Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution by TheBeautyofMath)
m (→‎Solution 4)
 
(14 intermediate revisions by 8 users not shown)
Line 4: Line 4:
 
<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==
+
==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 <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 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 <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>
 
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+{6}^2+6d+1</math> so it becomes <math>6d+253.</math> Setting the above equations equal to each other, we have  
 
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  
Line 20: Line 18:
 
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>.
 
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>.
  
 +
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>-29/3.</math> Because the base must be positive, <math>n=9.</math>
  
−
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> Solve using the quadratic formula to find that the solutions are 9 and -10. Because the base must be positive, <math>n=9.</math>
+
Adding 2 to 9 gets <math>\boxed{\textbf{(B)} ~11}</math>
−
 
 
−
Adding 2 to 9 gets <math>\boxed{\textbf{(B)}11}</math>
 
−
<br><br>
 
  
 
-Zeusthemoose (edited for readability)
 
-Zeusthemoose (edited for readability)
 +
-solution corrected by Billowingsweater
  
 
==Solution 2==
 
==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>.  
+
<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
 
-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*}
 +
d-4 &= 263 - (216 + 36 + 6d + 1) \\
 +
&= 263 - 253 - 6d \\
 +
&= 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
 +
 +
==Solution 4==
 +
Find that <math>d=2</math> using one of the methods above. Then we have that <math>3n^2 + 2n = 261</math>. We know that <math>n</math> is an integer, so we can solve the equation <math>n(3n+2)=261</math> (this is guaranteed to have a solution if we did this correctly). The prime factorization of <math>261</math> is <math>3^2 \cdot 29</math>, so the corresponding factor pairs are <math>(1, 261)</math>, <math>(3,87)</math>, and <math>(9,29)</math>. If <math>n = 9</math>, then the equation is true, so <math>n + d = 9 + 2 = \boxed{\textbf{(B)} ~11}</math>.
 +
 +
Note: This solution provides an alternate way to complete the final part of the problem if, like me, you really dislike quadratics.
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 +
 +
==Video Solution (🚀 Under 4 min 🚀)==
 +
https://youtu.be/jm2VbqRpFyI
 +
 +
~<i> Education, the Study of Everything </i>
  
 
== Video Solution by OmegaLearn (Bases and System of Equations) ==  
 
== Video Solution by OmegaLearn (Bases and System of Equations) ==  
Line 48: Line 75:
 
~Interstigation
 
~Interstigation
  
 +
==See Also==
 
{{AMC10 box|year=2021|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2021|ab=B|num-b=12|num-a=14}}
 +
{{MAA Notice}}

Latest revision as of 20:54, 12 May 2024

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

Solution 4

Find that $d=2$ using one of the methods above. Then we have that $3n^2 + 2n = 261$. We know that $n$ is an integer, so we can solve the equation $n(3n+2)=261$ (this is guaranteed to have a solution if we did this correctly). The prime factorization of $261$ is $3^2 \cdot 29$, so the corresponding factor pairs are $(1, 261)$, $(3,87)$, and $(9,29)$. If $n = 9$, then the equation is true, so $n + d = 9 + 2 = \boxed{\textbf{(B)} ~11}$.

Note: This solution provides an alternate way to complete the final part of the problem if, like me, you really dislike quadratics. ~ cxsmi

Video Solution (🚀 Under 4 min 🚀)

https://youtu.be/jm2VbqRpFyI

~ Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10B_Problems/Problem_13&oldid=219270"