Question

У нас имеются весы с одной чашей,которые показывают вес в граммах, и тримешка золотых монет. В одном мешке всемонеты фальшивые, весом55 граммкаждая,в двух других – монеты подлинные, вескаждой из них –50 граммов. Какоенаименьшее число взвешиваний потребуется,чтобы определить, в каком из мешковфальшивые монеты?

Answer 1

Наименьшее число - 2 раза. 1 раз Можно взять 2 мешка. Если их масса больше 100 грамм, то в одном из них фальшивые манеты, значит останится только положить один из мешочков ещё раз на весы. если они покажут 50 гр.-то манеты настоящие, а если нет-фальшивые

Answer 2

Problem Analysis

To determine which bag contains the counterfeit coins, we need to find a way to differentiate the bags based on the weight of the coins. We have a scale that shows the weight in grams and three bags: one with counterfeit coins weighing 55 grams each, and two with genuine coins weighing 50 grams each. We need to determine the minimum number of weighings required to identify the bag with the counterfeit coins.

Solution

To solve this problem, we can use a binary search approach. We will divide the coins into two groups and compare their weights. Based on the result, we can determine which bag contains the counterfeit coins. Here's a step-by-step solution:

1. Divide the coins into two groups: Bag A and Bag B. 2. Take one coin from Bag A and one coin from Bag B and place them on the scale. 3. There are three possible outcomes: - If the weights are equal, the counterfeit coins are in the third bag. - If the weight of the coins on the scale is less than 100 grams, the counterfeit coins are in Bag A. - If the weight of the coins on the scale is greater than 100 grams, the counterfeit coins are in Bag B. 4. If the counterfeit coins are in Bag A or Bag B, repeat the process with the corresponding bag. 5. Continue this process until the bag with the counterfeit coins is identified.

Minimum Number of Weighings

To determine the minimum number of weighings required, we can use the concept of binary search. Since we have three bags, we can divide them into two groups in the first weighing. If the first weighing does not reveal the bag with the counterfeit coins, we will need to perform additional weighings.

In the worst-case scenario, the counterfeit coins are in the last bag to be weighed. Therefore, we need to perform at least two weighings to identify the bag with the counterfeit coins.

Example

Let's go through an example to illustrate the solution. We have Bag A, Bag B, and Bag C. Bag C contains the counterfeit coins.

1. Weigh Bag A and Bag B. - If the weights are equal, Bag C contains the counterfeit coins. - If the weight of Bag A is less than Bag B, Bag C contains the counterfeit coins. - If the weight of Bag A is greater than Bag B, Bag C does not contain the counterfeit coins. 2. If Bag C does not contain the counterfeit coins, we repeat the process with Bag A and Bag B. 3. Weigh Bag A and Bag B again. - If the weights are equal, Bag C contains the counterfeit coins. - If the weight of Bag A is less than Bag B, Bag A contains the counterfeit coins. - If the weight of Bag A is greater than Bag B, Bag B contains the counterfeit coins.

In this example, we needed two weighings to identify the bag with the counterfeit coins.

Conclusion

To determine which bag contains the counterfeit coins, we can use a binary search approach. By dividing the coins into two groups and comparing their weights, we can identify the bag with the counterfeit coins in a minimum of two weighings.