2021 AMC 10B Problems/Problem 13

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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 $n + d ?$

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


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 $3{n}^2+2n-261 = 0.$ 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 (edited for readability)

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 $9+2=\boxed{\textbf{(B)}11}$.


Video Solution by OmegaLearn (Bases and System of Equations)


~ pi_is_3.14

Video Solution by TheBeautyofMath



Video Solution by Interstigation



See Also

2021 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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