- Translate the problem requirements: We need to find how many 3-digit numbers \(\mathrm{abc}\) (where \(\mathrm{a, b, c}\) are nonzero digits) have the property that \(\mathrm{abc - cba}\) is a positive multiple of 7, just like the given example \(942 - 249\).
- Express the difference algebraically: Convert \(\mathrm{abc - cba}\) into an algebraic expression using place values to identify the mathematical pattern.
- Apply divisibility by 7: Use the condition that the difference must be divisible by 7 to create constraints on the digits.
- Count valid combinations systematically: Determine which values of \(\mathrm{a}\) and \(\mathrm{c}\) satisfy our constraints, then count the total possibilities with nonzero \(\mathrm{b}\).
Execution of Strategic Approach
1. Translate the problem requirements
Let's start by understanding what we need to find. We have a 3-digit number that we can write as \(\mathrm{abc}\), where:
- \(\mathrm{a}\) is the hundreds digit (and must be nonzero since it's a 3-digit number)
- \(\mathrm{b}\) is the tens digit (must be nonzero as stated in the problem)
- \(\mathrm{c}\) is the units digit (must be nonzero as stated in the problem)
We need to find how many such numbers have the property that when we subtract \(\mathrm{cba}\) from \(\mathrm{abc}\), the result is a positive multiple of 7.
For example, in the given case: \(942 - 249 = 693\), and \(693\) is indeed a multiple of 7 (since \(693 = 7 \times 99\)).
Process Skill: TRANSLATE
2. Express the difference algebraically
Now let's write \(\mathrm{abc}\) and \(\mathrm{cba}\) in terms of their place values:
- \(\mathrm{abc} = 100\mathrm{a} + 10\mathrm{b} + \mathrm{c}\)
- \(\mathrm{cba} = 100\mathrm{c} + 10\mathrm{b} + \mathrm{a}\)
So the difference \(\mathrm{abc - cba}\) becomes:
\(\mathrm{abc - cba} = (100\mathrm{a} + 10\mathrm{b} + \mathrm{c}) - (100\mathrm{c} + 10\mathrm{b} + \mathrm{a})\)
Let's simplify this step by step:
\(= 100\mathrm{a} + 10\mathrm{b} + \mathrm{c} - 100\mathrm{c} - 10\mathrm{b} - \mathrm{a}\)
\(= 100\mathrm{a} - \mathrm{a} + 10\mathrm{b} - 10\mathrm{b} + \mathrm{c} - 100\mathrm{c}\)
\(= 99\mathrm{a} + 0 - 99\mathrm{c}\)
\(= 99\mathrm{a} - 99\mathrm{c}\)
\(= 99(\mathrm{a} - \mathrm{c})\)
So \(\mathrm{abc - cba} = 99(\mathrm{a} - \mathrm{c})\)
Process Skill: MANIPULATE
3. Apply divisibility by 7
We need \(\mathrm{abc - cba}\) to be a positive multiple of 7.
Since \(\mathrm{abc - cba} = 99(\mathrm{a} - \mathrm{c})\), we need \(99(\mathrm{a} - \mathrm{c})\) to be:
- Positive: This means \(\mathrm{a} - \mathrm{c} > 0\), so \(\mathrm{a} > \mathrm{c}\)
- A multiple of 7: This means \(99(\mathrm{a} - \mathrm{c})\) must be divisible by 7
Now, since \(99 = 9 \times 11\), and neither 9 nor 11 shares any common factors with 7, we need \((\mathrm{a} - \mathrm{c})\) itself to be divisible by 7.
So we need: \(\mathrm{a} - \mathrm{c} = 7\mathrm{k}\) for some positive integer \(\mathrm{k}\).
Since \(\mathrm{a}\) and \(\mathrm{c}\) are both nonzero digits (1 through 9), we have:
- The smallest value of \(\mathrm{a} - \mathrm{c}\) is \(1 - 9 = -8\)
- The largest value of \(\mathrm{a} - \mathrm{c}\) is \(9 - 1 = 8\)
For \(\mathrm{a} - \mathrm{c}\) to be positive and divisible by 7, the only possibility is \(\mathrm{a} - \mathrm{c} = 7\).
