No, 8 raised to the power of n (where n is a natural number) cannot end with zero.
To understand why, let’s break it down. A number ends with a zero if it is divisible by 10. For a number to be divisible by 10, it must have both 2 and 5 as its factors, since 10 = 2 × 5.
Now, consider the number 8. The prime factorization of 8 is 2 × 2 × 2, or 2³. When we raise 8 to the power of n, we get:
8n = (23)n = 23n.
This means 8n consists solely of the factor 2 and does not have the factor 5 in it, since we never introduced a 5 in any of our calculations.
Therefore, no matter what natural number n is, 8n will always be a power of 2 and will not include a factor of 5, which means it cannot be divisible by 10 and consequently cannot end with a zero.