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AB + CD = AAA, where AB and CD are two-digit numbers and AA

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AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 06 Aug 2012, 10:57
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AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1
(B) 3
(C) 7
(D) 9
(E) Cannot be determined
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 06 Aug 2012, 11:06
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navigator123 wrote:
AB + CD = AAA, where AB and CD are two-digit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C?

(A) 1
(B) 3
(C) 7
(D) 9
(E) Cannot be determined


Since AB and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.

So, A=1 and we have 1B+CD=111

Now, C can not be less than 9, because no two-digit integer with first digit 1 (1B<20) can be added to two-digit integer less than 90, so that to have the sum 111 (if CD<90, so if C<9, CD+1B<111).

Hence C=9.

Answer: D.
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 16 Aug 2012, 11:11
Hi Bunuel,

Doesn't 82+19=111. But you say no two digit number with 1st digit as 1 can be added to a number less than 90 to get a sum of 111. Can you elaborate??

One more thing is only 111 fits into this scenario as it is the sum of two digit numbers which cannot exceed 188 right??
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 16 Aug 2012, 11:14
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rajathpanta wrote:
Hi Bunuel,

Doesn't 82+19=111. But you say no two digit number with 1st digit as 1 can be added to a number less than 90 to get a sum of 111. Can you elaborate??

One more thing is only 111 fits into this scenario as it is the sum of two digit numbers which cannot exceed 188 right??


82+19=101, not 111.

As for 111: since AB and CD are two-digit integers, their sum can give us only one three digit integer of a kind of AAA: 111.
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 17 Aug 2012, 22:10
Hi Bunuel

How did u check that there is only one such number as AAA when AB and CD are added together..Please can you explain the logic behind it?
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 18 Aug 2012, 01:24
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ratinarace wrote:
Hi Bunuel

How did u check that there is only one such number as AAA when AB and CD are added together..Please can you explain the logic behind it?


AAA is a 3-digit number with all 3 digits alike, so it could be: 111, 222, 333, ..., 999. But the sum of any two 2-digit numbers cannot be more than 99+99=198, so AB+CD can only be 111.
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 10 Nov 2012, 06:47
Bunnuel, is there any algebraic approach to this?
Please help out
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 25 Feb 2014, 20:43
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AB+CD=AAA

10A+B+10C+D=100A+10A+A
10C+B+D=101A
(10C+B+D)/101=A

Just need to realize that A,B,C,D are integers that cannot exceed 9, as they are single digits

Keeping this in mind, C is at most 9, so (90+B+D)/101=A -->B+D=11 and A=1

For A = 2, it becomes obvious that either B, C, or D will have to exceed 9 and is thus not possible.

So C=9
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 06 Oct 2015, 03:19
the greatest 2 digit number is 99 so 99+99 =198...
thats shows that AAA can not be 222 or 333 or ...... 999, it has to be 111....
now that aaa is 111,ab+cd=aaa ...
a bit of logic, a=1 and c=9...
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 18 Mar 2017, 10:29
we can get there by trying numbers:
AB
CD
AAA
first what come to my mind is that A can be 1, hence B+D=11, then 1(A) + 9 (C)=10 + 1
Answer is D
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AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 20 Jul 2017, 03:45
Please ignore

One thing i am clear that the no has to be 111. But why is 24+87=111 wrong?
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 20 Jul 2017, 05:50
gps5441 wrote:
Please ignore

One thing i am clear that the no has to be 111. But why is 24+87=111 wrong?


AB + CD = AAA.

The tens digit in AB and the digits in AAA are the same. So, A = 1. All this is explained here: https://gmatclub.com/forum/ab-cd-aaa-wh ... l#p1110751
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 25 Nov 2017, 10:18
Bunuel

Is this a GMAT like question?
What is the actual source of this question?
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 25 Nov 2017, 10:48
mrinal0308 wrote:
Bunuel

Is this a GMAT like question?
What is the actual source of this question?


Don't know the source but the question is quite GMAT-like.

Check other Addition/Subtraction/Multiplication Tables problems from our Special Questions Directory
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 19 Jan 2018, 03:56
This is how I solved it.

There are 2 options:

Option 1: B + D = A, in that case : A+C = 10A+A => C = 10A (this is impossible knowing that max value of C is 9 and min value is 0)

Option 2: B+D = 10A (which means 1 carries over), in that case: A+C+1 = 10A+A => C= 10A- 1 (only 1 solution possible where A = 1 and C = 9)

Hope this helps !
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 19 Jan 2018, 06:38
Since the only possible value for AAA is 111, therefore A will always be 1. Now AB is in the range of 10 to 19 according to this fact but if we subtract 10 or 11 from 111, we get triple digit number. So lets just take max and minimum possible value of AB, that would be 12 and 19 respectively. Now if you subtract 12 from 111 you would get 99 and if you subtract 19 from 111 you get 92. (92 and 99 are max and min value of CD). Therefore C's value will always be 9.
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Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA  [#permalink]

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New post 17 Apr 2018, 10:09
Is this approach correct? Bunuel. Since we know that AAA is 111 Because is impossible for the sum of two digit number to be more than 198 (99+99), and A must be in the range of 0-9. Knowing this we can reflect the next table:

1_
A= 1 because of the limitation of the problems
B= FROM 0 TO 9 , EXCEPT FOR A

9_ ( IN ORDER TO SUM 111 C MUST BE 9 )

C=9
D= Any number between 0-9 excempt for A,B,C

19 / 92 = 111
18 / 93 = 111
17 / 94 = 111
16 / 95 = 111

Goes on ......
Re: AB + CD = AAA, where AB and CD are two-digit numbers and AA &nbs [#permalink] 17 Apr 2018, 10:09
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