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.

First D=1 is deduced to be the lowest integer, and it is 1. To me, it is easier to start from here and work upwards. With logic done correctly, the results should be no different than working from the other way.

If D=1, then C must be at least 3. There are 2 integers between C+C and D+D. D+D=2, so C+C can't be five.

Similar logic, implies B=5 and A=7. It also follows that when ever a lower value is increased, the next value must be increased as well to maintain a distance of at least 2 integers.

So A,B,C,D = 7,5,3,1. However, B+D is now = C+C, so the value of C must be increased. We now have A,B,C,D = 8,6,4,1.

Now A+C = B+B. Try to increase A, so we have. A,B,C,D = 9,6,4,1. But now, B+C = A+D. So C must be increased. We now have A,B,C,D = 9,7,5,1.

Still A+C = B+B.  A must be increased once again. Finally it works. A,B,C,D = 10,7,5,1

Posted by Michael on 2004-11-26 23:13:39