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Difference between revisions of "2021 AMC 10B Problems/Problem 13"
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-Zeusthemoose
 
-Zeusthemoose
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==Solution 2==
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<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>.
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-SmileKat32
  
 
{{AMC10 box|year=2021|ab=B|num-b=12|num-a=14}}
 
{{AMC10 box|year=2021|ab=B|num-b=12|num-a=14}}

Revision as of 06:29, 12 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

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

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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