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B
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Difficulty:
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Question Stats:
47% (01:59) correct
53%
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wrong
based on 265
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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
|
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M46-39
10 Jul 2026, 09:00
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GMAT Club 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.
General Discussion
10 Jul 2026, 08:12
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Here's my solution for this question
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Screenshot_2026-07-10-20-42-30-43_40deb401b9ffe8e1df2f1cc5ba480b12.jpg [ 731.48 KiB | Viewed 1683 times ]
vrinda6
Joined: 20 Jul 2020
Last visit: 05 Sep 2026
Posts: 65
Own Kudos:
Given Kudos: 17
Location: India
Schools: Booth '27 Kellogg '27 CBS '27 Duke '27 Haas '27 HBS '27 Ross '27 Stern '27 Stanford '27 Wharton '27 Anderson '27
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Schools: Booth '27 Kellogg '27 CBS '27 Duke '27 Haas '27 HBS '27 Ross '27 Stern '27 Stanford '27 Wharton '27 Anderson '27
Posts: 65
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10 Jul 2026, 08:17
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Start with the equation.
AB=10A+ B, BA= 10B+A
ALSO
CDC= 100C + 10D + C= 102C+10D
Solving for both, we have:
11(A+B)= 102C+ 10D
And considering A and B are ditinct digits
A+B<=9+8= 17
thus
11(A+B)<=11X17=187 ( if c>=2 then cdc>=202 which is not possible)
So we get c=1
Substituing C in 11(A+B)=102+10D we get D=3
Now the only thing remaining is finding the possible pairs.
sum needs to be 12, and digits need to be distinct which cannot be 1 or 3
A,B= (4,8) OR (8,4)
A,B= (5,7) OR (7,5)
Finding the product:
4x8x1x3= 96
5x7x1x3=105
hence correct answer a=2
AB=10A+ B, BA= 10B+A
ALSO
CDC= 100C + 10D + C= 102C+10D
Solving for both, we have:
11(A+B)= 102C+ 10D
And considering A and B are ditinct digits
A+B<=9+8= 17
thus
11(A+B)<=11X17=187 ( if c>=2 then cdc>=202 which is not possible)
So we get c=1
Substituing C in 11(A+B)=102+10D we get D=3
Now the only thing remaining is finding the possible pairs.
sum needs to be 12, and digits need to be distinct which cannot be 1 or 3
A,B= (4,8) OR (8,4)
A,B= (5,7) OR (7,5)
Finding the product:
4x8x1x3= 96
5x7x1x3=105
hence correct answer a=2
