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07 Aug 2026, 04:52
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AB
+BA
____
CDC
In the correctly worked addition problem shown, AB and BA are two-digit positive integers, CDC is a three-digit integer, and A, B, C, and D are distinct nonzero digits. How many different values can A * B * C * D take?
A. 2
B. 3
C. 4
D. 6
E. 8
+BA
____
CDC
In the correctly worked addition problem shown, AB and BA are two-digit positive integers, CDC is a three-digit integer, and A, B, C, and D are distinct nonzero digits. How many different values can A * B * C * D take?
A. 2
B. 3
C. 4
D. 6
E. 8
07 Aug 2026, 04:52
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Official Solution:
AB
+BA
____
CDC
In the correctly worked addition problem shown, AB and BA are two-digit positive integers, CDC is a three-digit integer, and A, B, C, and D are distinct nonzero digits. How many different values can A * B * C * D take?
A. 2
B. 3
C. 4
D. 6
E. 8
Since the sum of two two-digit numbers is a three-digit number, the hundreds digit \(C\) must be 1. The largest possible sum is \(99 + 99 = 198\), so \(C\) cannot be anything greater than 1.
Now look at the units column. Since the units digit of the sum is \(C = 1\), the sum \(A + B\) must have units digit 1. Also, since the result is three digits, \(A + B\) must be at least 10. Since \(A\) and \(B\) are digits, \(A + B\) cannot be 21 or more. Therefore:
\(A + B = 11\)
So the units column gives 1 and carries 1 to the tens column. Then the tens column is:
\(A + B + 1 = 12\)
Thus, \(D = 2\).
So \(C = 1\), \(D = 2\), and \(A + B = 11\).
Since \(A\), \(B\), \(C\), and \(D\) are distinct nonzero digits, \(A\) and \(B\) cannot be 1 or 2. The possible ordered pairs for \(A\) and \(B\) are:
\((3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3)\)
But reversed pairs give the same value of \(A * B * C * D\), so we only need:
\((3, 8), (4, 7), (5, 6)\)
Since \(C = 1\) and \(D = 2\), we get:
\(A * B * C * D = A * B * 1 * 2\)
The three possible values are:
\(3 * 8 * 1 * 2 = 48\)
\(4 * 7 * 1 * 2 = 56\)
\(5 * 6 * 1 * 2 = 60\)
So there are 3 different values.
Answer: B
AB
+BA
____
CDC
In the correctly worked addition problem shown, AB and BA are two-digit positive integers, CDC is a three-digit integer, and A, B, C, and D are distinct nonzero digits. How many different values can A * B * C * D take?
A. 2
B. 3
C. 4
D. 6
E. 8
Since the sum of two two-digit numbers is a three-digit number, the hundreds digit \(C\) must be 1. The largest possible sum is \(99 + 99 = 198\), so \(C\) cannot be anything greater than 1.
Now look at the units column. Since the units digit of the sum is \(C = 1\), the sum \(A + B\) must have units digit 1. Also, since the result is three digits, \(A + B\) must be at least 10. Since \(A\) and \(B\) are digits, \(A + B\) cannot be 21 or more. Therefore:
\(A + B = 11\)
So the units column gives 1 and carries 1 to the tens column. Then the tens column is:
\(A + B + 1 = 12\)
Thus, \(D = 2\).
So \(C = 1\), \(D = 2\), and \(A + B = 11\).
Since \(A\), \(B\), \(C\), and \(D\) are distinct nonzero digits, \(A\) and \(B\) cannot be 1 or 2. The possible ordered pairs for \(A\) and \(B\) are:
\((3, 8), (4, 7), (5, 6), (6, 5), (7, 4), (8, 3)\)
But reversed pairs give the same value of \(A * B * C * D\), so we only need:
\((3, 8), (4, 7), (5, 6)\)
Since \(C = 1\) and \(D = 2\), we get:
\(A * B * C * D = A * B * 1 * 2\)
The three possible values are:
\(3 * 8 * 1 * 2 = 48\)
\(4 * 7 * 1 * 2 = 56\)
\(5 * 6 * 1 * 2 = 60\)
So there are 3 different values.
Answer: B
