The difference 942 - 249 is a positive multiple of 7. If a, b, and c are nonzero digits, how...

1. 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 \(942 - 249\).
2. Express the difference algebraically: Convert \(\mathrm{abc - cba}\) into an algebraic expression using place values to identify the mathematical pattern.
3. Apply divisibility by 7: Use the condition that the difference must be divisible by 7 to create constraints on the digits.
4. 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.