MATH.11:5.1 - Derive weights for an exchange construction

States are triples (a,b,c) of nonnegative integer counts. Two rules are allowed:

• Replace two A units with three B units when a≥2: (a,b,c) → (a-2,b+3,c).
• Replace one B unit with one C unit when b≥1: (a,b,c) → (a,b-1,c+1).

Starting with four A units, can the construction end with exactly five C units and nothing else?

The unweighted count changes under the first rule. Try I(a,b,c)=α*a+β*b+γ*c. The two differences are -2*α+3*β and -β+γ. Their joint equations have the solution (α,β,γ)=(3,2,2), giving:

I(a,b,c)=3*a+2*b+2*c.

Every allowed step preserves this value. The start (4,0,0) has value 12; the target (0,0,5) has value 10. The target is impossible under the stated rules, however many steps are attempted.

The same invariant helps formulate a useful alternative: six C units have value 12. They are attainable. Apply the first rule twice to obtain (0,6,0), then the second rule six times to obtain (0,0,6). This sequence supplies the contribution that equality of the invariant alone left open.

Now start at (0,3,0) and ask for (2,0,0). Both values are 6, but the first rule cannot create A and the second only consumes B to create C. Thus the requested target is unreachable. Adding the reverse exchange (a,b,c) → (a+2,b-3,c) when b≥3 makes that target reachable in one step. Its invariant change is 2*3-3*2=0, so the old invariant survives while the reachability answer changes.

For a material or operational application, interpret the counted kinds and allowed exchanges before using this mathematical result. The calculated weights express preservation by these rules; identifying them with physical mass or monetary value requires the corresponding subject account.