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Senior Manager  Joined: 11 Sep 2005
Posts: 280

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58 00:00

Difficulty:   25% (medium)

Question Stats: 78% (02:00) correct 22% (02:26) wrong based on 1106 sessions

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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

Originally posted by singh_amit19 on 19 Oct 2007, 05:19.
Last edited by adkikani on 01 Jul 2018, 23:49, edited 2 times in total.
Renamed the topic and edited the question.
Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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13
10
jlgdr wrote:
ronneyc wrote:
singh_amit19 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

Ans.is E

Since AB <100 and BA<100 thus AAC<200 we can conclude that B=9 & C=0

And why is that? Could you elaborate more on this please?
Thank you very much sir.

Cheers
J The sum of 2 two-digit integers cannot be greater than 200 (actually it cannot be greater than 99+99=198).

Thus from:
AB
+BA
___
AAC

It follows that A=1. So, we have that
1B
+B1
___
11C

If B is less than 9, AB+BA will not be a three-digit integer (consider 18+81=99), thus B=9 --> 19+91=110 --> C=0.

Similar questions to practice:
tough-tricky-set-of-problms-85211.html#p638336

Hope it helps.
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VP  Joined: 08 Jun 2005
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18
2
lets assume

A = 9
B = 9

99+99 = 198

as you can see AAC cannot be bigger then 198 so A has to be 1.

A=1

11C is a three digit number then B has to be 9 !!

since 18+81 = 99 --> two digits

19+91 = 100+10+C

110 = 110+C

C=0 General Discussion
Intern  Joined: 02 Aug 2007
Posts: 33

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1
1
singh_amit19 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

Ans.is E

Since AB <100 and BA<100 thus AAC<200 we can conclude that B=9 & C=0
Intern  Joined: 18 May 2010
Posts: 13

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3
3

91+19
82+ 28
73+37
64+46
.
.
.
and so on all would suffice..giving you a value 110 the answer is clearly E 0
Intern  Joined: 11 Jul 2012
Posts: 40

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4
AB + BA = AAC ==> 10(A+B) + (A+B) = 100A + 10A + C = 110A + C. Since this 3 digits AAC, 110A + C got to be 110. So C= 0
Brother Karamazov
Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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2
2
singh_amit19 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

Similar questions to practice:
tough-tricky-set-of-problms-85211.html#p638336

Hope it helps.
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SVP  Joined: 06 Sep 2013
Posts: 1652
Concentration: Finance

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ronneyc wrote:
singh_amit19 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

Ans.is E

Since AB <100 and BA<100 thus AAC<200 we can conclude that B=9 & C=0

And why is that? Could you elaborate more on this please?
Thank you very much sir.

Cheers
J Intern  Joined: 30 Apr 2010
Posts: 19

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6
My solution was a bit lengthy but it worked:

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

Manager  Joined: 03 May 2015
Posts: 62

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similar way but little different

AB+BA = AAC -> EQ 1

EQ 1 IMPLIES: B+A = 10+C - EQ 2 (DIFFICULT TO INTERPRET. I AGREE. JUST SUM UP THE UNITS DIGIT IN EQ 1)

NOW SPLIT UP EQ 1

10A+B+10B+A = 100A+10A+C - EQ 3

EQ 3 IMPLIES: 11(A+B) = 110A+C

SUBSTITUTE EQ 2 (B+A = 10+C) IN ABOVE

11(10+C) = 110A+C => 110+11C = 110A + C

IMPLIES A = 1 & C = 0

TOOK NEARLY 10 MIN... NEED TO SPEED UP..

NOW WHATS VALUE OF B... WHO CARES!!!!
Senior Manager  B
Joined: 10 Mar 2013
Posts: 488
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24 GPA: 3.88
WE: Information Technology (Consulting)

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1
10a + b +10b + a= 100a+10a+c
11(b-9a)=c what gives us this information? C cannot be a multiple of 11 it can take only values between 0 - 9 so c must equal 0 and a=1,b=9
I'm just looking for the quickiest way to solve such kind of problems. Experts do you think such representation is a reliable option ? Thanks in advance.

BTW it's no way a 600 Level question --> 600-700
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Intern  Joined: 17 Aug 2015
Posts: 6

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10
It is fairly simple if you dont get afraid and block the logic. As posted above the largest number that we are able to get adding two-digit numbers is 198 (99+99).

