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18 is not enough (Posted on 2017-06-06) Difficulty: 1 of 5
Let A be any set of 19 distinct integers chosen from the arithmetic progression 1, 4, 7,..., 100.

Prove that there must be two distinct integers in A whose sum is 104.

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.0000 (1 votes)

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Solution Proof (spoiler) | Comment 1 of 2
You can safely pick the integers 1 and 52.  You can also pick 1 integer from each of the 16 pairs that sums to 104: (4,100), (7,97), (10,94) ... (49,55).  But then you are stuck.  Your 19th pick must necessarily be the other half of a pair that sums to 104.
  Posted by Steve Herman on 2017-06-06 10:33:08
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