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Obviously, D should be 1, the smallest positive integer. If D was bigger than 1 then 1 could be subtracted from A,B,C, and D to make a smaller solution.
To try to make things as small as possible, start with C=2. Then the inequality is
A+A > A+B > A+2 > B+B > B+2 > A+1 > 4 > B+1 > 3 > 2. B+1 can not both be an integer and be between 3 and 4, so C must be larger than 2.
If C=3 then the inequality is A+A > A+B > A+3 > B+B > B+3 > A+1 > 6 > B+1 > 4 > 2. B must be 4. Then we have A+A > A+4 > A+3 > 8 > 7 > A+1 > 6 > 5 > 4 > 2. A+1 can not both be an integer and be between 6 and 7, so B can not be 4 and C can not be 3 and B-C must be greater than 1.
If C=4 then the inequality is A+A > A+B > A+4 > B+B > B+4 > A+1 > 8 > B+1 > 5 > 2. B must be 6. Then A+A > A+6 > A+4 > 12 > 10 > A+1 > 8 > 7 > 5 > 2, A=8 is implied by 10 > A+1 > 8 and A>=9 is implied by A+4 > 12, since 8<9 B can not be 6 and C can not be 4.
If C=5 then the inequality is A+A > A+B > A+5 > B+B > B+5 > A+1 > 10 > B+1 > 6 > 2. B is 7 or 8. First, if B=8 then B+5 > A+1 > 10 would force A=10 or 11. But A+5 > B+B forces A>11, so B=8 is not a solution. Now try B=7. Then A+A > A+7 > A+5 > 14 > 12 > A+1 > 10 > 8 > 6 > 2. A=10 is forced by 12 > A+1 > 10 which agrees with A+5 > 14.
The smallest solution is A=10,B=7,C=5,D=1. This is the smallest solution since if C is larger, then A and B must also be larger. |
