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

Revision as of 15:04, 16 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 $3{n}^2+2n+d.$ 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 $9+2=\boxed{\textbf{(B)}11}$ .

-SmileKat32

Video Solution by OmegaLearn (Bases and System of Equations)

https://youtu.be/oAc3GdAm6lk

~ pi_is_3.14

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

@@ Line 29: / Line 29: @@
 ==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

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