Art of Problem Solving
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

2024 AMC 12B Problems/Problem 6: Difference between revisions

← Older editNewer edit →
Cyantist (talk | contribs)
editor
252 edits
Cyantist (talk | contribs)
editor
252 edits
Line 22: Line 22:


~sidkris
~sidkris
==Solution 2a (Base Conversion) ==
To convert the number \(8192\) from base 10 to base 5, we follow these steps:
1. Divide the number by 5 repeatedly, noting the quotient and remainder each time.
2. Stop when the quotient becomes 0, then read the remainders from bottom to top.
<cmath>
8192 \div 5 = 1638 \text{ remainder } 2
</cmath>
<cmath>
1638 \div 5 = 327 \text{ remainder } 3
</cmath>
<cmath>
327 \div 5 = 65 \text{ remainder } 2
</cmath>
<cmath>
65 \div 5 = 13 \text{ remainder } 0
</cmath>
<cmath>
13 \div 5 = 2 \text{ remainder } 3
</cmath>
<cmath>
2 \div 5 = 0 \text{ remainder } 2
</cmath>
Now, reading the remainders from bottom to top: \( 2, 3, 0, 2, 3, 2 \).
Thus, \(8192\) in base 5 is:
<cmath>
\boxed{230232_5}
</cmath>
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]


==See also==
==See also==
{{AMC12 box|year=2024|ab=B|num-b=5|num-a=7}}
{{AMC12 box|year=2024|ab=B|num-b=5|num-a=7}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 21:59, 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 1

The number of digits is just $\lfloor \log_{5} 5\times 10^{13}+1\rfloor$. 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 (small modification by notknowanything)

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 $\fbox{\textbf{(B)} 20}$.

~sidkris

See also

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_6&oldid=234237"