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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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Solution Spoiler | Comment 3 of 10 |

Breaking down the inequalities:

A>B>C>D
A+C>2B
B+C>A+D
A+D>2C
2C>B+D

Interpreting the inequalities:

Let ab=A-B, bc=B-C and cd=C-D.  Note that all three must be positive integers

A+C>2B
A-B>B-C
ab>bc

B+C>A+D
C-D>A-B
cd>ab

A+D>2C
A-C>C-D
ab+bc>cd

2C>B+D
This inequality isn't needed.

Therefore:
ab+bc>cd>ab>bc

The values that minimize A, B, C, and D, trivially proven:
ab=3
bc=2
cd=4

The solution:
D=1
C=5
B=7
A=10


  Posted by Tristan on 2004-11-06 18:54:36
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