Answer to August 31, 1998 Problem |
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The A B C D Problem Find the lowest integral values for A, B, C, and D where: A + B = CA + D = B 2C = 3D B > 0 |
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Solution to Problem:
Solving simultaneously, you can get D = (4/5) B and D = 4A Then A = (1/5) B and C = (6/5) B Since B > 0, the smallest number for B would be 5 (to make it an integer). Then D = 4, A = 1, and C = 6. |
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