https://file.aiquickdraw.com/mathbot/file/17263900365568zsheesg.jpg

The question in the image is written in Indonesian and translates to:

"Find the smallest positive integer whose cube ends in 2017."

To solve this problem, we are looking for a positive integer \( n \) such that the last four digits of \( n^3 \) are 2017.

### Steps:
1. We need to find \( n \) such that:
   \[
   n^3 \equiv 2017 \ (\text{mod} \ 10000)
   \]
   This means that the cube of \( n \) leaves a remainder of 2017 when divided by 10000.

2. We can attempt to check values of \( n \) and calculate \( n^3 \) until we find a cube that ends in 2017.

Let me calculate the smallest such integer \( n \).The smallest positive integer \( n \) whose cube ends in 2017 is \( n = 9073 \).

Would you like further explanation on how this result was obtained, or do you have any other questions? 

Here are some related questions for further exploration:

1. How is modular arithmetic used to solve such problems?
2. Can we generalize this method to find cubes ending in any specific number?
3. What is the significance of the modulus 10000 in this problem?
4. How does the cube function change the properties of numbers in modular arithmetic?
5. Are there other numbers \( n \) whose cubes also end in 2017?

**Tip**: Modular arithmetic is especially useful in number theory and cryptography, where you deal with patterns in the remainders of division.

Temukan bilangan bulat positif terkecil yang kubusnya berakhir pada 2017.

This problem involves finding the smallest positive integer whose cube ends in 2017. It covers key concepts like modular arithmetic and number theory, and includes steps to solve n^3 ≡ 2017 (mod 10000). Suitable for students in Grades 10-12.

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