Question

У Васи и Пети по 55 гирь весом 1, 2,, 55 кг. Они по очереди подкладывают свои гири каждый на свою чашу двухчашечных весов. Первым ходит Вася. Петя выигрывает, если разность масс гирь на чашах окажется равной 50 кг. Сможет ли он этого добиться?

Answer 1

Да он сможет
так что не волнуйся

Answer 2

Петя может повторять ходы Васи. В определенный  момент Вася будет вынужден  сходить гирей 50 кг и  проиграет.

Answer 3

Problem Analysis

Vasya and Petya have 55 weights, each weighing 1, 2, ..., 55 kg. They take turns placing their weights on their respective sides of a two-pan balance scale. Vasya goes first. Petya wins if the difference in weight between the two pans is exactly 50 kg. The question is whether Petya can achieve this outcome.

Solution

To determine if Petya can win, we need to analyze the possible combinations of weights that Vasya and Petya can place on the scale. Let's consider the following:

1. Vasya starts by placing a weight on his side of the scale. Petya can then respond by placing a weight on his side. 2. The maximum weight that Vasya can place is 55 kg, and the minimum weight is 1 kg. 3. The maximum weight that Petya can place is also 55 kg, and the minimum weight is 1 kg.

To simplify the analysis, we can consider the difference in weight between the two pans as a variable. Let's call this variable "diff." Initially, diff is equal to 0 kg.

When Vasya places a weight on his side, the value of diff increases by the weight Vasya placed. When Petya places a weight on his side, the value of diff decreases by the weight Petya placed.

If Petya wants to win, the value of diff should be exactly 50 kg. This means that the sum of the weights placed by Petya should be equal to the sum of the weights placed by Vasya plus 50 kg.

Let's analyze the possible combinations of weights that can lead to a diff of 50 kg:

1. If Vasya places a weight of 55 kg, Petya needs to place weights with a total weight of 5 kg (55 kg - 50 kg). 2. If Vasya places a weight of 54 kg, Petya needs to place weights with a total weight of 6 kg (54 kg - 50 kg). 3. If Vasya places a weight of 53 kg, Petya needs to place weights with a total weight of 7 kg (53 kg - 50 kg). 4. ...

We can continue this analysis until Vasya places a weight of 1 kg.

By examining the possible combinations, we can conclude that Petya can win if and only if the sum of the weights from 1 kg to 5 kg is present among the weights available to Vasya. This is because Petya needs to place weights with a total weight of 5 kg to achieve a diff of 50 kg.

Let's check if the sum of the weights from 1 kg to 5 kg is present among the weights available to Vasya:

1 + 2 + 3 + 4 + 5 = 15 kg

According to the given information, Vasya and Petya have weights ranging from 1 kg to 55 kg. Since the sum of the weights from 1 kg to 5 kg (15 kg) is present among the available weights, Petya can achieve a diff of 50 kg and win the game.

Therefore, Petya can achieve a diff of 50 kg and win the game.