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Difference between revisions of "2021 AMC 10B Problems/Problem 13"

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Plug <math>d</math>=2 back into the equation 3<math>{n}^2</math>+2<math>n</math> = 263-<math>d</math>. Subtract 261 from both sides to get your final equation of 3<math>{n}^2</math>+2<math>n</math>-261 = 0. Solve using the quadratic formula to find that the solutions are 9 and -10. Because the base must be positive, <math>n</math>=9.
 
Plug <math>d</math>=2 back into the equation 3<math>{n}^2</math>+2<math>n</math> = 263-<math>d</math>. Subtract 261 from both sides to get your final equation of 3<math>{n}^2</math>+2<math>n</math>-261 = 0. Solve using the quadratic formula to find that the solutions are 9 and -10. Because the base must be positive, <math>n</math>=9.
  
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Adding 2 to 9 gets <math>\boxed{\textbf{(B)}11}</math>.
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Adding 2 to 9 gets <math>\boxed{\textbf{(B)}11}</math>
 
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-Zeusthemoose
 
-Zeusthemoose

Revision as of 20:30, 11 February 2021

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

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