AB + CD = AAA
and AB<100 and CD<100 => AB+CD < 200 => AAA is 111.
=> 10A + B + 10C + D = 111
=> B+D+10C = 101
B+D form the units digits so B+C should be 1 or B+C = 10+1. B+C !=1, So B+C = 10+1
=> 10C = 101-11 = 90
=> C = 9
IMO D
Addition
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neelgandham
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user123321
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should be D.gmatblood 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
IMO: B
when two digit numbers are added then the three digit +ve integers formed should be always less than or equal to 198. So the sum should be 111. So we have 1A+CD = 111.
above equation follows given conditions only when we have C as 9.
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[email protected]
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How you guys are saying AAA should be equal to 111. Please elaborate.
Ashish
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shankar.ashwin
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You can't add two 2-digit numbers to get a value more than 198.
Max possible is 99+99. (Again in this problem this would not be valid as A,B,C and D are given to be distinct)
Sum of AB+CD is of the form AAA. AAA can only be 111. (<198)
Max possible is 99+99. (Again in this problem this would not be valid as A,B,C and D are given to be distinct)
Sum of AB+CD is of the form AAA. AAA can only be 111. (<198)
[email protected] wrote:How you guys are saying AAA should be equal to 111. Please elaborate.
