Three Weighings That Do Not Fully Order the Tokens
The math club has four brass tokens labeled A, B, C, and D. Each token has a different whole-number weight in grams, and none weighs less than 1 g. A balance scale produced only three results, recorded in the club notebook:
- A and B together are heavier than C and D together.
- A and C together balance exactly against B and D.
- A and D together are lighter than B and C.
A visitor glances at the notebook and announces, "So the order from heaviest to lightest is B, A, C, D." The club president is not convinced. Some members start hunting for sample weights, others start building a chain of inequalities, and one member insists that "different whole-number weights" must pin down a single ordering. The president asks the group to determine exactly what the three results force, what they leave open, and whether the visitor's announcement is a conclusion or a guess.
1. Using the three comparisons, prove which token must be heaviest and which must be lightest. State clearly which comparisons between the middle two tokens are still undecided.
Read the worked answer
Let the weights be a, b, c, d. Write t = a − b.
From result 2: a + c = b + d, so a − b = d − c, giving a = b + t and d = c + t.
From result 1: a + b > c + d. Substituting: (b + t) + b > c + (c + t), so 2b > 2c, hence b > c.
From result 3: a + d < b + c. Substituting: (b + t) + (c + t) < b + c, so 2t < 0, hence t < 0. Therefore a < b and d < c.
Finally, a − d = (b + t) − (c + t) = b − c > 0, so a > d.
Putting the forced relations together: b > a > d and b > c > d. So B is certainly heaviest and D certainly lightest. The relation between A and C is not fixed by the three results - a could exceed c or fall below it. The visitor's full ordering is a guess in its middle two positions.
2. Demonstrate that the order of A and C cannot be determined by giving two assignments of distinct whole-number weights that both satisfy all three comparisons but order A and C differently.
Read the worked answer
Assignment 1 - A heavier than C: a = 9, b = 10, c = 8, d = 7. Check: A + B = 19 > C + D = 15 ✓; A + C = 17 = B + D = 17 ✓; A + D = 16 < B + C = 18 ✓. All weights distinct whole numbers. Here a > c.
Assignment 2 - C heavier than A: a = 5, b = 10, c = 6, d = 1. Check: A + B = 15 > C + D = 7 ✓; A + C = 11 = B + D = 11 ✓; A + D = 6 < B + C = 16 ✓. All distinct whole numbers. Here c > a.
Both assignments obey every recorded comparison and the stated conditions on the weights, yet they reverse the middle pair. That is the definition of underdetermination: the data admit two models with different answers to the question "which of A and C is heavier?" No amount of searching for the "right" numbers will close the gap, because the gap is in the data, not in the arithmetic.
Note also that B stays heaviest and D stays lightest in both models, confirming that those two conclusions were genuinely forced.
3. A member argues that requiring distinct whole-number weights should force a unique order. Explain why that requirement does not help, and state the single additional weighing that would settle the A-versus-C question.
Read the worked answer
Distinctness is a constraint on how far apart the weights can be, not a constraint that links A to C. The three comparisons only ever involve A and C in ways that cancel out: result 2 sets a − b equal to d − c, and results 1 and 3 fix only the sign of b − c and of t = a − b. Nothing in the algebra produces a − c. Varying the gap between b and c while keeping t negative generates valid assignments on both sides of a = c, which is why the two models in the previous answer both survive; distinctness simply rules out a = c itself, narrowing the field without picking a side.
One weighing settles it: put A in one pan and C in the other. If A sinks, a > c; if C sinks, c > a; a balance is impossible under the stated distinctness condition, which is itself a useful check that the original three results were recorded correctly.