Choose two weights to compare for the first weighing. Choose two weights other than the first two to compare for the second weighing. There are three possible cases:
Case 1:Both weighings are equal
Case 2:One weighing is equal and one is not
Case 3:Both weighings are not equal
Case 1:Both weighings are equal
Choose one weight from weighing 1 and one weight from weighing two and weigh them. There are two possible results: they are equal or they are not equal
- Subcase 1: the third weighing is equal
Label the weights so that the weighings were A=B, B=C, C=D and E is the last weight. Since this is a broken scale and none of the weights have the same weight, the weights are in consecutive order of size, either A>B>C>D or D>C>B>A. Weigh A vs D to find out which is the correct order. Then weigh C vs E to find out which end E belongs.
Total weighings: 5
- Subcase 2: the third weighing is not equal
Label the weights so that the weighings were A=B, C=D, A>C and E is the last weight. A,B,C and D must be one of the three orders: A>B>D>C, B>A>D>C, or A>B>C>D. Weigh (C+A) vs (D+B). If there is a difference, that difference is at least 2g so the broken scale will be able to sense it. If (C+A)=(D+B) then the order is A>B>D>C. If (C+A)>(D+B) then the order is A>B>C>D. If (D+B)>(C+A) then the order is B>A>D>C. Relabel A,B,C,D so that the inequality is W>X>Y>Z. Weigh E vs X. If X>E then weigh E vs Y to determine if the order is W>X>Y>E>Z or W>X>Y>Z>E. If X=E then weigh E vs Y to determine if the order is W>X>E>Y>Z or W>E>X>Y>Z. If E>X then the order is E>W>X>Y>Z.
Total weighings: 5
Case 2:One weighing is equal and one is not
Label the weights so that the weighings were A>B, D=E and C is the last weight. Weigh (A+B) vs (C+D). There are three possible results: (A+B)>(C+D), (A+B)=(C+D) and (C+D)>(A+B).
- Subcase 1: (A+B)>(C+D)
A is the heaviest (A=55). B and C are consecutive and C is not consecutive to D or E. C is either consecutive to A or the smallest (C=54 or 51). Weigh A vs C and B vs D. If A=C and B=D then the order is A>C>B>D>E. If A=C and D>B then the order is A>C>B>E>D. If A>C and B=D then the order is A>E>D>B>C. If A>C and D>B then the order is A>D>E>B>C.
Total weighings: 5
- Subcase 2: (A+B)=(C+D)
A is the heaviest and B is the lightest (A=55 and B=51). C is not the middle weight (C!=53). Weigh A vs C and C vs D. There are four possible results. If A=C and C=D then the order is A>C>D>E>B. If A=C and C>D then the order is A>C>E>D>B. If A>C and C=D then the order is A>E>D>C>B. If A>C and D>C then the order is A>D>E>C>B.
Total weighings: 5
- Subcase 3: (C+D)>(A+B)
B is the lightest (B=51). A and C are consecutive and C is not consecutive to D or E. C is either consecutive to B or the largest (C=55 or 52). Weigh B vs C and A vs D. If B=C and A=D then the order is E>D>A>C>B. If B=C and A>D then the order is D>E>A>C>B. If C>B and A=D then the order is C>A>D>E>B. If C>B and A>D then the order is C>A>E>D>B.
Total weighings: 5
Case 3:Both weighings are not equal
Take the two weights from the first weighing as a group and take the two weights from the second weighing as a group. Weigh those two groups of two. There are two possible results: they are equal or they are not equal.
- Subcase 1: the third weighing is not equal
Label the weights so that the weighings were W>Y, X>Z, (W+Y)>(X+Z), and V is the last weight. W,X,Y and Z must be in the order W>X>Y>Z. Weigh V vs X. If X>V then weigh V vs Y to determine if the order is W>X>Y>V>Z or W>X>Y>Z>V. If X=V then weigh V vs Y to determine if the order is W>X>V>Y>Z or W>V>X>Y>Z. If V>X then the order is V>W>X>Y>Z.
Total weighings: 5
- Subcase 2:the third weighing is equal
Label the weights so that the weighings were A>B, C>D, (A+B)=(C+D), and E is the last weight. E must be the middle weight (E=53). The order of the weights is either A>C>E>D>B or C>A>E>B>D. Weigh A vs E. If A=E then the order is C>A>E>B>D, if A>E then the order is A>C>E>D>B.
Total weighings: 4
In every case, at most five weighings were made. It is impossible to solve by making at most four weighings. There are 120 possible ways for the five weights to be mixed up at the beginning but since the scale has only three outputs, only 3^4=81 different cases could be distinguished, and that is not enough. |
