Answer to August 31, 1998 Problem |
---|
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 |
---|
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. |
1. Devin Grim | Winchester, VA |
2. Matthew Hurst | Winchester, VA |
3. Angie Cunsolo | Winchester, VA |
4. Ginger Anderson | Winchester, VA |
5. Krista Ramey | Winchester, VA |
6. Barrett Waybright | Winchester, VA |
7. Ryan Dutton | Winchester, VA |
8. John Beavers | Winchester, VA |