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In the correctly worked addition problem shown

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In the correctly worked addition problem shown

by rajeet123 » Tue Jun 11, 2019 8:52 am

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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
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by deloitte247 » Fri Jun 14, 2019 11:44 pm
AB = sum of 2 digits integers
Sum of two - digit integer cannot be greater than 99 + 99 = 198
AB + BA = AAC where A = 1
IB + BI = IIC
If B < 9 AB + BA will only be a two digit integer e.g 17 + 71 = 88
Therefore,
B = 9 and C = 0
$$AB+BA=AAC\ \left(19+91=110=AAC\right)$$
$$C=0$$

$$answer\ is\ Option\ C$$
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by swerve » Sun Jun 16, 2019 3:01 pm
AB + BA = AAC
10A + B + 10B + A = 100A + 10A + C (cancel out the 10A's)
B + 10B + A - 100A = C
B(1+10) + A(1 - 100) = C

11B - 99A = C

The only value for C that satisfies this equation is C = 0. Then B = 9 and A = 1.

Therefore, the correct option is __E__
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Re: In the correctly worked addition problem shown

by Brent@GMATPrepNow » Mon Nov 08, 2021 4:08 pm
rajeet123 wrote: ↑
Tue Jun 11, 2019 8:52 am
 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
First recognize that, if two 2-digit numbers have a sum that is a 3-digit number, then the hundredth digit of the sum must be 1.
In other words, A = 1.
So we now have the following sum:
_1B
+B1
11C

If the sum 1B and B1 is a 3-digit number, then it must be the case that B = 9 (since 18 + 81 = 99, and 99 is not a 3-digit number)
If B = 9, the sum becomes:
_19
+91
11C

As we can see, C = 0

Answer: E
Brent Hanneson - Creator of GMATPrepNow.com
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