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In the correctly worked computation above A, B, C and D repr

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In the correctly worked computation above A, B, C and D repr [#permalink]

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14 May 2013, 08:31
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AB
+BA
---------
CDC

In the correctly worked computation above A, B, C and D represent distinct nonzero digits. What is the value of A+B+C+D

A. 14
B. 16
C. 18
D. 20
E. It cannot be determined from given information
[Reveal] Spoiler: OA

Last edited by Bunuel on 23 Jul 2013, 01:36, edited 3 times in total.
Renamed the topic and edited the question.

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Re: in the correctly worked computation above A, B, C and D repr [#permalink]

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14 May 2013, 08:54
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clciotola wrote:
AB
+BA
---------
CDC

in the correctly worked computation above A, B, C and D represent distinct nonzero digits. What is the value of A+B+C+D

a 14
b 16
c 18
d 20

e it cannot be determined from given information

Sum of two double digit integers is 198 at the maximum.
So as CDC is a three digit number we have C = 1

CDC = 1D1

now A+B = some number which gives a carry and with units place as one.It must give a carry because if we add units ie A+B we are getting C and then if we add tens we get CD .So there must be a carry for tens place.

So by adding A and B we must get 11.(As the maximum carry we can get by adding two single digit numbers is 1 and also C = 1)

now D will be 2 as A+B + 1 = 12

so we will get CDC as 121.

Sum of the digits will be 11 + 1 + 2

ie 14

correct me if I am wrong
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Last edited by SrinathVangala on 14 May 2013, 09:03, edited 2 times in total.

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Re: in the correctly worked computation above A, B, C and D repr [#permalink]

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14 May 2013, 09:01
SrinathVangala wrote:
clciotola wrote:
AB
+BA
---------
CDC

in the correctly worked computation above A, B, C and D represent distinct nonzero digits. What is the value of A+B+C+D

a 14
b 16
c 18
d 20

e it cannot be determined from given information

Sum of two double digit integers is 198 at the maximum.
So as CDC is a three digit number we have C = 1

CDC = 1D1

now A+B = some number which gives a carry and with units place as one.It must give a carry because if we add units ie A+B we are getting three and then if we add tens we get CD .So there must be a carry for tens place.

So by adding A and B we must get 11.(As the maximum carry we can get by adding two single digit numbers is 1)

now D will be 2 as A+B + 1 = 12

so we will get CDC as 121.

Sum of the digits will be 11 + 1 + 2

ie 14

correct me if I am wrong

IT's C but i don't understand how is calculate

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Re: In the correctly worked computation above A, B, C and D repr [#permalink]

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06 Jun 2016, 20:33
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Re: In the correctly worked computation above A, B, C and D repr [#permalink]

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18 Jul 2017, 11:17
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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In the correctly worked computation above A, B, C and D repr [#permalink]

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19 Jul 2017, 13:47
clciotola wrote:
AB
+BA
---------
CDC

In the correctly worked computation above A, B, C and D represent distinct nonzero digits. What is the value of A+B+C+D

A. 14
B. 16
C. 18
D. 20
E. It cannot be determined from given information

From tens' digit D which does not equal C we know that A + B > 9 (i.e., we have to carry).

We know that AB and BA have inverted digits.

We know that CDC has to be 1_1 because

--In order to make D different from C, we have to carry.

--But you can't add two single digits in the ones' column and get anything greater than 17 (8+9), so the most you can carry is 10.

--If the most you can carry is 10, then in the tens' place, the most you can get is 180 (8_ + 9_ + 1_ = 18_) --

-- which means that C, in the hundreds' place, HAS to be 1.

We also know, where O = Odd and E = Even, that the equation must be

OE
EO
OEO --->

C = 1 is odd. Only E + O = O

Similarly, for tens' column, we know we have to carry 1 (10). O + 1 = E. Then E + E = E

So C = 1, AB and BA must be EO and OE (or vice-versa), and A and B must sum to 11 .

First I tried

92
29
121 - no, D must be different

BUT

83
38
121 - bingo - and

74
47
121 - and

65
56
121

No matter which set you choose A + B + C + D = 14

Something in this problem seems to be connected to 9: all the numbers that work are separated by 9 (38, 47, 56...). And all the nonzero digits show up as addend digits or sum digits - except 9.

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Re: In the correctly worked computation above A, B, C and D repr [#permalink]

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24 Jul 2017, 02:41
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KUDOS
It can be done in a more simpler way as follows:

1. AB + BA = (10A+B)+(10B+A) = 11(A+B) , i.e the sum is always a multiple of 11
2. Since on the other side CDC = AB+BA , therefore CDC is also a multiple of 11
3. The only 3 digit multiple of 11 in the form of CDC is 121
4. Now for 11(A+B) = 121, A+B = 11 therefore A and B can be 6 and 5

So the sum is A+B+C+D= 6+5+1+2= 14

Kudos [?]: 1 [1], given: 11

Re: In the correctly worked computation above A, B, C and D repr   [#permalink] 24 Jul 2017, 02:41
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