Solution to the Problem:
To solve #1, let A = 1st weight, B = 2nd weight, and C = 3rd weight.
Then you can write the equations:
A + B + C = 21
3A = 2B + 4C
Solving for A in the first equation and substituting in the second:
3(21 - B - C) = 2B + 4c
Then, 63 = 5B + 7C
Solving for B:
B = (63 - 7C) / 5
Now substitute values for C until you find an integral solution for B:
The first value for C that yields an integral value for B is C = 4.
Then B = 7
So, A = 10
To solve #2, let A = 1st weight, B = 2nd weight, C = 3rd weight, and D = 4th weight.
Then you can write the equations:
A + C = 21
B + D = 21
3A + B = 4C + 5D
Then, 3(21 - C) + (21 - D) = 4C + 5D
84 = 7C + 6D
So, C = 12 - 6D/7
Then substitute values for D until you get an integral value for C.
The first one that works is D = 7
So, C = 6
A = 15
B = 14
Use a similar approach for #3 and #4.
Here is the big equation for example #4 where A, B, C, D, E, and F represent the six weights going across:
6*A + 3 * (B + C) = 2 * (D + E) + 5 * F
Then in the first secondary weight (on the left), you would have 2 * B = 1 * C.
In the secondary weight on the right, you would have: 3 * D = 1 * E.
You would also have two equations that add up to 21:
A + B + C = 21 and D + E + F = 21.
You have 5 equations and 6 variables so you must try several numbers to find the solution.
Here is Veena Mg's solution:
Correctly solved by:
| 1. Veena Mg ** | Bangalore, Karnataka, India |
| 2. Kelly Stubblefield ** | Mobile, Alabama |
| 3. Colin (Yowie) Bowey ** | Beechworth, Victoria, Australia |
** solved the extra credit