一些逻辑训练的谜题：娜塔莎、十二点钟和茶
我的同事，娜塔莎喜欢小孩子，就给他们出了一些题。而我在十二点钟和娜塔莎闲谈时，总会为她补充一些，而她也觉得这还不错。到了下午，我们一起喝茶时，我们会把饼干放在鸟笼边，猫头鹰则会将其啄成碎片。我们的生活就是这样悠闲地度过的，和这所大学里的本科生的终日忙碌仿佛处在两个世界，他们没有心情欣赏校园里的建筑多么有艺术感，也看不见柳树和池塘蕴含着十四行诗——就算你不是莎士比亚也可以发现——只等着被抓出来和记录下来，但我们是刚好反过来的。
在这里，我会归纳一些我们设计的，助益于逻辑能力的谜题。
在这里，我会归纳一些我们设计的，助益于逻辑能力的谜题。
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>>2024年6月29日阿列克谢的好朋友S先生最近拜访了一个波兰餐厅“Krzysiak小屋”在密歇根州，...
A. 鸭血汤（15元）
B. 花园沙拉（10元）
C. 牛肝（25元）
D. 炸虾（20元）
E. 鱼肉三明治（15元）
Let the amount of Czernina be A, so the price is 15A.
Let the amount of Salad be B, so the price is 10B.
Let the amount of Liver be C, so the price is 25C.
Let the amount of Shrimp be D, so the price is 20D.
Let the amount of Sandwich be E, so the price is 15E.
S先生点了其中4种食物，最终花了总共120元。
<*> 15A + 10B + 25C + 20D + 15E = 120. One and only one of the unknowns is 0.
而且他点的牛肝的数量多于鱼肉三明治（若某个菜品没有被点，则数量应视为0，下同），
<**> C > E
-> C > 0
他点的炸虾的数量多于花园沙拉，
<***> D > B
-> D > 0
他点的鸭血汤的数量不同于鱼肉三明治，
<****> A != E
花园沙拉的数量不同于牛肝。
<*****> B != C
Therefore, S's order is C + D + 3C2 (from A, B, and E), as one of the following:
A + B + C + D
A + E + C + D
B + E + C + D
Given <*> and the circumstances, A, B, C, D, and E must be positive integers in each of the following cases:
Case 1: 15A + 10B + 25C + 20D = 120
Case 2: 15A + 15E + 25C + 20D = 120
Case 3: 10B + 15E + 25C + 20D = 120
Case 1:
15A + 10B + 25C + 20D = 120
5 * (3A + 2B + 5C + 4D) = 120
3A + 2B + 5C + 4D = 24
Given <***>, D > B, and B >= 1
-> D >= 2
The minimum order would be A = 1, B = 1, C = 1, and D = 2.
Let the unknown remainder be X:
3(1) + 2(1) + 5(1) + 4(2) + X = 24
18 + X = 24
X = 6
For all unknowns A, B, C, and D to be integers,
X = 6 can only come from:
- 3A when A = 2,
- 2B when B = 3, or
- 4D + 2B when both = 1.
IF X = 3A and A = 2, the equation solves as:
3(1) + 2(1) + 5(1) + 4(2) + 3(2) = 24
A = 3
B = 1
C = 1
D = 2
The solution is FALSE because it contradicts <*****> B != C.
IF X = 2B and B = 3, the equation solves as:
3(1) + 2(1) + 5(1) + 4(2) + 2(3) = 24
A = 1
B = 4
C = 1
D = 2
The solution is also FALSE because it contradicts <***> D > B.
IF X = 4D + 2B when both = 1, the equation solves as:
3(1) + 2(1) + 5(1) + 4(2) + 2(1) + 4(1) = 24
A = 1
B = 2
C = 1
D = 3
The solution meets all given conditions.
Therefore, Mr. S ordered:
A. 鸭血汤（15元） x 1 = 15
B. 花园沙拉（10元） x 2 = 20
C. 牛肝（25元） x 1 = 25
D. 炸虾（20元） x 3 = 60
Total: 120
QED
By the same reasoning, the following cases
Case 2: 15A + 15E + 25C + 20D = 120
Case 3: 10B + 15E + 25C + 20D = 120
can be falsified as they do not satisfy the given conditions.
Falsification of Case 2:
15A + 15E + 25C + 20D = 120
3A + 3E + 5C + 4D = 24
Given <**> C > E, E >= 1,
-> C >= 2
The minimum order would be A = 1, E = 1, C = 2, and D = 1. Let the unknown remainder be X:
3(1) + 3(1) + 5(2) + 4(1) + X = 24
20 + X = 24
X = 4
The only possible integer solution is X = 4(1).
3(1) + 3(1) + 5(2) + 4(1) + 4(1) = 24
Then:
A = 1
E = 1
C = 2
D = 2
The solution is FALSE because it contradicts <****> A != E.
Falsification of Case 3:
10B + 15E + 25C + 20D = 120
2B + 3E + 5C + 4D = 24
Given <***> D > B and <**> C > E,
The minimum order would be B = 1, E = 1, C = 2, D = 2. Let the unknown remainder be X:
2(1) + 3(1) + 5(2) + 4(2) + X = 24
23 + X = 24
X = 1
There is no possible integer solution for X to be divisible by any of the 4 items, therefore the case is falsified.
