How many even 3 digit integers greater than 700 with distinct non zero digits are there ?

1. Translate the problem requirements: We need 3-digit numbers ABC where \(\mathrm{A} \geq 7\), all digits are different, all digits are non-zero (1-9), and the number is even (C must be even)
2. Identify the constraint hierarchy: Determine which digit constraint is most restrictive and should drive our counting approach
3. Case-by-case systematic counting: Count valid combinations for each possible value of the hundreds digit, considering the even requirement
4. Apply multiplication principle: For each case, multiply the number of choices for each digit position to get the total count

Execution of Strategic Approach

1. Translate the problem requirements

We need to find 3-digit numbers that meet ALL of these conditions at once:

• The number must be greater than 700 (so it starts with 7, 8, or 9)
• The number must be even (so it ends with 2, 4, 6, or 8)
• All three digits must be different from each other
• All three digits must be non-zero (we can only use digits 1, 2, 3, 4, 5, 6, 7, 8, 9)

Let's call our 3-digit number ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.

So we need: \(\mathrm{A} \geq 7\), C is even, \(\mathrm{A} \neq \mathrm{B} \neq \mathrm{C}\), and A, B, C are all from \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\)

Process Skill: TRANSLATE - Converting all problem constraints into clear mathematical conditions

2. Identify the constraint hierarchy

Looking at our constraints, let's see which one limits us the most:

• A can be 7, 8, or 9 (3 choices)
• C can be 2, 4, 6, or 8 (4 choices)
• But A and C must be different!

This means the hundreds digit A will determine how many choices we have for the units digit C. Since A is the most restrictive starting point (only 3 possibilities), we should organize our counting by considering each possible value of A separately.

This gives us a clear path: Count the valid numbers for \(\mathrm{A} = 7\), then \(\mathrm{A} = 8\), then \(\mathrm{A} = 9\), and add them up.

3. Case-by-case systematic counting

Case 1: A = 7 (number starts with 7)

• C must be even and different from 7, so C can be: 2, 4, 6, or 8 (4 choices)
• B must be different from both A = 7 and whatever we pick for C
• B must be non-zero, so B is chosen from \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\)
• Since B ≠ 7 and B ≠ C, we have \(9 - 2 = 7\) choices for B
• Total for Case 1: \(1 \times 7 \times 4 = 28\)

Case 2: A = 8 (number starts with 8)

• C must be even and different from 8
• Since 8 is even, C can only be: 2, 4, or 6 (3 choices)
• B must be different from A = 8 and different from C
• So B has \(9 - 2 = 7\) choices
• Total for Case 2: \(1 \times 7 \times 3 = 21\)

Case 3: A = 9 (number starts with 9)

• C must be even and different from 9
• Since 9 is odd, C can be: 2, 4, 6, or 8 (4 choices)
• B must be different from A = 9 and different from C
• So B has \(9 - 2 = 7\) choices
• Total for Case 3: \(1 \times 7 \times 4 = 28\)

Process Skill: CONSIDER ALL CASES - Systematically examining each possible starting digit

4. Apply multiplication principle

For each case, we used the multiplication principle:
Number of valid combinations = (choices for A) × (choices for B) × (choices for C)

Where the choices for each position take into account all the constraints.

Final Count:
Case 1 (A = 7): 28 numbers
Case 2 (A = 8): 21 numbers
Case 3 (A = 9): 28 numbers

Total = \(28 + 21 + 28 = 77\)

Process Skill: APPLY CONSTRAINTS - Ensuring each digit choice respects all problem conditions simultaneously

Final Answer

There are 77 even 3-digit integers greater than 700 with distinct non-zero digits.

This matches answer choice E.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the "greater than 700" constraint
Students often think "greater than 700" means the number can start with any digit from 7-9, but they forget that numbers starting with 7 must actually be greater than 700 (not equal to 700). However, since we need 3-digit numbers with distinct non-zero digits, any number starting with 7 will automatically be greater than 700, so this constraint is correctly handled in the solution.

2. Overlooking the "distinct digits" requirement
Many students initially count choices for each position independently (3 choices for hundreds place × 8 choices for tens place × 4 choices for units place) without realizing that the digits must all be different from each other. This leads to overcounting and incorrect answers like 96 instead of 77.

3. Forgetting that zero is not allowed
Students might include 0 as a possible digit for the tens or units place, thinking they have 10 total digits to work with. Since the problem specifies "distinct non-zero digits," all positions must use digits from {1,2,3,4,5,6,7,8,9} only.

Errors while executing the approach

1. Incorrectly counting choices for the middle digit
When calculating choices for digit B (tens place), students often forget that B must be different from BOTH A and C. They might calculate B as having 8 choices (9 total digits minus A) instead of 7 choices (9 total digits minus both A and C), leading to overcounting in each case.

2. Making errors in case-by-case arithmetic
Students may correctly identify the three cases (A=7, A=8, A=9) but make computational mistakes. For example, in Case 2 where A=8, they might incorrectly count C as having 4 choices instead of 3, forgetting that C cannot be 8 since it must be different from A=8.

3. Missing the constraint interaction between A and C
Students might not realize that when A=8 (which is even), the number of choices for C (which must also be even but different from A) becomes restricted. They may use 4 choices for C in all cases instead of recognizing that Case 2 has only 3 choices for C.

Errors while selecting the answer

1. Addition errors in final calculation
After correctly calculating each case (28 + 21 + 28), students might make simple arithmetic errors and arrive at incorrect totals like 76 or 78, potentially selecting answer choice D (88) if they double-count one of the cases.

2. Selecting an intermediate result instead of the final answer
Students might calculate one case correctly (like Case 1 = 28) and mistakenly think this represents the total count, potentially selecting a wrong answer choice that matches one of their intermediate calculations rather than the sum of all three cases.