2019 AMC 12A Problem 13

The primes $5$ and $7$ have no proper divisors here, giving $3$ choices each. Along the chain $2 \to 4 \to 8,$ there are $3 \cdot 2 \cdot 1 = 6$ colorings. Number $9$ must differ from $3,$ giving $2$ choices once $3$ is set. Number $6$ must differ from both $2$ and $3.$ Summing over the colors of $2$ and $3$ (equal in $3$ pairs, unequal in $6$ pairs), the combined factor for $4, 8, 9, 6$ totals $2 \cdot 2 \cdot (3 \cdot 2 + 6 \cdot 1) = 48.$ Multiplying by the $9$ ways for $5$ and $7$ gives $48 \cdot 9 = 432.$ Thus, the correct answer is E.