2024 AMC 10A Problem 18

1. Turn the base-$b$ number into ordinary arithmetic. The digits $2, 0, 2, 4$ in base $b$ mean
$$2024_b = 2b^3 + 0 \cdot b^2 + 2b + 4.$$
For example, in base $10$ this is $2000 + 20 + 4$. (The digit $4$ is only allowed if $b \ge 5$, which is why the problem starts at $5$.)
2. Pull out the common factor $2$. $2b^3 + 2b + 4 = 2(b^3 + b + 2)$. So $2024_b$ is a multiple of $16$ exactly when $b^3 + b + 2$ is a multiple of $8$.
3. Only the remainder of $b$ matters. If you add $8$ to $b$, then $b^3 + b + 2$ changes by a multiple of $8$ (every term in $(b+8)^3 - b^3$ has a factor of $8$, and so does $(b + 8) - b$). So whether $b^3 + b + 2$ is a multiple of $8$ depends only on the remainder when $b$ is divided by $8$. Test the eight possible remainders:

So the good values of $b$ are those that leave remainder $3$, $6$ or $7$ when divided by $8$. (In congruence notation, $b \equiv 3, 6, 7 \pmod 8$, which just means those remainders.)
4. Count them from $5$ to $2024$. Since $2024 = 8 \cdot 253$, the numbers $1$ to $2024$ contain exactly $253$ numbers with each remainder.
- Remainder $3$: $3, 11, 19, \ldots$ The first one, $3$, is too small ($b$ must be at least $5$), so this gives $252$.
- Remainder $6$: $6, 14, \ldots$ All $253$ count.
- Remainder $7$: $7, 15, \ldots$ All $253$ count.
5. Add: $K = 252 + 253 + 253 = 758$. Its digits add to $7 + 5 + 8 = 20$.

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

A number written in base $b$ is really an expression in $b$, so a question about dividing it becomes a question about $b$. Pull out common factors first (here the $2$ turned $16$ into $8$). Then only $b$'s remainder matters, so a short table of $8$ cases settles everything. At the end, watch the edges of the range: the starting value $5$ removes one number.

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 $\boxed{\textbf{(D)}\ 20}$.