Recall the physics of the problem, which is that each point where a string is tied to a rod exerts a torsional force equal to the sum of all the weights hanging from it times the distance between the string and the point at which the rod is suspended from above, and that a rod hanging straight means that the sum of the clockwise forces is equal to the sum of the counterclockwise forces.
|
H is three times J, and the sum of these two weights plus double C's weight is equal to D's weight. There are three possibilities for this: (J, H, C, D) = (1, 3, 2, 8) or (1, 3, 4, 12) or (2, 6, 1, 10), for sums of 14, 20, or 19 respectively. This sum must be equal to E+F+G, which also satisfies 2E+F = G. The fact that the small numbers are already mostly taken up by J, H, and C eliminate all of the possibilities except that (J, H, C, D) = (1, 3, 4, 12) and (E, F, G) = (2, 7, 11) |
|
The unused weights are 5, 6, 8, 9, 10. Considering the top bar, we have 40 + 3A = B + 3(38-A-B), or 74 = 6A + 2B or 37 = 3A + B. The only solution is A=9, B=10. The remaining weights satisfy 3K = L+2M, which means that (K,L,M) = (6,8,5) |
|
The answer is 9 10 4 12 2 7 11 3 1 6 8 5. |
Notes: These puzzles come together quickly if you see the key, and are impossible if you don't. I didn't see it in the test, and this was one of the last problems that I solved afterward.