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In the correctly worked addition problem shown, where the

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In the correctly worked addition problem shown, where the [#permalink]

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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 10 Nov 2012, 04:29, edited 1 time in total.
Renamed the topic and edited the question.

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Re: PS - SET25 Q37 [#permalink]

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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

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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

the answer is (E)

:)

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Re: PS - SET25 Q37 [#permalink]

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Please note the numbers...

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

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Re: AB +BA ----------- AAC In the correctly worked [#permalink]

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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
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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
in-the-correctly-worked-addition-problem-above-a-b-c-d-128953.html

Hope it helps.
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Re: PS - SET25 Q37 [#permalink]

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New post 17 Oct 2013, 16:47
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 :)

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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.

Answer: E.

Similar questions to practice:
tough-tricky-set-of-problms-85211.html#p638336
in-the-correctly-worked-addition-problem-above-a-b-c-d-128953.html

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Re: In the correctly worked addition problem shown, where the [#permalink]

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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

Answer: E

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In the correctly worked addition problem shown, where the [#permalink]

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New post 20 Aug 2015, 07:23
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!!!!

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In the correctly worked addition problem shown, where the [#permalink]

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New post 24 Sep 2015, 06:54
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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Re: In the correctly worked addition problem shown, where the [#permalink]

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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.

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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.

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I don't understand the solution!? [#permalink]

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New post 01 Apr 2016, 22:48
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! :)

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Re: In the correctly worked addition problem shown, where the [#permalink]

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New post 02 Apr 2016, 01:06
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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New post 02 Apr 2016, 01:07
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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Problem Solving question that i was not able to understand [#permalink]

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New post 10 Aug 2016, 15:50
Hi everyone,

My name is Marcos Ramalho and i am a Portuguese Industrial Engineer.

Yesterday after dinner and during my study time, i was faced with the following PS question that i was not able to understand besides book's proposed resolution.

Whats the logic behind the problem as a hole?
Whats the logic behind the proposed equations?
Is there any other way to achieve the solution?

Thank you in advance for your help.

Best regards,

Marcos Image[img][IMG]
Attachment:
IMG_20160809_214555.jpg
[/img][/IMG]Ramalho

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In the correctly worked addition problem shown, where the [#permalink]

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New post 10 Aug 2016, 23:48
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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
adding them =11A+11B

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
zero is the only answer

Last edited by ShravyaAlladi on 11 Aug 2016, 00:00, edited 2 times in total.

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New post 10 Aug 2016, 23:56
Marcos Ramalho wrote:
Hi everyone,

My name is Marcos Ramalho and i am a Portuguese Industrial Engineer.

Yesterday after dinner and during my study time, i was faced with the following PS question that i was not able to understand besides book's proposed resolution.

Whats the logic behind the problem as a hole?
Whats the logic behind the proposed equations?
Is there any other way to achieve the solution?

Thank you in advance for your help.

Best regards,

Marcos Image[img][IMG]
Attachment:
IMG_20160809_214555.jpg
[/img][/IMG]Ramalho

Sent from my FDR-A01w using GMAT Club Forum mobile app


Merging topics. Please refer to the discussion above.

Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Thank you.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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