Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 11 Sep 2005
Posts: 302

In the correctly worked addition problem shown, where the [#permalink]
Show Tags
Updated on: 10 Nov 2012, 04:29
6
This post received KUDOS
24
This post was BOOKMARKED
Question Stats:
80% (01:48) correct 20% (02:01) wrong based on 815 sessions
HideShow timer Statistics
AB +BA  AAC In the correctly worked addition problem shown, where the sum of the twodigit positive integers AB and BA is the threedigit 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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by singh_amit19 on 19 Oct 2007, 05:19.
Last edited by Bunuel on 10 Nov 2012, 04:29, edited 1 time in total.
Renamed the topic and edited the question.



Intern
Joined: 02 Aug 2007
Posts: 36

Re: PS  SET25 Q37 [#permalink]
Show Tags
19 Oct 2007, 05:38
1
This post received KUDOS
1
This post was BOOKMARKED
singh_amit19 wrote: AB +BA  AAC
In the correctly worked addition problem shown, where the sum of the twodigit positive integers AB and BA is the threedigit 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



VP
Joined: 08 Jun 2005
Posts: 1140

11
This post received KUDOS
2
This post was BOOKMARKED
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)



Intern
Joined: 18 May 2010
Posts: 14

Re: PS  SET25 Q37 [#permalink]
Show Tags
02 Jun 2010, 22:09
2
This post was BOOKMARKED
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



Intern
Joined: 11 Jul 2012
Posts: 41

Re: AB +BA  AAC In the correctly worked [#permalink]
Show Tags
09 Nov 2012, 13:31
2
This post received KUDOS
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
Joined: 02 Sep 2009
Posts: 44586

Re: In the correctly worked addition problem shown, where the [#permalink]
Show Tags
10 Nov 2012, 04:31
1
This post received KUDOS
Expert's post
2
This post was BOOKMARKED



Current Student
Joined: 06 Sep 2013
Posts: 1919
Concentration: Finance

Re: PS  SET25 Q37 [#permalink]
Show Tags
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 twodigit positive integers AB and BA is the threedigit 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



Math Expert
Joined: 02 Sep 2009
Posts: 44586

Re: PS  SET25 Q37 [#permalink]
Show Tags
18 Oct 2013, 05:21
7
This post received KUDOS
Expert's post
3
This post was BOOKMARKED
jlgdr wrote: ronneyc wrote: singh_amit19 wrote: AB +BA  AAC
In the correctly worked addition problem shown, where the sum of the twodigit positive integers AB and BA is the threedigit 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 twodigit 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 threedigit integer (consider 18+81=99), thus B=9 > 19+91=110 > C=0. Answer: E. Similar questions to practice: toughtrickysetofproblms85211.html#p638336inthecorrectlyworkedadditionproblemaboveabcd128953.htmlHope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 30 Apr 2010
Posts: 21

Re: In the correctly worked addition problem shown, where the [#permalink]
Show Tags
19 Oct 2013, 08:13
2
This post received KUDOS
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



Manager
Joined: 03 May 2015
Posts: 99

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



Director
Joined: 10 Mar 2013
Posts: 563
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

In the correctly worked addition problem shown, where the [#permalink]
Show Tags
24 Sep 2015, 06:54
10a + b +10b + a= 100a+10a+c 11(b9a)=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 > 600700
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Intern
Joined: 17 Aug 2015
Posts: 7

Re: In the correctly worked addition problem shown, where the [#permalink]
Show Tags
27 Sep 2015, 02:08
4
This post received KUDOS
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 twodigit 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
Posts: 4

In the correctly worked addition problem shown, where the [#permalink]
Show Tags
21 Jan 2016, 03:15
2
This post received KUDOS
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
Posts: 17

I don't understand the solution!? [#permalink]
Show Tags
01 Apr 2016, 22:48
In the correctly worked out addition problem shown, where the sum of the twodigit positive integers AB and BA is the threedigit 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
Joined: 02 Sep 2009
Posts: 44586

Re: In the correctly worked addition problem shown, where the [#permalink]
Show Tags
02 Apr 2016, 01:06
Flexxice wrote: In the correctly worked out addition problem shown, where the sum of the twodigit positive integers AB and BA is the threedigit 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: numberpropertiestipsandhints174996.html
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 44586

Re: In the correctly worked addition problem shown, where the [#permalink]
Show Tags
02 Apr 2016, 01:07



Intern
Joined: 29 Jun 2016
Posts: 48

In the correctly worked addition problem shown, where the [#permalink]
Show Tags
Updated on: 11 Aug 2016, 00:00
1
This post received KUDOS
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 =>11B99A=C =>11(B9A)=C
It is also given that A, B, C are different digits digits can vary only from 0 to 9
From 11(B9A)=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
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.



Intern
Joined: 14 Jul 2016
Posts: 6

Re: In the correctly worked addition problem shown, where the [#permalink]
Show Tags
11 Aug 2016, 11:10
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



Senior Manager
Joined: 09 Mar 2016
Posts: 428

In the correctly worked addition problem shown, where the [#permalink]
Show Tags
05 Mar 2018, 04:33
What does +B mean ?



Math Expert
Joined: 02 Sep 2009
Posts: 44586

Re: In the correctly worked addition problem shown, where the [#permalink]
Show Tags
05 Mar 2018, 04:54




Re: In the correctly worked addition problem shown, where the
[#permalink]
05 Mar 2018, 04:54






