2024 AMC 10A Problem 18

In base $b$, $2024_b = 2b^3 + 0\cdot b^2 + 2b + 4 = 2(b^3 + b + 2)$. So $16 \mid 2024_b$ exactly when $8 \mid b^3 + b + 2$.

The value of $b^3 + b + 2 \bmod 8$ depends only on $b \bmod 8$, so test the eight residues:

So the condition holds exactly when $b \equiv 3, 6, 7 \pmod 8$.

Now count such $b$ with $5 \le b \le 2024$. Since $2024 = 8\cdot 253$, each residue class contains $253$ numbers in $1, \dots, 2024$. The class $3 \bmod 8$ starts at $b = 3$, which is excluded (we need $b \ge 5$), leaving $252$; the classes $6$ and $7$ start at $6$ and $7$, so they contribute $253$ each. Hence
$$K = 252 + 253 + 253 = 758,$$
and the digit sum is $7 + 5 + 8 = 20$.

The answer is $\boxed{\textbf{(D)}\ 20}$.

A base-$b$ numeral is a polynomial in $b$, so a divisibility question about it is a congruence in $b$. Pull out common factors first ($2024_b = 2(\ldots)$ reduces the modulus from $16$ to $8$), then a polynomial modulo $m$ only depends on $b \bmod m$, so a table of $m$ residues settles everything. The final count is an arithmetic-progression count with attention to the endpoints.

• Forgetting to drop b = 3 (below the required b ≥ 5) from the residue class 3 mod 8, which gives K = 759 and digit sum 21.
• Working mod $16$ without factoring out the $2$, which makes the residue table twice as long and error-prone.