We are given three numbers (1, 7, 6) and three operations that are repeatedly applied to them. The operations are as follows: 1. Replace the first number with the sum of the second and third numbers minus the first number. Keep the other two numbers the same.

1. Problem Description

We are given three numbers (1, 7, 6) and three operations that are repeatedly applied to them. The operations are as follows:

1. Replace the first number with the sum of the second and third numbers minus the first number. Keep the other two numbers the same.

2. Replace the second number with the sum of the first and third numbers minus the second number. Keep the other two numbers the same.

3. Replace the third number with the sum of the first and second numbers minus the third number. Keep the other two numbers the same.

These three operations are repeated in a cycle. We need to find the numbers after 2025 operations.

2. Solution Steps

Let the three numbers be $a$, $b$, and $c$. The operations are:

1. $(a, b, c) \rightarrow (b+c-a, b, c)$

2. $(a, b, c) \rightarrow (a, a+c-b, c)$

3. $(a, b, c) \rightarrow (a, b, a+b-c)$

We start with (1, 7, 6).

Operation 1: $(1, 7, 6) \rightarrow (7+6-1, 7, 6) = (12, 7, 6)$

Operation 2: $(12, 7, 6) \rightarrow (12, 12+6-7, 6) = (12, 11, 6)$

Operation 3: $(12, 11, 6) \rightarrow (12, 11, 12+11-6) = (12, 11, 17)$

Operation 4: $(12, 11, 17) \rightarrow (11+17-12, 11, 17) = (16, 11, 17)$

Operation 5: $(16, 11, 17) \rightarrow (16, 16+17-11, 17) = (16, 22, 17)$

Operation 6: $(16, 22, 17) \rightarrow (16, 22, 16+22-17) = (16, 22, 21)$

Operation 7: $(16, 22, 21) \rightarrow (22+21-16, 22, 21) = (27, 22, 21)$

Operation 8: $(27, 22, 21) \rightarrow (27, 27+21-22, 21) = (27, 26, 21)$

Operation 9: $(27, 26, 21) \rightarrow (27, 26, 27+26-21) = (27, 26, 32)$

Operation 10: $(27, 26, 32) \rightarrow (26+32-27, 26, 32) = (31, 26, 32)$

Operation 11: $(31, 26, 32) \rightarrow (31, 31+32-26, 32) = (31, 37, 32)$

Operation 12: $(31, 37, 32) \rightarrow (31, 37, 31+37-32) = (31, 37, 36)$

Since the operations are repeated every 3 steps, we need to find the remainder when 2025 is divided by

3. $2025 \div 3 = 675$ with a remainder of

0. This means that after 2025 operations, we will have completed the same sequence as after 3 operations * 675 = 2025 operations, and the result will be the same as after 3 operations.

So after 2025 operations, the numbers will be the same as after 3 operations, which is (12, 11, 17).

3. Final Answer

(12, 11, 17)