The problem states:

The first digit of a four-digit natural number is 7. If this digit is moved to the last position, the resulting four-digit number is 864 less than the original number. Find the initial number.

Let's solve this step by step:

1. Let the original number be $N$: Since the first digit is 7, the number can be written as: $N = 7000 + x$ where $x$ is the rest of the number (a three-digit number).

2. When the digit 7 is moved to the last position: The new number becomes: $M = 10x + 7$

3. The condition is that the new number is 864 less than the original: $N - M = 864$

4. Substitute $N$ and $M$ into the equation: $(7000 + x) - (10x + 7) = 864$

5. Simplify the equation: $7000 + x - 10x - 7 = 864$ $6993 - 9x = 864$

6. Solve for $x$: $6993 - 864 = 9x$ $6129 = 9x$ $x = \frac{6129}{9} = 681$

7. Find $N$: Substitute $x = 681$ back into $N$: $N = 7000 + 681 = 7681$

Final Answer: The initial number is $\mathbf{7681}$.

Would you like further clarification or additional details?

Related Questions:

1. How would the solution change if the number had five digits instead of four?
2. What if the first digit was not fixed, and any digit could be moved?
3. How can we generalize this solution for $n$-digit numbers?
4. What is the impact on the calculation if the difference was not 864 but another number?
5. Can this problem be solved using modular arithmetic for quicker computation?

Tip: Always check your solution by substituting it back into the problem's conditions to ensure accuracy!