for the second digit.

Third digit constraint: Can be any digit from 0 to 9. That gives us 10 valid options for the third digit.

Special restriction: The second and third digits cannot both be 0 in the same code. This means we cannot have codes that end with "00".

Process Skill: TRANSLATE - Converting the word constraints into clear mathematical understanding of our options

2. Calculate total unrestricted combinations

Let's imagine we didn't have the special restriction about both second and third digits being 0. How many codes would be possible?

Using the basic counting principle, we multiply the number of options for each position:

• First digit: 8 options
• Second digit: 2 options
• Third digit: 10 options

Total unrestricted combinations = \(8 \times 2 \times 10 = 160\) codes

This represents all possible three-digit codes following the first three constraints, ignoring the restriction about "00" endings.

3. Identify and subtract prohibited cases

Now we need to find how many of these 160 codes violate our special restriction.

The prohibited cases are those where:

• Second digit = 0 AND third digit = 0

For these prohibited codes:

• First digit: still has 8 options (2, 3, 4, 5, 6, 7, 8, 9)
• Second digit: must be 0 (only 1 option in this case)
• Third digit: must be 0 (only 1 option in this case)

Number of prohibited codes = \(8 \times 1 \times 1 = 8\) codes

These are the codes: 200, 300, 400, 500, 600, 700, 800, 900

Process Skill: APPLY CONSTRAINTS - Systematically identifying which combinations violate our restrictions

Final count = Total unrestricted - Prohibited cases = \(160 - 8 = 152\)

4. Verify the final count

Let's double-check our logic:

• We correctly identified 8 options for the first digit
• We correctly identified 2 options for the second digit
• We correctly identified that when second digit is 0, we lose 8 codes (one for each first digit option)
• Our final answer of 152 matches option B

Alternative verification: We can think of this as two separate cases:

Case 1: Second digit is 1

• First digit: 8 options, Second digit: 1, Third digit: 10 options
• Subtotal: \(8 \times 1 \times 10 = 80\) codes

Case 2: Second digit is 0

• First digit: 8 options, Second digit: 0, Third digit: 9 options (cannot be 0)
• Subtotal: \(8 \times 1 \times 9 = 72\) codes

Total: \(80 + 72 = 152\) codes ✓

Final Answer

The number of different codes possible is 152, which corresponds to answer choice B.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint "second and third digits cannot both be 0"
Students often misread this as "second and third digits cannot be 0" (meaning neither can be 0), rather than understanding it correctly as "they cannot both be 0 simultaneously" (meaning one can be 0, but not both together). This leads to incorrectly eliminating valid codes like 210, 310, etc.

2. Overlooking the special restriction entirely
Many students focus on the individual digit constraints (first digit ≠ 0,1; second digit = 0,1; third digit = any) but completely miss the additional restriction about the "00" ending. They would then calculate \(8 \times 2 \times 10 = 160\) and select answer choice C without considering the prohibited cases.

3. Confusing which digits the constraints apply to
With multiple constraints involving 0 and 1, students may mix up which position has which restriction. For example, they might think the first digit must be 0 or 1 instead of cannot be 0 or 1, leading to a completely different calculation.

Errors while executing the approach

1. Arithmetic errors in basic multiplication
Students may make simple calculation mistakes when computing \(8 \times 2 \times 10 = 160\) or \(8 \times 1 \times 1 = 8\), or when subtracting \(160 - 8 = 152\). These errors are especially common under time pressure.

2. Incorrectly counting prohibited cases
When identifying codes ending in "00", students might miscalculate how many such codes exist. They may forget that the first digit still has 8 options (not 10) even in prohibited cases, or they might double-count certain scenarios.

3. Adding instead of subtracting prohibited cases
Some students correctly identify that there are 8 prohibited cases but then add them to the total instead of subtracting: 160 + 8 = 168, leading them to select answer choice D.

Errors while selecting the answer

No likely faltering points - once students have correctly calculated 152, the answer selection is straightforward as it directly matches option B.