Question

Let $n$ be a positive integer and $d$ be a digit such that the value of the numeral $\underline{32d}$ in base $n$ equals $263,$ and the value of the numeral $\underline{324}$ in base $n$ equals the value of the numeral $\underline{11d1}$ in base six. What is $n + d ?$

$10$

$11$

$13$

$15$

$16$

Answer: B

Concepts:number base system of equations

Difficulty rating: 1280

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Written solution:

The first statement means $3n^2 + 2n+d = 263.$

Similarly, the second statement means $3n^2 +2n +4$ $= 6^3 +6^2 +6d+1$ $= 253 + 6d.$

Subtracting these shows us that $4-d = 6d-10$ $7d = 14$ $d=2.$ Therefore, $3n^2 + 2n + 2 = 263,$ so $n(3n+2) = 261.$

This implies $n=9.$ Therefore, $n+d = 11.$

Thus, the answer is B .

2021 AMC 10B Spring Problem 13

Answer: B

Difficulty rating: 1280

Video solution:

Written solution:

Problem 13 in Other Years