2021 AMC 10B Spring Problem 13

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

Translate each base numeral into base ten

Subtract the two equations to solve for the digit first

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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 .

Problem 13 in other years