2021 AMC 10B Problems/Problem 13

Problem

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 $n + d ?$

$\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16$

Solution

We can start by setting up an equation to convert $\underline{32d}$ base $n$ to base 10. To convert this to base 10, it would be 3 ${n}^2$ +2 $n$ +d. Because it is equal to 263, we can set this equation to 263. Finally, subtract $d$ from both sides to get 3 ${n}^2$ +2 $n$ = 263- $d$ .

We can also set up equations to convert $\underline{324}$ base $n$ and $\underline{11d1}$ base 6 to base 10. The equation to covert $\underline{324}$ base $n$ to base 10 is 3 ${n}^2$ +2 $n$ +4. The equation to convert $\underline{11d1}$ base 6 to base 10 is ${6}^3$ + ${6}^2$ +6 $d$ +1.

Simplify ${6}^3$ + ${6}^2$ +6 $d$ +1 so it becomes 6 $d$ +253. Setting the above equations equal to each other, we have 3 ${n}^2$ +2n+4 = 6d+253. Subtracting 4 from both sides gets 3 ${n}^2$ +2n = 6d+249.

We can then use 3 ${n}^2$ +2 $n$ = 263- $d$ and 3 ${n}^2$ +2 $n$ = 6 $d$ +249 to solve for $d$ . Set 263- $d$ equal to 6 $d$ +249 and solve to find that $d$ =2.

Plug $d$ =2 back into the equation 3 ${n}^2$ +2 $n$ = 263- $d$ . Subtract 261 from both sides to get your final equation of 3 ${n}^2$ +2 $n$ -261 = 0. Solve using the quadratic formula to find that the solutions are 9 and -10. Because the base must be positive, $n$ =9.

Adding 2 to 9 gets $\boxed{\textbf{(B)}11}$

-Zeusthemoose

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2021 AMC 10B Problems/Problem 13

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