In the correctly worked computation above A, B, C and D repr : GMAT Problem Solving (PS)
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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, 07:31
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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, 00:36, 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, 07:54
2
KUDOS
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, 08:03, 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, 08:01
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, 19:33
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Re: In the correctly worked computation above A, B, C and D repr   [#permalink] 06 Jun 2016, 19:33
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# In the correctly worked computation above A, B, C and D repr

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