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 "2024 AMC 12B Problems/Problem 6"

(→‎Solution 2)
m
Line 20: Line 20:
  
 
We see that <math>5\times 10^{13} = 2^{13} \cdot 5^{14}</math> and <math>2^{13} = 8192</math>. Converting this to base <math>5</math> gives us <math>230232</math> (trust me it doesn't take that long). So the final number in base <math>5</math> is <math>230232</math> with <math>14</math> zeroes at the end, which gives us <math>6 + 14 = 20</math> digits. So the answer is <math>\fbox{\textbf{(B)} 20}</math>.
 
We see that <math>5\times 10^{13} = 2^{13} \cdot 5^{14}</math> and <math>2^{13} = 8192</math>. Converting this to base <math>5</math> gives us <math>230232</math> (trust me it doesn't take that long). So the final number in base <math>5</math> is <math>230232</math> with <math>14</math> zeroes at the end, which gives us <math>6 + 14 = 20</math> digits. So the answer is <math>\fbox{\textbf{(B)} 20}</math>.
−
 
−
~sidkris
 

Revision as of 02:25, 14 November 2024

Problem 6

The national debt of the United States is on track to reach $5\times10^{13}$ dollars by $2023$. How many digits does this number of dollars have when written as a numeral in base 5? (The approximation of $\log_{10} 5$ as $0.7$ is sufficient for this problem)

$\textbf{(A) } 18 \qquad\textbf{(B) } 20 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 24 \qquad\textbf{(E) } 26$

Solution

The number of digits is just $\lceil \log_{5} 5\times 10^{13} \rceil$. Note that \[\log_{5} 5\times 10^{13}=1+\frac{13}{\log_{10} 5}\] \[\approx 1+\frac{13}{0.7}\] \[\approx 19.5\]

Hence, our answer is $\fbox{\textbf{(B) } 20}$

~tsun26

Solution 2

We see that $5\times 10^{13} = 2^{13} \cdot 5^{14}$ and $2^{13} = 8192$. Converting this to base $5$ gives us $230232$ (trust me it doesn't take that long). So the final number in base $5$ is $230232$ with $14$ zeroes at the end, which gives us $6 + 14 = 20$ digits. So the answer is $\fbox{\textbf{(B)} 20}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_6&oldid=233692"