August 31, 1998
Problem of the Month

The A B C D Problem

Find the lowest integral values for A, B, C, and D where:

A + B = C
A + D = B
2C = 3D
B > 0

Solution to the 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.

Correctly solved by:

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

Send any comments or questions to: David Pleacher