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

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Director
Joined: 30 Nov 2006
Posts: 591
Location: Kuwait
AB CD + AAA = where AB and CD are two-digit numbers [#permalink]

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06 May 2007, 03:39
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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
Senior Manager
Joined: 03 May 2007
Posts: 270

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06 May 2007, 22:55
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

Director
Joined: 26 Feb 2006
Posts: 899

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11 May 2007, 12:37
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
Joined: 08 Jun 2005
Posts: 1145

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11 May 2007, 12:57

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

11 May 2007, 12:57
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