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Ordering Pairs (Posted on 2004-11-06) Difficulty: 3 of 5
What are the smallest positive integers A, B, C, and D such that A+A > A+B > A+C > B+B > B+C > A+D > C+C > B+D > C+D > D+D ?

Note: Of all solutions, choose the one with the smallest A, then smallest B if there are more than one with the smallest A, etc.

See The Solution Submitted by Brian Smith    
Rating: 3.7500 (4 votes)

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My Solution | Comment 5 of 10 |

It is obvious that D has to be 1. The minimum number that satisfies C is 5 (gotten from trying numbers 2, 3, and 4). The same idea is applied to B and A to yield the family of numbers that satisfy the question. The numbers are

A = 10 +a

B = 7 + a

C = 5 + a

D = 1 + a. where a > 0 and "a" should be a postive integer

The lowest numbers occur when a = 0, therefore the minimum numbers are 10, 7, 5, and 1 for A, B, C, and D respectively.  

Edited on November 7, 2004, 1:56 pm
  Posted by Osi on 2004-11-07 09:44:19

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