1. **State the problem:** We have a number $n$ with prime factorization $$n = 2^5 \times x \times y^4,$$ where $x$ and $y$ are distinct primes greater than 2. 2. **Recall the rule for factors:** A number $m$ is a factor of $n$ if and only if the prime factorization of $m$ has powers of primes less than or equal to those in $n$. 3. **Analyze each candidate factor:** - $x^2$: Since $n$ has $x^1$, $x^2$ requires more $x$s than $n$ has, so $x^2$ is **not** a factor. - $4y^3$: $4 = 2^2$, so $4y^3 = 2^2 \times y^3$. Since $n$ has $2^5$ and $y^4$, powers of 2 and $y$ in $4y^3$ are less or equal to those in $n$, so $4y^3$ **is** a factor. - $xy$: $x^1 y^1$ is less than or equal to $x^1 y^4$, so $xy$ **is** a factor. - $8$: $8 = 2^3$, and $n$ has $2^5$, so $8$ **is** a factor. - $2y^5$: $2 = 2^1$ is fine, but $y^5$ requires more $y$s than $n$ has ($y^4$), so $2y^5$ is **not** a factor. - $64x$: $64 = 2^6$, but $n$ has only $2^5$, so $64x$ requires more 2s than $n$ has, so it is **not** a factor. 4. **Final answer:** The factors of $n$ among the options are $$4y^3, xy, 8.$$