Question

В первой коробке имеется 11 карандашей, во второй 7 карандашей и в третьей 6. Разрешается удвоить количество карандашей в любой коробке, переложив в неё карандаши из другой коробки. Ни в одной коробке не поместится больше, чем 12 карандашей.

Answer 1

11+7+6=24 всего карандашей
24:4=8 карандашей в каждой каробке

Answer 2

Надо ко второй коробке прибавить 6 кар. из 3 коробки (7+6=13) потом из полученного отнять 1 и переложить в 1 коробку.

Answer 3

Problem Analysis

We have three boxes with different numbers of pencils: 11 in the first box, 7 in the second box, and 6 in the third box. We are allowed to double the number of pencils in any box by transferring pencils from another box. However, no box can contain more than 12 pencils.

We need to determine the maximum number of pencils that can be in each box after performing the allowed operations.

Solution

To solve this problem, we can follow these steps:

1. Start with the initial number of pencils in each box: 11, 7, and 6. 2. Check if any box has fewer than 12 pencils. If so, we can double the number of pencils in that box by transferring pencils from another box. 3. Repeat step 2 until no box has fewer than 12 pencils or we cannot transfer any more pencils. 4. Output the final number of pencils in each box.

Let's go through the steps in detail.

Step 1: Initial Number of Pencils

The initial number of pencils in each box is as follows: - First box: 11 pencils - Second box: 7 pencils - Third box: 6 pencils

Step 2: Doubling the Number of Pencils

We need to check if any box has fewer than 12 pencils. If so, we can double the number of pencils in that box by transferring pencils from another box.

In this case, the third box has 6 pencils, which is less than 12. We can transfer pencils from the first box to double the number of pencils in the third box.

After transferring pencils, the number of pencils in each box becomes: - First box: 11 - 6 = 5 pencils - Second box: 7 pencils - Third box: 6 + 6 = 12 pencils

Step 3: Repeating the Process

We need to check if any box has fewer than 12 pencils again. If so, we can double the number of pencils in that box by transferring pencils from another box.

In this case, the first box has 5 pencils, which is less than 12. We can transfer pencils from the second box to double the number of pencils in the first box.

After transferring pencils, the number of pencils in each box becomes: - First box: 5 + 7 = 12 pencils - Second box: 7 - 7 = 0 pencils - Third box: 12 pencils

Now, the second box has 0 pencils, so we cannot transfer any more pencils.

Step 4: Final Number of Pencils

The final number of pencils in each box is as follows: - First box: 12 pencils - Second box: 0 pencils - Third box: 12 pencils

Therefore, after performing the allowed operations, the maximum number of pencils in each box is 12 for the first and third boxes, and 0 for the second box.

Note: It's important to note that the solution provided above is based on the given problem statement and the assumption that we can only transfer pencils between boxes. If there are any additional constraints or rules, please let me know, and I'll be happy to provide a revised solution.