2021 AMC 10A Problem 11

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 $\boxed{\textbf{(E)}\ 8}$.

A base-$b$ numeral is a polynomial in $b$, so a difference of two numerals is a polynomial too, and factoring it reveals divisibility for every base at once. Because $b$ and $b - 1$ are consecutive, the product misses a factor of $3$ only when $b \equiv 2 \pmod 3$.

• Converting each choice numerically and making an arithmetic slip, or forgetting that b - 1 divisible by 3 also works.
• Misaligning the digits when subtracting (treating $221_b$ as $2210_b$), which destroys the cancellation.