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Posts: 181
Updated on: 02 Jan 2021, 02:42
Originally posted by singh_amit19 on 19 Oct 2007, 04:19.
Last edited by Bunuel on 02 Jan 2021, 02:42, edited 3 times in total.
Last edited by Bunuel on 02 Jan 2021, 02:42, edited 3 times in total.
Renamed the topic and edited the question.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
81% (02:04) correct
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(02:31)
wrong
based on 5721
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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
+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
Solving this question helps.
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is even better.
jlgdr
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
Hope it helps.
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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.
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.
Posts: 6,727
singh_amit19
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
General Discussion
singh_amit19
Ans.is E
Since AB <100 and BA<100 thus AAC<200 we can conclude that B=9 & C=0
Posts: 520
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)
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)
Posts: 13
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
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
petercao
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.
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pinchharmonic
Joined: 03 Aug 2011
Last visit: 15 Aug 2013
Posts: 188
Location: United States
Concentration: General Management, Entrepreneurship
GMAT 1: 750 Q49 V44
GPA: 3.38
WE:Engineering (Computer Software)
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Posts: 188
VeritasPrepKarishma
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.
pinchharmonic
You can absolutely do that and I actually like it better. We solve the entire question using the same approach.
pinchharmonic
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.
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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
Brother Karamazov
Brother Karamazov
singh_amit19
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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jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,296
Concentration: Finance
Schools: Wharton '17 (A) HBS '17 (WA$)
Schools: Wharton '17 (A) HBS '17 (WA$)
Posts: 1,296
ronneyc
And why is that? Could you elaborate more on this please?
Thank you very much sir.
Cheers
J
Posts: 14
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
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
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.
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.
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!
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!
Flexxice
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
Check other Addition/Subtraction/Multiplication Tables problems from our Special Questions Directory
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Posts: 35
Updated on: 10 Aug 2016, 23:00
Originally posted by ShravyaAlladi on 10 Aug 2016, 22:48.
Last edited by ShravyaAlladi on 10 Aug 2016, 23:00, edited 2 times in total.
Last edited by ShravyaAlladi on 10 Aug 2016, 23:00, edited 2 times in total.
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
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
What does +B mean ?
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