2021 AMC 10B Problem 13

Translate the numerals:
$$3n^2 + 2n + d = 263, \qquad 3n^2 + 2n + 4 = 216 + 36 + 6d + 1 = 253 + 6d.$$

Bound $n$ from the first equation: $3n^2 \le 263$, so $n^2 \le 87$ and $n \le 9$.

Bound $n$ from the second: the right side is at least $253$, so $3n^2 + 2n \ge 249$. For $n = 8$, $3 \cdot 64 + 16 = 208 < 249$, so $n \ge 9$.

Hence $n = 9$. Then $3 \cdot 81 + 18 + d = 263$ gives $d = 2$. Check the second equation: $243 + 18 + 4 = 265$ and $253 + 12 = 265$. Consistent, and $d = 2$ is a valid digit in both bases.

So $n + d = 9 + 2 = 11$.

The answer is $\boxed{\textbf{(B)}\ 11}$.

Base-$n$ numerals are polynomials in $n$ with small coefficients, so a numeral's value pins $n$ down tightly: the leading term $3n^2$ alone squeezes $n$ into a narrow window, and two-sided bounds close it completely. Bound first, then solve for the remaining digit; it beats solving a quadratic.

Subtract the two equations: $d - 4 = 263 - 253 - 6d = 10 - 6d$, so $7d = 14$ and $d = 2$. Then $3n^2 + 2n - 261 = 0$, which factors as $(n - 9)(3n + 29) = 0$, giving $n = 9$.

• Expanding 11d1 in base six as 216 + 36 + d + 1, forgetting the factor of 6 on the digit d.
• Solving the quadratic $3n^2 + 2n - 261 = 0$ by formula and making an arithmetic slip; the discriminant is $4 + 3132 = 3136 = 56^2$.