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AB +BA ----------- AAC
In the correctly worked addition problem shown, where the sum of the two-digit positive integers \(\mathrm{AB}\) and \(\mathrm{BA}\) is the three-digit integer \(\mathrm{AAC}\), and \(\mathrm{A}\), \(\mathrm{B}\), and \(\mathrm{C}\) are different digits, what is the units digit of the integer \(\mathrm{AAC}\)?
Let's start by understanding what we have here. We're looking at an addition problem where two 2-digit numbers add up to a 3-digit number:
\(\mathrm{AB + BA = AAC}\)
This means we have a number where the tens digit is A and units digit is B, plus another number where the tens digit is B and units digit is A. When we add them together, we get a 3-digit number that starts with A, then A again, then ends with C.
The key constraints are:
- A, B, and C are all different digits
- \(\mathrm{AB}\) and \(\mathrm{BA}\) are positive 2-digit integers (so A and B cannot be 0)
- We need to find the units digit C
Process Skill: TRANSLATE - Converting the visual addition problem into mathematical constraints
Now let's write this using place values to make the math clearer:
The number \(\mathrm{AB}\) means: \(\mathrm{10A + B}\)
The number \(\mathrm{BA}\) means: \(\mathrm{10B + A}\)
The number \(\mathrm{AAC}\) means: \(\mathrm{100A + 10A + C = 110A + C}\)
So our equation becomes:
\(\mathrm{(10A + B) + (10B + A) = 110A + C}\)
Simplifying the left side:
\(\mathrm{11A + 11B = 110A + C}\)
\(\mathrm{11(A + B) = 110A + C}\)
Let's think about what happens when we add the units digits. In the addition \(\mathrm{AB + BA}\), we're adding \(\mathrm{B + A}\) in the units column.
This sum \(\mathrm{B + A}\) will give us some result. If \(\mathrm{B + A}\) is less than 10, then C would simply equal \(\mathrm{A + B}\). But if \(\mathrm{B + A}\) is 10 or greater, then we get a carry to the tens column, and C would be \(\mathrm{(A + B) - 10}\).
Let's call the result of \(\mathrm{A + B = S}\) for now. We have two cases:
- If \(\mathrm{S < 10}\): \(\mathrm{C = S}\) and no carry
- If \(\mathrm{S \geq 10}\): \(\mathrm{C = S - 10}\) and we carry 1 to the tens column
In the tens column, we're adding \(\mathrm{A + B}\), plus any carry from the units column.
The result in the tens column should give us the middle digit of \(\mathrm{AAC}\), which is A.
Case 1: If no carry (\(\mathrm{A + B < 10}\))
Tens column: \(\mathrm{A + B = A}\)
This means \(\mathrm{B = 0}\), but B can't be 0 since \(\mathrm{BA}\) must be a 2-digit number.
Case 2: If there's a carry (\(\mathrm{A + B \geq 10}\))
Tens column: \(\mathrm{A + B + 1 = 10 + A}\) (the 1 goes to hundreds, A stays in tens)
This gives us: \(\mathrm{A + B + 1 = 10 + A}\)
Simplifying: \(\mathrm{B + 1 = 10}\)
Therefore: \(\mathrm{B = 9}\)
Process Skill: CONSIDER ALL CASES - Systematically checking both carry scenarios
We've determined that \(\mathrm{B = 9}\), and we know that \(\mathrm{A + B \geq 10}\) (so there's a carry).
Since \(\mathrm{A + B \geq 10}\) and \(\mathrm{B = 9}\):
\(\mathrm{A + 9 \geq 10}\)
So \(\mathrm{A \geq 1}\)
Since A must be a single digit and \(\mathrm{A \geq 1}\), we need \(\mathrm{A + 9}\) to produce a carry. The hundreds digit of our result \(\mathrm{AAC}\) is A, which confirms that \(\mathrm{A + B + 1}\) produces exactly one carry (giving us A in the hundreds place).
For the units digit C:
\(\mathrm{C = (A + B) - 10 = (A + 9) - 10 = A - 1}\)
But wait! We also know from our original equation:
\(\mathrm{11(A + B) = 110A + C}\)
\(\mathrm{11(A + 9) = 110A + C}\)
\(\mathrm{11A + 99 = 110A + C}\)
\(\mathrm{99 = 99A + C}\)
\(\mathrm{C = 99 - 99A = 99(1 - A)}\)
Since C must be a single digit (0-9), and \(\mathrm{C = 99(1 - A)}\):
For C to be a valid digit, we need \(\mathrm{A = 1}\).
When \(\mathrm{A = 1}\): \(\mathrm{C = 99(1 - 1) = 99(0) = 0}\)
Let's verify: \(\mathrm{A = 1, B = 9, C = 0}\)
- All digits are different ✓
- \(\mathrm{AB = 19, BA = 91}\)
- \(\mathrm{19 + 91 = 110 = AAC}\) ✓
Process Skill: APPLY CONSTRAINTS - Using the requirement that A, B, C are different digits to find the unique solution
The units digit of the integer \(\mathrm{AAC}\) is 0.
Verification: \(\mathrm{19 + 91 = 110}\), where \(\mathrm{A = 1, B = 9, C = 0}\), and all digits are different.
The answer is E. 0
1. Misinterpreting the visual representation: Students often struggle to correctly translate the visual addition \(\mathrm{AB + BA = AAC}\) into mathematical expressions. They might incorrectly assume that A, B, and C represent the actual numbers rather than individual digits, or fail to recognize that \(\mathrm{AB}\) means a two-digit number with tens digit A and units digit B.
2. Overlooking the constraint that digits must be different: The problem states that A, B, and C are different digits, but students frequently ignore this critical constraint when setting up their equations. This oversight can lead them to accept invalid solutions where two or more digits are the same.
3. Missing the constraint that AB and BA are positive 2-digit integers: Students may forget that for \(\mathrm{AB}\) and \(\mathrm{BA}\) to be valid 2-digit numbers, neither A nor B can be zero. This constraint is crucial for eliminating impossible cases but is often overlooked during the initial problem setup.
1. Incorrectly handling the carry operation: When analyzing what happens in the units and tens columns, students often make errors with carry logic. They might assume there's no carry when there actually is one, or incorrectly calculate what happens when \(\mathrm{A + B \geq 10}\), leading to wrong values for the digits.
2. Algebraic manipulation errors: Students frequently make mistakes when simplifying the equation \(\mathrm{11(A + B) = 110A + C}\). Common errors include incorrect distribution, sign errors when rearranging terms, or failing to properly isolate variables to solve for specific digit values.
3. Case analysis confusion: The problem requires considering different cases (with and without carry), but students often get confused about which case applies or fail to systematically check all possibilities. They might pursue only one case or incorrectly conclude which case is valid based on faulty reasoning.
1. Failing to verify the solution: Students may arrive at values for A, B, and C but forget to check that their solution actually works by substituting back into the original addition problem. This verification step is crucial to catch computational errors, but students often skip it and select an incorrect answer.
2. Confusing which digit the question asks for: The question specifically asks for the units digit of \(\mathrm{AAC}\), which is C. However, students who have calculated values for A, B, and C might accidentally select the value of A or B instead of C, especially if they're rushing or not carefully reading the question stem.