2024 AMC 10A Problem 18

$2024_b = 2b^3 + 2b + 4;$ dividing by $16$ requires $b^3 + b + 2$ to be divisible by $8$

Check $b \bmod 8;$ the residues $3, 6, 7$ work, so count those $b$ in $[5, 2024]$

In base $b,$ $2024_b = 2b^3 + 2b + 4$ $= 2(b^3 + b + 2),$ so $2024_b$ is divisible by $16$ exactly when $b^3 + b + 2$ is divisible by $8.$ Test the residues modulo $8:$ this holds precisely for $b \equiv 3, 6, 7 \pmod 8.$ Counting the $b$ with $5 \le b \le 2024$ in those three classes gives $K = 758,$ whose digit sum is $7 + 5 + 8 = 20.$ Therefore, the answer is D.