Case 24 · investigation · Hard · About 6–10 minutes
Three sealed museum parcels, A, B, and C, have reached the station freight office. Someone altered exactly one line of the weighing record. The parcels themselves were not changed.
Vale has four recorded totals and a trustworthy ordering note. Determine the false line and recover the three individual weights.
The four recorded lines
- A and B together: 9 units.
- B and C together: 11 units.
- A and C together: 8 units.
- A, B, and C together: 15 units.
Verified conditions
- Each parcel weighs a whole number of units, from 1 through 9 inclusive.
- A is lighter than B, and B is lighter than C. No two weights are equal.
- Exactly three recorded lines are correct. The remaining line gives an incorrect total.
- Every weighing uses the same unit and the same unchanged parcels. No container weight or hidden packing adjustment needs to be added.
Which numbered line was altered? Give A’s, B’s, and C’s weights, then calculate all four totals to check your answer.
You may solve by algebra or systematically test increasing triples. A contradiction among the totals does not tell you which line is false by itself. The trustworthy light-to-heavy ordering is an essential additional constraint.
The illustration is atmospheric. The sizes or number of drawn parcels and the empty balance do not supply measured evidence.
Hint 1 - reveal when you want a nudge
Temporarily suppose the first two pair totals and the grand total are correct. Subtract each pair total from the grand total.
Hint 2 - reveal when you want a nudge
Those three lines give C=15−9 and A=15−11. Use either pair total to find B, then test the remaining line.
Reveal the solution and explanation
Line 3 is false. A weighs 4 units, B weighs 5, and C weighs 6. Their totals are A+B=9, B+C=11, A+C=10, and A+B+C=15.
To prove uniqueness, consider the alternatives. If line 1 were false, the other lines would give A=4, B=7, C=4, violating the strict order. If line 2 were false, they would give A=2, B=7, C=6, again violating the order.
If line 4 were false, adding the three pair totals gives twice the combined weight: 28. The total would be 14, producing A=3, B=6, C=5, which also fails B<C.
Only treating line 3 as false yields an increasing triple in the allowed range and exactly three correct lines. Its recorded 8 should be 10.
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