2021 AMC 10A Problem 11

1. In base $b$, the digits count powers of $b$, just as in base $10$ they count powers of $10$. For example, in base $10$, $2021 = 2 \times 1000 + 0 \times 100 + 2 \times 10 + 1$.
2. So in base $b$:
$$2021_b = 2b^3 + 0 \cdot b^2 + 2b + 1, \qquad 221_b = 2b^2 + 2b + 1.$$
3. Subtract. The $2b + 1$ parts cancel:
$$2021_b - 221_b = 2b^3 - 2b^2.$$
4. Factor out $2b^2$:
$$2b^3 - 2b^2 = 2 \times b \times b \times (b - 1).$$
5. Since $3$ is prime, a product is a multiple of $3$ exactly when one of its factors is. The factor $2$ is not, so we need $b$ or $b - 1$ to be a multiple of $3$.
6. Check each choice:

1. Check $b = 8$ directly: $2 \times 64 \times 7 = 896$, and its digit sum $8 + 9 + 6 = 23$ is not a multiple of $3$.

The answer is $\boxed{\textbf{(E)}\ 8}$.

A number written in base $b$ is really an expression in $b$, so a difference of two such numbers can be simplified and factored like algebra. Factoring answers the question for every base at once. Since $b - 1$ and $b$ are next to each other, the only bases that fail are those one less than a multiple of $3$, like $8$.

Use the factored form to compute each value: $b = 3$ gives $2 \times 9 \times 2 = 36$, $b = 4$ gives $96$, $b = 6$ gives $360$, $b = 7$ gives $588$, and $b = 8$ gives $896$. Their digit sums are $9, 15, 9, 21, 23$, and only $23$ is not a multiple of $3$.

Write both numbers as polynomials in $b$:
$$2021_b = 2b^3 + 0b^2 + 2b + 1,\qquad 221_b = 2b^2 + 2b + 1.$$
Subtracting, the $2b$ and $1$ terms cancel:
$$2021_b - 221_b = 2b^3 - 2b^2 = 2b^2(b - 1).$$

Since $3$ is prime and does not divide $2$, this is divisible by $3$ exactly when $3 \mid b$ or $3 \mid b - 1$, that is, when $b \equiv 0$ or $1 \pmod 3$.

Check the choices: $3 \equiv 0$, $4 \equiv 1$, $6 \equiv 0$, $7 \equiv 1$ all work, but $8 \equiv 2 \pmod 3$ does not. Indeed for $b = 8$ the value is $2 \cdot 64 \cdot 7 = 896$, whose digit sum $23$ is not a multiple of $3$.

The answer is $\boxed{\textbf{(E)}\ 8}$.