Process Skill: APPLY CONSTRAINTS
4. Count valid combinations systematically
We need \(\mathrm{a} - \mathrm{c} = 7\), where both \(\mathrm{a}\) and \(\mathrm{c}\) are nonzero digits (1 through 9).
Let's find all valid pairs \((\mathrm{a}, \mathrm{c})\):
- If \(\mathrm{c} = 1\), then \(\mathrm{a} = 1 + 7 = 8\) ✓
- If \(\mathrm{c} = 2\), then \(\mathrm{a} = 2 + 7 = 9\) ✓
- If \(\mathrm{c} = 3\), then \(\mathrm{a} = 3 + 7 = 10\) ✗ (not a single digit)
So we have exactly 2 valid pairs: \((\mathrm{a}, \mathrm{c}) = (8, 1)\) and \((\mathrm{a}, \mathrm{c}) = (9, 2)\)
For each valid pair \((\mathrm{a}, \mathrm{c})\), the middle digit \(\mathrm{b}\) can be any nonzero digit from 1 to 9.
Therefore:
- For \((\mathrm{a}, \mathrm{c}) = (8, 1)\): we have 9 choices for \(\mathrm{b}\), giving us 9 numbers
- For \((\mathrm{a}, \mathrm{c}) = (9, 2)\): we have 9 choices for \(\mathrm{b}\), giving us 9 numbers
Total: \(9 + 9 = 18\) possible 3-digit numbers.
Process Skill: CONSIDER ALL CASES
Final Answer
The number of possible 3-digit numbers \(\mathrm{abc}\) such that \(\mathrm{abc - cba}\) is a positive multiple of 7 is 18.
This matches answer choice E.
Common Faltering Points
Errors while devising the approach
- Misunderstanding the constraint requirements: Students may overlook that ALL digits a, b, and c must be nonzero, thinking that only the hundreds digit a needs to be nonzero (as is typical for 3-digit numbers). This leads to incorrectly including cases where b=0 or c=0 in their counting.
- Misinterpreting "positive multiple of 7": Students might focus only on the divisibility by 7 requirement and forget that the difference abc - cba must be positive, missing the critical constraint that a > c.
- Incorrect algebraic setup: Students may struggle with the place value representation, either writing the numbers incorrectly (mixing up the order of digits) or making errors when setting up the subtraction abc - cba.
Errors while executing the approach
- Arithmetic errors in simplification: When simplifying \(\mathrm{abc - cba} = (100\mathrm{a} + 10\mathrm{b} + \mathrm{c}) - (100\mathrm{c} + 10\mathrm{b} + \mathrm{a})\), students commonly make sign errors or incorrectly combine like terms, potentially getting \(99(\mathrm{c} - \mathrm{a})\) instead of \(99(\mathrm{a} - \mathrm{c})\).
- Incorrect divisibility analysis: Students may incorrectly conclude that since \(99 = 9 \times 11\), they need to find when \(9(\mathrm{a}-\mathrm{c})\) or \(11(\mathrm{a}-\mathrm{c})\) is divisible by 7, rather than recognizing that since gcd(99,7) = 1, the factor \((\mathrm{a}-\mathrm{c})\) itself must be divisible by 7.
- Missing valid cases: When finding pairs where \(\mathrm{a} - \mathrm{c} = 7\), students might only check a few values systematically or forget to verify that both a and c remain valid single digits, potentially missing one of the two valid pairs (8,1) or (9,2).
Errors while selecting the answer
- Forgetting to account for all digit choices: Students may correctly find the 2 valid pairs (a,c) but forget that b can take any of the 9 nonzero values, leading them to answer 2 instead of 18.
- Incorrect final counting: Students might correctly identify that b has 9 choices for each valid (a,c) pair but then incorrectly calculate the total, perhaps multiplying \(2 \times 9 = 18\) but then selecting a different answer choice due to second-guessing their work.
