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

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Director
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AB CD + AAA = where AB and CD are two-digit numbers [#permalink] New post 06 May 2007, 02:39
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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
Senior Manager
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 [#permalink] New post 06 May 2007, 21:55
ok that's how i solved this problem

AAA is 111 (we can't use 222 because 222/2 is a three digit number)

so now we know that 111=1B+CD B can be 3 to 8 => CD is 98 to 93

C=9

Answer is D
Director
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Re: PS = Addition [#permalink] New post 11 May 2007, 11:37
Mishari 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


is not it too easy to guess AAA as 111 cux no 2 digit integers sum up more than 198. therefore AAA cannot be other than 111 and to have 111, C must be 9.
VP
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 [#permalink] New post 11 May 2007, 11:57
adding my two cents:

The max value that AAA can get is when

AB = 99 and CD = 99

AB+CD = 198

so A has to be 1. when A=1 then AAA=111

1B+CD = 111

If C = 8 then the max value is 19+89 = 108 which is less then 111

so C has to be 9 !

13+98 = 111 or 18+93 = 111

:-D
  [#permalink] 11 May 2007, 11:57
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