proof

來源: Windows8 於 2009-06-15 15:13:58 [] [舊帖] [給我悄悄話] 閱讀數 : (353 bytes)

First, S={1005, 1006, ..., 2009} is a solution.
If there is another subset S1, then there is one number smaller than 1005 in the S1, and because none of the two can add to 2009, 1010, 2 of the members of S are not in S1, but S1 has 1005 members, so there is another in S1

請您先登陸，再發跟帖！