The prime factorization of $216$ is $2^3\times 3^3$, and since it contains $2^3$ and at least one $3$, it must be divisible by $24$.

To analyze the divisibility of $216$ by $24$, we must compare their prime factorizations.

1. Prime factorization of 24: $24=2^3\times 3$. For a number to be divisible by $24$, its prime factorization must contain at least $2^3$ (which is $8$) and at least one factor of $3$.
2. Prime factorization of 216: $216=6^3={(2\times 3)}^3=2^3\times 3^3$.
3. Comparison: The prime factorization of $216$ ($2^3\times 3^3$) contains $2^3$ and it contains $3^3$ (which is more than the required single factor of $3$). Since all the necessary prime factors of $24$ are present in the prime factorization of $216$, $216$ is divisible by $24$. In fact, $216÷24=9$.

Analyzing the options:

• The statement that the prime factorization of $216$ is $2^3\times 3^3$ and that it contains the necessary factors ($2^3$ and $3$) is the correct analysis.
• The prime factorization $2^2\times 3^4$ is incorrect; it equals $4\times 81=324$, not $216$.
• Checking for divisibility by $4$ and $6$ is explicitly stated as an unreliable test.
• While the sum of digits being a multiple of $3$ does prove divisibility by $3$, it says nothing about divisibility by $8$, which is also required for divisibility by $24$.

Ok