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## digit issue

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mjmehta81 Just gettin' started!
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amit.trivedi@ymail.com GMAT Destroyer!
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Thu Mar 22, 2012 1:54 am
Yes as per the trial and error method, I got the answer as D. But just be careful in one of the cases...

19 + 92 = 111 cannot be the case that can be suited into our given equation...

We need A B C and D as distinct positive integers...

All others are genuine cases present...

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ronnie1985 GMAT Destroyer!
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Thu Mar 29, 2012 8:29 am
A has to be 1 as sum of 9+9 = 18 < 20.
Hence B+D = 11
and C+A+1 = 11 => C = 9

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klmehta03 Rising GMAT Star
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Sat Mar 31, 2012 5:00 am
Would this be 700-800 level question?

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gughanbose Really wants to Beat The GMAT!
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Tue Sep 04, 2012 7:19 am
The only three digit possibility among 111,222,333 etc is 111

That can be obtained by

18+93=111
17+94=111
16+95=111
15+96=111
14+97=111
13+98=111

Hence 9

chris558 Rising GMAT Star
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Mon Oct 01, 2012 7:56 am
AB+CD=AAA
A(10)+B+C(10)+D=A(100)+A(10)+A=111A
C10+B+D=101A

At Maximum, C10+B+D must equal less than 90+8+7=105 since all integers are different. Therefore A must equal 1 because it is the only value that will make it less than 105.

If A=1.. Then C must equal 9. If C equaled 8, then 8(10)+9+7=96 does not put it at 101. Nor does anything less than 8. And because C is an single digit integer, then it can't be greater than 10. Therefore, 8<C<10... C=9.

mparakala Rising GMAT Star
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Fri Jan 18, 2013 11:43 am
. AB
+ CD
AAA

if we separately evaluate the left half, A+C must give AA
Logically, take the highest single digit number for ex: 9 =>9+9 = 18 that means, when two sngle numbers are added, the sum can be 18 or less than 18.

so, out of the many possibilities of AA i.e., 11, 22,33 etc, AA can only take the value "11" which is less than 18

in the options, check which value can produce AA = 11.
Ans: 9

rajeshsinghgmat Really wants to Beat The GMAT!
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Fri Apr 12, 2013 1:49 am
(D) 9

A=1

A+C+1=11

A+C=10

C=10-1
C=9

B+D=11
B/D=6/5,7/4,8/3

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