Question

Сто орехов разложены на пять кучек. В первой и второй в сумме - 51 орех, во второй и третьей- 44, в третьей и четвертой- 31, в четвертой и пятой -33. Найдите число орехов в каждой кучке и докажите это

Answer 1

Для краткой понятной записи условия и решения обозначим орехи в кучках: К1, К2, К3, К4, К5.
1) К1+К2+К3+К4+К5=100.
2) К1+К2=51.
3) К2+К3=44.
4) К3+К4=31.
5) К4+К5=33.
Подставим выражения 2) и 4) в выражение 1) и найдем К5: 51+31+К5=100; К5=100-82=18 орехов.
Из выражения 5) найдем К4=33-К5=33-18=15 орехов.
Из выражения 4) найдем К3=31-К4=31-15=16 орехов.
Из выражения 3) найдем К2=44-К3=44-16=28 орехов.
Из выражения 2) найдем К1=51-К2=51-28=23 ореха.
Ответ: 23, 28, 16, 15, 18.

Answer 2

1) 100-51-31=18  ( 5 кучка)
2) 33-18=15 (4кучка)
3) 31-15=16 (3кучка)
4) 44-16=28 (2кучка)
5) 51-28=23 (1кучка)

Answer 3

Problem Analysis

We are given that there are 100 nuts divided into five piles. The sum of the first and second piles is 51, the sum of the second and third piles is 44, the sum of the third and fourth piles is 31, and the sum of the fourth and fifth piles is 33. We need to find the number of nuts in each pile and prove it.

Solution

Let's assume the number of nuts in the first pile is x. Then, the number of nuts in the second pile is also x. The sum of the first and second piles is 51, so we can write the equation:

x + x = 51

Simplifying the equation, we get:

2x = 51

Solving for x, we find that x = 25.5. However, since we are dealing with whole numbers of nuts, this is not a valid solution. Therefore, we can conclude that there is no solution for the given conditions.

Proof

To prove that there is no solution, we can analyze the other sums given in the problem.

The sum of the second and third piles is 44. If we assume the number of nuts in the second pile is y, then the number of nuts in the third pile is also y. We can write the equation:

y + y = 44

Simplifying the equation, we get:

2y = 44

Solving for y, we find that y = 22. However, this contradicts our assumption that the number of nuts in the second pile is x, which we found to be 25.5. Therefore, there is no valid solution for the given conditions.

We can apply the same logic to the other sums given in the problem, and we will find that there is no valid solution for any of them. Hence, we can conclude that there is no solution for the given conditions.

Conclusion

The problem of dividing 100 nuts into five piles with the given sums does not have a valid solution.