Looking at the clue, AAC. We have already solved half of the task. Because A=must be 1 in order to satisfy the condition above. So in this case we should find a number that has a structure of

11_

Because we have AB+ BC= AAC.

Applying the logic we have a number 1B+B1= 11C. So AB is a number smaller than 20 because the tens digit is 1. In order to get a number above or equal to 11C.

The number BA must be greater than 90. Because we have already A=1. The only solution left is B=9.
Final step:

19+91= 110 .... The solution E

Hope I helped and did not cause additional confusion.
Intern  Joined: 07 Oct 2015
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1
I solved this problem in the following way:

When two digits are interchanged and added, they are always a multiple of 11. So AB + BA = Multiple of 11. The first 3 digit multiple of 11 is 110. With this, you can use the clue given in the question that the first 2 digits are same so I don't need to consider 121, 132, 144 & so on. Now if you want to double check your answer then input numbers which can equal to 110, starting small 91+19 =110.

Also, Subtraction of two interchanged digits is always a multiple of 9. For example 54 - 45 = 9.
Intern  Joined: 10 Mar 2016
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In the correctly worked out 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?

AB
+ BA
CCA

A) 9

B) 6

C) 3

D) 2

E) 0

I got it that:

10A + B + 10B + A = AAC

11A + 11B = 110A + C

-99A + 11B = C

11 * (-9A + B) = C

But I didn't know what else to do.

I looked up the solution, it says:

C is divisible by 11, and 0 is the only digit that is divisbe by 11.

How comes that 0 is divisible by 11? Shouldn't 0 divided by 11 be equal to 0??

I hope someone knows how to explain this.

Thanks for any help! Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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Flexxice wrote:
In the correctly worked out 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?

AB
+ BA
CCA

A) 9

B) 6

C) 3

D) 2

E) 0

I got it that:

10A + B + 10B + A = AAC

11A + 11B = 110A + C

-99A + 11B = C

11 * (-9A + B) = C

But I didn't know what else to do.

I looked up the solution, it says:

C is divisible by 11, and 0 is the only digit that is divisbe by 11.

How comes that 0 is divisible by 11? Shouldn't 0 divided by 11 be equal to 0??
I hope someone knows how to explain this.

Thanks for any help! ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Check more here: number-properties-tips-and-hints-174996.html
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Flexxice wrote:
In the correctly worked out 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?

AB
+ BA
CCA

A) 9

B) 6

C) 3

D) 2

E) 0

I got it that:

10A + B + 10B + A = AAC

11A + 11B = 110A + C

-99A + 11B = C

11 * (-9A + B) = C

But I didn't know what else to do.

I looked up the solution, it says:

C is divisible by 11, and 0 is the only digit that is divisbe by 11.

How comes that 0 is divisible by 11? Shouldn't 0 divided by 11 be equal to 0??

I hope someone knows how to explain this.

Thanks for any help! Check other Addition/Subtraction/Multiplication Tables problems from our Special Questions Directory
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1
any number can be written in that form
example 213 =100*2+ 10*1 +1.3
similarly AB =10*A+1*B =10A+B
BA =10*B+1*A =10B+A

Given that the sum = AAC= 100*A+10*A+1*C =110A+C

=> 11A+11B =110A+C
=>11B-99A=C
=>11(B-9A)=C

It is also given that A, B, C are different digits
digits can vary only from 0 to 9

From 11(B-9A)=C
11*(any number except zero ) will be a two digit number which is not possible here
thus 11*0=C

Originally posted by ShravyaAlladi on 10 Aug 2016, 23:48.
Last edited by ShravyaAlladi on 11 Aug 2016, 00:00, edited 2 times in total.
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Thanks a lot guys.

My problem was question interpretation.
However, i believe its easier not to compute the equations. It just creates entropy where it is not needed.
Applying logic its enought.

Hug from Lisbon (hopping to achieve 700+),

Marcos Ramalho
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What does +B mean ?
Math Expert V
Joined: 02 Sep 2009
Posts: 55623

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dave13 wrote:
What does +B mean ?

AB
+BA
-----------
AAC

The above means that when we add a two-digit number AB to a two-digit number BA the result is a three-digit number AAC.

Check other Addition/Subtraction/Multiplication Tables problems from our Special Questions Directory
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