AB
+BA
-----
AAC
In the correctly worked addition problem shown, where
the sum of the two-digit positive integers AB and BA is
the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the
integer AAC?
A. 9
B. 6
C. 3
D. 2
E. 0
ab ba gmat number prop
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- rommysingh
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AB + BA = AACrommysingh wrote:AB
+BA
-----
AAC
In the correctly worked addition problem shown, where
the sum of the two-digit positive integers AB and BA is
the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the
integer AAC?
A. 9
B. 6
C. 3
D. 2
E. 0
10A + B + 10B + A = 100A + 10A + C
11A + 11B = 110A + C
C = 11(A + B - 10A)
Since C is a digit, the only time 11x = digit, is when x=0
Therefore C = 0
ans = e
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Here's another one to practice with - https://www.beatthegmat.com/a-and-b-repr ... 11247.html
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An easier way:
Since A and B are different digits, the maximum value of our sum is 98 + 89, or 187. So A must = 1. (It has to be three digits, and it must be ≤ 187, so it must begin with 1.)
That means our result is the three digit number 11C. (Remember that this is NOT 11*C.)
Since A = 1, we know AB + BA is 1B + B1. For 1B + B1 to be a three digit number, B must be 9.
So A = 1, B = 9, and A + B = 10, which ends in 0.
Since A and B are different digits, the maximum value of our sum is 98 + 89, or 187. So A must = 1. (It has to be three digits, and it must be ≤ 187, so it must begin with 1.)
That means our result is the three digit number 11C. (Remember that this is NOT 11*C.)
Since A = 1, we know AB + BA is 1B + B1. For 1B + B1 to be a three digit number, B must be 9.
So A = 1, B = 9, and A + B = 10, which ends in 0.
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An alternate approach is the PLUG IN THE ANSWERS, which represent the value of C.rommysingh wrote:AB
+BA
-----
AAC
In the correctly worked addition problem shown, where
the sum of the two-digit positive integers AB and BA is
the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the
integer AAC?
A. 9
B. 6
C. 3
D. 2
E. 0
When the correct answer choice is plugged in, AB + BA = AAC.
Note the following:
A=0 is not possible, since 0B is not a viable integer.
B=0 is not possible, since 0A is not a viable integer.
C=9:
Let A=1 and B=8, with the result that AB + BA = 18+81 = 99.
Doesn't work.
C=6:
Let A=1 and B=5, with the result that AB + BA = 15+51 = 66.
Let A=7 and B=9, with the result that AB + BA = 79+97 = 176.
Doesn't work.
C=3:
Let A=1 and B=2, with the result that AB + BA = 12+21 = 33.
Let A=4 and B=9, with the result that AB + BA = 49+94 = 143.
Doesn't work.
C=2:
A=1 and B=1 is not viable, since A and B must be different digits.
Let A=3 and B=9, with the result that AB + BA = 39+93 = 132.
Doesn't work.
The correct answer is E.
C=0:
Let A=3 and B=7, with the result that AB + BA = 37+73 = 110.
Success!
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Solution:rommysingh wrote: ↑Sun Oct 25, 2015 4:41 pmAB
+BA
-----
AAC
In the correctly worked addition problem shown, where
the sum of the two-digit positive integers AB and BA is
the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the
integer AAC?
A. 9
B. 6
C. 3
D. 2
E. 0
Since the sum of two 2-digit numbers is less than 200 (since each is less than 100), we see that A must be 1. Therefore, we have:
1B
+B1
11C
At this point, we see that B must be 9 so that the sum of the two 2-digit numbers can be a 3-digit number. Therefore, we have:
19
+91
110
Therefore, the units digit of AAC (or 110) is 0.
Answer: E
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