Last visit was: 23 Jul 2024, 19:54 It is currently 23 Jul 2024, 19:54
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# In the correctly worked addition problem shown, where the

SORT BY:
Tags:
Show Tags
Hide Tags
Manager
Joined: 11 Sep 2005
Posts: 181
Own Kudos [?]: 3489 [329]
Given Kudos: 0
Math Expert
Joined: 02 Sep 2009
Posts: 94589
Own Kudos [?]: 643389 [108]
Given Kudos: 86728
Intern
Joined: 17 Aug 2015
Posts: 4
Own Kudos [?]: 62 [48]
Given Kudos: 1
GMAT Club Legend
Joined: 12 Sep 2015
Posts: 6804
Own Kudos [?]: 30854 [13]
Given Kudos: 799
7
Kudos
6
Bookmarks
Top Contributor
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

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

General Discussion
Intern
Joined: 02 Aug 2007
Posts: 12
Own Kudos [?]: 134 [2]
Given Kudos: 0
1
Kudos
1
Bookmarks
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
Director
Joined: 08 Jun 2005
Posts: 523
Own Kudos [?]: 575 [31]
Given Kudos: 0
26
Kudos
5
Bookmarks
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

Intern
Joined: 18 May 2010
Posts: 13
Own Kudos [?]: 8 [6]
Given Kudos: 14
3
Kudos
3
Bookmarks

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
Tutor
Joined: 16 Oct 2010
Posts: 15140
Own Kudos [?]: 66816 [11]
Given Kudos: 436
Location: Pune, India
8
Kudos
3
Bookmarks
petercao 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

Let's see what the addition should look like:

A B
+B A
--------
A A C

Look at the big picture first. Two 2 digit numbers get added to give you a 3 digit number, AAC. A has to be 1. It cannot be 0 because then AAC is not a three digit number and it cannot be anything more than 1 because 2 two digit numbers cannot make 200 or more (To make 200, at least 1 number must be a 3 digit number. The maximum you can make out of the sum of two digit numbers is 99+99 = 198)

So A = 1.

A B
+B A
--------
A A C

Next we see that B+A gives us something ending in C but A + B gives us something ending in A. This means that B+A must have given us a carryover 1.

(1)1 B
+ B 1
-------
1 1 C

To get a carryover 1, B has to be 9 so that sum = 9 + 1 = 10. Therefore, C must be 0.
Manager
Joined: 03 Aug 2011
Posts: 188
Own Kudos [?]: 56 [1]
Given Kudos: 12
Location: United States
Concentration: General Management, Entrepreneurship
GMAT 1: 750 Q49 V44
GPA: 3.38
WE:Engineering (Computer Software)
1
Kudos
VeritasPrepKarishma wrote:
petercao 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

Let's see what the addition should look like:

A B
+B A
--------
A A C

Look at the big picture first. Two 2 digit numbers get added to give you a 3 digit number, AAC. A has to be 1. It cannot be 0 because then AAC is not a three digit number and it cannot be anything more than 1 because 2 two digit numbers cannot make 200 or more (To make 200, at least 1 number must be a 3 digit number. The maximum you can make out of the sum of two digit numbers is 99+99 = 198)

So A = 1.

A B
+B A
--------
A A C

Next we see that B+A gives us something ending in C but A + B gives us something ending in A. This means that B+A must have given us a carryover 1.

(1)1 B
+ B 1
-------
1 1 C

To get a carryover 1, B has to be 9 so that sum = 9 + 1 = 10. Therefore, C must be 0.

karishma, thanks for your answer. Just a question about the thought process. I started off thinking AA must be 11,22,33,44,55, etc. then realizing the sum can only be in the 100s as you did yourself.

then I end up with :

1B
+ B1
11C

From here you were talking about how 1+B should = B +1 unless there is a carryover. But shouldn't we have taken a similar, more heuristic approach like we did earlier with the < 200 to think that 1B and B1 cannot be anything less than 9 because 18 + 81 (where b is 8) is already less than 100. then b must be larger and thus 9.

I'm just curious if when you do problems like this, you're actually thinking of the carry, etc. because that seems very difficult to replicate when you are really taking a test.
Tutor
Joined: 16 Oct 2010
Posts: 15140
Own Kudos [?]: 66816 [3]
Given Kudos: 436
Location: Pune, India
2
Kudos
pinchharmonic wrote:

karishma, thanks for your answer. Just a question about the thought process. I started off thinking AA must be 11,22,33,44,55, etc. then realizing the sum can only be in the 100s as you did yourself.

then I end up with :

1B
+ B1
11C

From here you were talking about how 1+B should = B +1 unless there is a carryover. But shouldn't we have taken a similar, more heuristic approach like we did earlier with the < 200 to think that 1B and B1 cannot be anything less than 9 because 18 + 81 (where b is 8) is already less than 100. then b must be larger and thus 9.

You can absolutely do that and I actually like it better. We solve the entire question using the same approach.

pinchharmonic wrote:
I'm just curious if when you do problems like this, you're actually thinking of the carry, etc. because that seems very difficult to replicate when you are really taking a test.

Ok, I will tell you my thought process. When I solve questions, I almost never use a pen and paper. So when I solved it, I didn't write
1B
+ B1
11C
down. So I didn't see the big picture in the next step. After figuring out that A = 1, I still focused on the original addition and saw that B+A in the right column was giving C while in the left column, the same addition was giving A. This should not have been the case since I expect both additions to give the same result. I expected symmetry there but didn't find it. So obviously, the two additions were not the same. How is it possible? Only if the second addition had a carry over from the first one! This was possible only if B = 9 since I knew that A = 1. I was not thinking of the carry over, I was looking for relations among the digits.
Intern
Joined: 11 Jul 2012
Posts: 35
Own Kudos [?]: 24 [6]
Given Kudos: 0
6
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: 94589
Own Kudos [?]: 643389 [5]
Given Kudos: 86728
2
Kudos
3
Bookmarks
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.
VP
Joined: 06 Sep 2013
Posts: 1335
Own Kudos [?]: 2451 [0]
Given Kudos: 355
Concentration: Finance
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: 14
Own Kudos [?]: 76 [13]
Given Kudos: 2
12
Kudos
1
Bookmarks
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

Intern
Joined: 07 Oct 2015
Posts: 2
Own Kudos [?]: 5 [5]
Given Kudos: 66
2
Kudos
2
Bookmarks
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: 15
Own Kudos [?]: 80 [1]
Given Kudos: 0
1
Bookmarks
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
Joined: 02 Sep 2009
Posts: 94589
Own Kudos [?]: 643389 [1]
Given Kudos: 86728
1
Bookmarks
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
Math Expert
Joined: 02 Sep 2009
Posts: 94589
Own Kudos [?]: 643389 [0]
Given Kudos: 86728
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
Intern
Joined: 29 Jun 2016
Posts: 37
Own Kudos [?]: 26 [3]
Given Kudos: 5
1
Kudos
2
Bookmarks
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.
VP
Joined: 09 Mar 2016
Posts: 1142
Own Kudos [?]: 1028 [0]
Given Kudos: 3851