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Alright, we need to figure out the minimum number of weighings to identify the pile with fake coins. Each pile has 100 coins, and one pile has all fake coins. The scale can weigh a maximum of 50kg (50,000 grams).
Hmm, let’s explore: I could choose a set amount from each pile, say 1 coin from the first pile, 2 from the second, and so on, adding up a total weight and then calculating which pile is different.
Let’s think—this might only take one weighing!
I’m considering the single weighing strategy: take different amounts of coins from each pile. The extra weight from the fake pile would be the number of coins times 1 gram, but the total weight needs to be under 50kg.
If I take coins from piles in increasing amounts (1 from pile 1, 2 from pile 2, etc.), the combined weight would exceed 50,000 grams (too much!). I need to reduce the number of coins selected in a way that still allows for a clear pattern to identify the fake pile. Let’s figure out how many coins I can actually take while keeping everything under the limit.
Alright, I’m trying to figure out how to balance the total weight with the need for unique coins from each pile so we can use one weighing to identify the fake pile. We learned that the sum of coins from 1 to 100 exceeds the weight limit of the scale, so using simply “a_i = i” doesn’t work. Powers of two seem too large.
Instead, I might need to use multiple weighings with different combinations of coins, creating a multi-digit code to encode which pile is fake. Each weighing must still be under the 50kg limit.
After rethinking it, I realize that one weighing might actually work for the 100 piles. By assigning coefficients 0 to 99 to the piles, I can take coins according to those numbers, with the total number of coins being 4950. If all piles were genuine, the weight would be 49,500 grams. If the fake pile has a coefficient of “a”, then I would see an extra “a” grams. So, tracking the extra weight confirms the fake pile. The total weight stays within the 50kg limit, so one weighing seems solid.
我们可以设计一种取法，使得只用一次称重就能确定哪一堆是假币，同时保证取出的金币总重不超过50公斤。
一种常见的方法是给每一堆赋予一个不同的系数，然后一次性称重。具体步骤如下：
1. 给100堆金币依次标号1到100，并预先为每一堆指定一个不同的取币数量。注意：由于每堆只有100个金币，我们不能取超过100个金币。如果我们直接取1, 2, 3, …, 100个金币，取出的总数为
1+2+\cdots+100=\frac{100\cdot101}{2}=5050 \text{个}
总重量为 5050\times10=50500 克，超过了50,000克的上限。
2. 解决方案是让其中一堆取0个金币，其他堆依次取1, 2, …, 99个金币。这样，100堆金币的取法就是：
a_1=0,\, a_2=1,\, a_3=2,\,…,\, a_{100}=99.
3. 这样取出的金币总数为
0+1+2+\cdots+99=\frac{99\cdot100}{2}=4950 \text{个}.
若所有金币均为真币，总重量为 4950\times10=49500 克，低于50公斤的上限。
4. 由于只有一堆是假币（每个假币比真币重1克），如果假币堆的取币数为 a_i，实际称得的总重量就会比预期多 a_i 克。称重结果为
49500 + a_i \text{ 克}.
根据多出的克数 a_i（可能为0到99之间的某个数），我们就可以确定假币堆的编号：例如如果多出7克，则假币堆就是那个我们预先安排取7个金币的堆；如果没有多出额外克数（即结果正好49500克），则假币堆就是取0个金币的那一堆。
因此，只需要一次称重就能确定哪一堆是假币。 |
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发表于 2025-2-18 17:01
