# 2021 AMC 10B Problems/Problem 13

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## 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 1

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 $3{n}^2+2n+d.$ Because it is equal to 263, we can set this equation to 263. Finally, subtract $d$ from both sides to get $3{n}^2+2n = 263-d$.

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 $3{n}^2+2n+4.$ The equation to convert $\underline{11d1}$ base 6 to base 10 is ${6}^3+{6}^2+6d+1.$

Simplify ${6}^3+{6}^2+6d+1$ so it becomes $6d+253.$ 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 equations $$3{n}^2+2n = 263-d$$ $$3{n}^2+2n = 6d+249$$ to solve for $d$. Set $263-d$ equal to $6d+249$ and solve to find that $d=2$.

Plug $d=2$ back into the equation $3{n}^2+2n = 263-d$. Subtract 261 from both sides to get your final equation of $3{n}^2+2n-261 = 0.$ We solve using the quadratic formula to find that the solutions are $9$ and $-29/3.$ Because the base must be positive, $n=9.$

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

-Zeusthemoose (edited for readability) -solution corrected by Billowingsweater

## Solution 2 $32d$ is greater than $263$ when both are interpreted in base 10, so $n$ is less than $10$. Some trial and error gives $n=9$. $263$ in base 9 is $322$, so the answer is $9+2=\boxed{\textbf{(B)} ~11}$.

-SmileKat32

## Solution 3

We have $$3n^2 + 2n + d = 263$$ $$3n^2 + 2n + 4 = 6^3 + 6^2 + 6d + 1$$ Subtracting the 2nd from the 1st equation we get \begin{align*} d-4 &= 263 - (216 + 36 + 6d + 1) \\ &= 263 - 253 - 6d \\ &= 10 - 6d \end{align*} Thus we have $d=2.$ Substituting into the first, we have $3n^2 + 2n + 2 = 263 \Rightarrow 3n^2 + 2n - 261 = 0.$ Factoring, we have $(n-9)(3n+29)=0.$ A digit cannot be negative, so we have $n=9.$ Thus, $d+n=2+9=\boxed{\textbf{(B)} ~11}$

mathboy282 signing off ~ $\LaTeX$ fixed by Lamboreghini

~ pi_is_3.14

~IceMatrix

## Video Solution by Interstigation

~Interstigation

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