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logi さんへ。

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とりあえず。
私が事前に用意していたのが以下です。
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Let C_u = {(i, j, k) | 1 ≤ i ≤ 5, 1 ≤ j ≤ 5, 1 ≤ k ≤ 5} be the set of all possible combinations of indices for the gold, silver, and bronze coins; this set has 125 elements. If we let g ∈ {1, 2, 3, 4, 5} be the index of the counterfeit gold coin, s ∈ {1, 2, 3, 4, 5} the index of the counterfeit silver coin, and c ∈ {1, 2, 3, 4, 5} the index of the counterfeit bronze coin, then the set representing the triplet of counterfeit coins is the singleton set
F = {(g, s, c)}.
We now define the following subsets of C_u:
C_1 = {(i, j, k) | 2 ≤ i ≤ 5, 1 ≤ j ≤ 4, 2 ≤ k ≤ 5}.
The number of elements in C_1 is 64.
C_21 = {(i, j, k) | i = 1, 2 ≤ j ≤ 5, 2 ≤ k ≤ 5}.
The number of elements in C_21 is 16.
C_22 = {(i, j, k) | 2 ≤ i ≤ 5, j = 5, 2 ≤ k ≤ 5}.
The number of elements in C_22 is 16.
Define C_2 as the union:
C_2 = C_21 ∪ C_22.
Thus, the number of elements in C_2 is 32.
Also, let
T_2 = {(i, j, k) | 1 ≤ i ≤ 5, 2 ≤ j ≤ 5, 2 ≤ k ≤ 5}.
Then we have:
C_2 = T_2 ∩ (C_u - C_1).
Next, define
C_3 = {(i, j, k) | 2 ≤ i ≤ 5, 1 ≤ j ≤ 4, k = 1}.
The number of elements in C_3 is 16.
Define
C_41 = {(i, j, k) | i = 1, 2 ≤ j ≤ 5, k = 1}.
The number of elements in C_41 is 4.
And
C_42 = {(i, j, k) | 2 ≤ i ≤ 5, j = 5, k = 1}.
The number of elements in C_42 is 4.
Let
C_4 = C_41 ∪ C_42.
The number of elements in C_4 is 8.
Also, let
T_4 = {(i, j, k) | 1 ≤ i ≤ 5, 2 ≤ j ≤ 5, k = 1}.
Then:
C_4 = T_4 ∩ (C_u - (C_1 ∪ C_2 ∪ C_3))).
Next, define
C_5 = {(i, j, k) | i = 1, j = 1, 2 ≤ k ≤ 5}.
The number of elements in C_5 is 4.
Finally, define
C_6 = {(i, j, k) | i = 1, j = 1, k = 1}.
The number of elements in C_6 is 1.
It is clear that for any two different sets X and Y chosen from {C_1, C_2, C_3, C_4, C_5, C_6}, we have X ∩ Y = ∅. Moreover,
C_u = C_1 ∪ C_2 ∪ C_3 ∪ C_4 ∪ C_5 ∪ C_6.
In terms of element counts, we have:
125 = |C_u| = |C_1| + |C_2| + |C_3| + |C_4| + |C_5| + |C_6| = 64 + 32 + 16 + 8 + 4 + 1.
To determine the indices g, s, and c of the counterfeit coins, we proceed with the following operations:
First Spell:
We test whether F = {(g, s, c)} is a subset of C_1.
If the result is true (i.e., the bag glows), then it is almost evident that g, s, and c can be determined with the remaining 6 spells.
If the result is false, we move on to the next step.
Second Spell:
We test whether F = {(g, s, c)} is a subset of T_2.
If the result is true, then it immediately follows that F ⊂ C_2 = C_21 ∪ C_22.
Then, with the remaining 5 spells, g, s, and c can be determined almost trivially.
If the result is false, we proceed to the next step.
Third Spell:
We test whether F = {(g, s, c)} is a subset of C_3.
If the result is true, then with the remaining 4 spells g, s, and c can be determined almost trivially.
If false, we continue to the next step.
Fourth Spell:
We test whether F = {(g, s, c)} is a subset of T_4.
If the result is true, then it immediately follows that F ⊂ C_4 = C_41 ∪ C_42.
Then, with the remaining 3 spells, g, s, and c can be determined almost trivially.
If the result is false, we move on to the next step.
Fifth Spell:
We test whether F = {(g, s, c)} is a subset of C_5.
If the result is true, then with the remaining 2 spells, g, s, and c can be determined almost trivially.
If the result is false, then F = {(g, s, c)} must be equal to C_6,
which implies that g = s = c = 1.