To determine whether $8^n$ (where $n$ is a positive integer) ends with zero, we need to check if it is divisible by 10. A number ends with zero if and only if it is divisible by 10. For a number to be divisible by 10, it must be divisible by both 2 and 5.

Now, consider $8^n$:

• $8 = 2^3$, so $8^n = (2^3)^n = 2^{3n}$.
• $2^{3n}$ is clearly divisible by 2, but it does not have any factors of 5.

Since $8^n$ does not contain the factor 5, it cannot be divisible by 10. Therefore, $8^n$ does not end with zero for any positive integer $n$.

Would you like more details on this concept or have any questions?

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Here are some related questions:

1. What are the conditions for a number to end in a different digit, like 5 or 6?
2. How do you determine the last digit of a power like $7^n$?
3. What are some examples of numbers that always end with a specific digit?
4. How does the prime factorization of a number affect its divisibility by 10?
5. Can any power of a prime number end with zero?

Tip: A number ends in zero if it is divisible by both 2 and 5, meaning it must have at least one factor of 10.