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The number 2011 has the property that one of its digits is the sum of its other digits, i.e., 0+1+1=2. Compute the sum of the two largest integers less than 2011 with this property.

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2002+1980? There must be a catch. –  Henning Makholm Oct 27 '11 at 2:17
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-1: I don't think that a contest problem, posted by itself without further motivation, is an appropriate question for this site. –  Nate Eldredge Oct 27 '11 at 2:40
    
At the very least, OP should have shown what he tried to solve this problem... –  J. M. Oct 27 '11 at 2:42
    
I have tried to answer the problem but I answered it incorrectly... But I have found my mistake. ;) –  Tayyeb Din Oct 27 '11 at 3:05
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closed as not constructive by Nate Eldredge, Hans Lundmark, t.b., Asaf Karagila, J. M. Oct 30 '11 at 13:56

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1 Answer

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Simply searching by hand down from 2011 turns up 2002, but there are no others close by (greater than 2000). So just below 2000, we have numbers of the form 19XX. For three of the digits to add up to the last one, they must add up to the 9 in the hundreds spot (because otherwise the sum would be more than one digit long). We already have a 1, so the other digits must add up to 8; this leaves us with (8, 0), (7, 1), (6, 2), (5, 3), and (4, 4) as candidate pairs. We want to find the largest, so we put the largest possible number in the tens spot and end up with 1980. So the two largest integers less than 2011 with that property are 2002 and 1980, and their sum is 3982.

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you missed 1980. –  Soarer Oct 27 '11 at 2:19
    
Why not try 1980? –  AMPerrine Oct 27 '11 at 2:20
    
@Soarer and AMPerrine many thanks, corrected. How silly of me. –  smackcrane Oct 27 '11 at 2:22
    
The same mistake I made^^ Sadly I lost a point on the competition cause of this careless error. –  Tayyeb Din Oct 27 '11 at 3:07
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