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Difficulty: 555-605 Level,   Arithmetic,                           
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Re: In the correctly worked addition problem shown, where the [#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


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

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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Re: In the correctly worked addition problem shown, where the [#permalink]
What does +B mean ?
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