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

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<math>\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16</math>
 
<math>\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16</math>
  
−
==Solution==
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==Solution 1==
−
We can start by setting up an equation to convert <math>\underline{32d}</math> base <math>n</math> to base 10. To convert this to base 10, it would be 3<math>{n}^2</math>+2<math>n</math>+d. Because it is equal to 263, we can set this equation to 263. Finally, subtract <math>d</math> from both sides to get 3<math>{n}^2</math>+2<math>n</math> = 263-<math>d</math>.
+
We can start by setting up an equation to convert <math>\underline{32d}</math> base <math>n</math> to base 10. To convert this to base 10, it would be <math>3{n}^2+2n+d.</math> Because it is equal to 263, we can set this equation to 263. Finally, subtract <math>d</math> from both sides to get <math>3{n}^2+2n = 263-d</math>.
  
−
We can also set up equations to convert <math>\underline{324}</math> base <math>n</math> and <math>\underline{11d1}</math> base 6 to base 10. The equation to covert <math>\underline{324}</math> base <math>n</math> to base 10 is 3<math>{n}^2</math>+2<math>n</math>+4. The equation to convert <math>\underline{11d1}</math> base 6 to base 10 is <math>{6}^3</math>+<math>{6}^2</math>+6<math>d</math>+1.
+
We can also set up equations to convert <math>\underline{324}</math> base <math>n</math> and <math>\underline{11d1}</math> base 6 to base 10. The equation to covert <math>\underline{324}</math> base <math>n</math> to base 10 is <math>3{n}^2+2n+4.</math> The equation to convert <math>\underline{11d1}</math> base 6 to base 10 is <math>{6}^3+{6}^2+6d+1.</math>
  
−
Simplify <math>{6}^3</math>+<math>{6}^2</math>+6<math>d</math>+1 so it becomes 6<math>d</math>+253. Setting the above equations equal to each other, we have 3<math>{n}^2</math>+2n+4 = 6d+253. Subtracting 4 from both sides gets 3<math>{n}^2</math>+2n = 6d+249.
+
Simplify <math>{6}^3+{6}^2+6d+1</math> so it becomes <math>6d+253.</math> Setting the above equations equal to each other, we have  
 +
<cmath>3{n}^2+2n+4 = 6d+253.</cmath>
 +
Subtracting 4 from both sides gets <math>3{n}^2+2n = 6d+249.</math>
  
−
We can then use 3<math>{n}^2</math>+2<math>n</math> = 263-<math>d</math> and 3<math>{n}^2</math>+2<math>n</math> = 6<math>d</math>+249 to solve for <math>d</math>. Set 263-<math>d</math> equal to 6<math>d</math>+249 and solve to find that <math>d</math>=2.
+
We can then use equations
 +
<cmath>3{n}^2+2n = 263-d</cmath>
 +
<cmath>3{n}^2+2n = 6d+249</cmath>
 +
to solve for <math>d</math>. Set <math>263-d</math> equal to <math>6d+249</math> and solve to find that <math>d=2</math>.
  
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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=2</math> back into the equation <math>3{n}^2+2n = 263-d</math>. Subtract 261 from both sides to get your final equation of <math>3{n}^2+2n-261 = 0.</math> We solve using the quadratic formula to find that the solutions are <math>9</math> and <math>-10.</math> Because the base must be positive, <math>n=9.</math>
  
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Adding 2 to 9 gets <math>\boxed{\textbf{(B)}11}</math>.
+
Adding 2 to 9 gets <math>\boxed{\textbf{(B)} ~11}</math>
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<br><br>
 
  
 +
-Zeusthemoose (edited for readability)
  
−
-Zeusthemoose
+
==Solution 2==
 +
<math>32d</math> is greater than <math>263</math> when both are interpreted in base 10, so <math>n</math> is less than <math>10</math>. Some trial and error gives <math>n=9</math>. <math>263</math> in base 9 is <math>322</math>, so the answer is <math>9+2=\boxed{\textbf{(B)} ~11}</math>.
 +
 
 +
-SmileKat32
 +
 
 +
==Solution 3==
 +
We have
 +
<cmath>3n^2 + 2n + d = 263</cmath>
 +
<cmath>3n^2 + 2n + 4 = 6^3 + 6^2 + 6d + 1</cmath>
 +
Subtracting the 2nd from the 1st equation we get
 +
<cmath>\begin{align*}
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d-4 &= 263 - (216 + 36 + 6d + 1) \\
 +
&= 263 - 253 - 6d \\
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&= 10 - 6d
 +
\end{align*}</cmath>
 +
Thus we have <math>d=2.</math>
 +
Substituting into the first, we have <math>3n^2 + 2n + 2 = 263 \Rightarrow 3n^2 + 2n - 261 = 0.</math>
 +
Factoring, we have <math>(n-9)(3n+29)=0.</math>
 +
A digit cannot be negative, so we have <math>n=9.</math>
 +
Thus, <math>d+n=2+9=\boxed{\textbf{(B)} ~11}</math>
 +
 
 +
mathboy282 signing off
 +
~<math>\LaTeX</math> fixed by Lamboreghini
 +
 
 +
== Video Solution by OmegaLearn (Bases and System of Equations) ==
 +
https://youtu.be/oAc3GdAm6lk
 +
 
 +
~ pi_is_3.14
 +
 
 +
==Video Solution by TheBeautyofMath==
 +
https://youtu.be/L1iW94Ue3eI?t=880
 +
 
 +
~IceMatrix
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/X86a7-pSSSY
 +
 
 +
~Interstigation
 +
 
 +
==See Also==
 +
{{AMC10 box|year=2021|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Revision as of 22:30, 8 October 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 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 $-10.$ 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 $9+2=\boxed{\textbf{(B)} ~11}$.

-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

Video Solution by OmegaLearn (Bases and System of Equations)

https://youtu.be/oAc3GdAm6lk

~ pi_is_3.14

Video Solution by TheBeautyofMath

https://youtu.be/L1iW94Ue3eI?t=880

~IceMatrix

Video Solution by Interstigation

https://youtu.be/X86a7-pSSSY

~Interstigation

See Also

2021 AMC 10B (Problems • Answer Key • Resources)
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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