Prove that $8^{n}$ cannot end in zero for any natural number $n \in N$.
(N/A) For a number to end with the digit $0$,its prime factorization must contain the factors $2$ and $5$.
The prime factorization of $8$ is $2^3$.
Therefore,$8^n = (2^3)^n = 2^{3n}$.
The only prime factor in the prime factorization of $8^n$ is $2$.
Since the factor $5$ is not present in the prime factorization of $8^n$,it is impossible for $8^n$ to end with the digit $0$ for any natural number $n \in N$.
The prime factorization of $8$ is $2^3$.
Therefore,$8^n = (2^3)^n = 2^{3n}$.
The only prime factor in the prime factorization of $8^n$ is $2$.
Since the factor $5$ is not present in the prime factorization of $8^n$,it is impossible for $8^n$ to end with the digit $0$ for any natural number $n \in N$.
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