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In the correctly worked addition problem shown, where the
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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
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Originally posted by singh_amit19 on 19 Oct 2007, 05:19.
Last edited by adkikani on 01 Jul 2018, 23:49, edited 2 times in total.
Renamed the topic and edited the question.




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Re: In the correctly worked addition problem shown, where the
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18 Oct 2013, 05:21
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.
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Re: In the correctly worked addition problem shown, where the
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19 Oct 2007, 05:42
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: In the correctly worked addition problem shown, where the
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19 Oct 2007, 05:38
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



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Re: In the correctly worked addition problem shown, where the
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02 Jun 2010, 22:09
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: In the correctly worked addition problem shown, where the
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09 Nov 2012, 13:31
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



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Re: In the correctly worked addition problem shown, where the
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10 Nov 2012, 04:31



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Re: In the correctly worked addition problem shown, where the
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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



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Re: In the correctly worked addition problem shown, where the
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19 Oct 2013, 08:13
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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Re: In the correctly worked addition problem shown, where the
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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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Re: In the correctly worked addition problem shown, where the
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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
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Re: In the correctly worked addition problem shown, where the
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27 Sep 2015, 02:08
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.



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Re: In the correctly worked addition problem shown, where the
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21 Jan 2016, 03:15
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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Re: In the correctly worked addition problem shown, where the
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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!



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Re: In the correctly worked addition problem shown, where the
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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
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Re: In the correctly worked addition problem shown, where the
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Re: In the correctly worked addition problem shown, where the
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Updated on: 11 Aug 2016, 00:00
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.



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Re: In the correctly worked addition problem shown, where the
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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.
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Re: In the correctly worked addition problem shown, where the
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05 Mar 2018, 04:33
What does +B mean ?
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