- Translate the problem requirements: We need three-digit numbers where all digits are different, nonzero, odd, and greater than 700
- Identify the structural constraints: Determine what values each digit position can take based on the given conditions
- Apply the counting principle systematically: Count valid options for each digit position, considering the dependencies between positions
- Calculate the total combinations: Multiply the number of valid choices for each position to get the final count
Execution of Strategic Approach
1. Translate the problem requirements
Let's break down what we're looking for step by step:
- We need three-digit numbers (so numbers from 100 to 999)
- All three digits must be different from each other
- All three digits must be nonzero (so we can only use digits 1, 2, 3, 4, 5, 6, 7, 8, 9)
- The number must be odd (so the last digit must be 1, 3, 5, 7, or 9)
- The number must be greater than 700 (so it starts with 7, 8, or 9)
So we're counting three-digit numbers of the form ABC where:
- \(\mathrm{A} \in \{7, 8, 9\}\) (first digit is 7, 8, or 9)
- \(\mathrm{C} \in \{1, 3, 5, 7, 9\}\) (last digit is odd)
- B can be any nonzero digit except A and C
- All digits A, B, C are different
Process Skill: TRANSLATE - Converting the word problem into clear mathematical constraints
2. Identify the structural constraints
Now let's think about what values each position can take. Since the digits must all be different, the choice for each position depends on what we've already chosen.
For the first digit (hundreds place): We need the number to be greater than 700, so we can choose 7, 8, or 9. That's 3 choices.
For the last digit (units place): We need the number to be odd, so we can choose 1, 3, 5, 7, or 9. That's 5 choices initially, but it depends on what we picked for the first digit.
For the middle digit (tens place): We can choose any nonzero digit (1 through 9) except the ones we already used for the first and last positions.
The key insight is that we need to be systematic about this because the choices interact with each other.
Process Skill: CONSIDER ALL CASES - We need to think about how the constraint that all digits are different affects our counting
3. Apply the counting principle systematically
Let's count by considering cases based on the first digit:
Case 1: First digit is 7
- First digit: 7 (fixed)
- Last digit: Must be odd and different from 7, so we can choose from \(\{1, 3, 5, 9\}\). That's 4 choices.
- Middle digit: Must be different from both 7 and our chosen last digit. We have 9 total nonzero digits, minus the 2 we've used, so 7 choices.
- Total for Case 1: \(1 \times 4 \times 7 = 28\)
Case 2: First digit is 8
- First digit: 8 (fixed)
- Last digit: Must be odd and different from 8. Since 8 is even, we can choose from all odd digits: \(\{1, 3, 5, 7, 9\}\). That's 5 choices.
- Middle digit: Must be different from both 8 and our chosen last digit. We have 9 total nonzero digits, minus the 2 we've used, so 7 choices.
- Total for Case 2: \(1 \times 5 \times 7 = 35\)
Case 3: First digit is 9
- First digit: 9 (fixed)
- Last digit: Must be odd and different from 9, so we can choose from \(\{1, 3, 5, 7\}\). That's 4 choices.
- Middle digit: Must be different from both 9 and our chosen last digit. We have 9 total nonzero digits, minus the 2 we've used, so 7 choices.
- Total for Case 3: \(1 \times 4 \times 7 = 28\)
4. Calculate the total combinations
Adding up all our cases:
- Case 1 (starts with 7): 28 numbers
- Case 2 (starts with 8): 35 numbers
- Case 3 (starts with 9): 28 numbers
Total: \(28 + 35 + 28 = 91\)
Final Answer
We found that there are 91 three-digit positive integers whose three digits are all different and nonzero, and which are odd integers greater than 700.
This matches answer choice (B) 91.
Common Faltering Points
Errors while devising the approach
- Misunderstanding the "greater than 700" constraint: Students often interpret "greater than 700" to mean the first digit must be exactly 7, forgetting that 8 and 9 also make numbers greater than 700. This leads them to only consider numbers starting with 7, missing two-thirds of the valid cases.
- Overlooking the "all digits different" requirement: Students may initially count as if any digit can be used in any position without restriction, failing to recognize that once a digit is used in one position, it cannot be used again in the remaining positions.
- Confusion about which digits are available: Students might include 0 as a possible digit choice, even though the problem explicitly states all digits must be "nonzero." This could lead to overcounting in their systematic approach.
Errors while executing the approach
- Incorrect case analysis when first and last digits overlap: When the first digit is 7 or 9 (both odd), students often forget that these digits cannot also be used as the last digit. They might count 5 choices for the last digit instead of 4, leading to overcounting in Cases 1 and 3.
- Wrong calculation of middle digit choices: Students frequently miscalculate how many choices remain for the middle digit. They might use 8 choices (forgetting one constraint) or 6 choices (double-counting a constraint) instead of the correct 7 choices (9 total nonzero digits minus 2 already used).
Errors while selecting the answer
No likely faltering points
