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"

m
(→‎Solution 2)
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>.
 +
 +
==Solution 3==
 +
 +
<cmath>
 +
5 \times 10^{13} = 5 \times (2^{13} \times 5^{13}) = 2^{13} \times 5^{14}
 +
</cmath>
 +
<cmath>
 +
2^{10} = 1024 \approx 10^3
 +
</cmath>
 +
<cmath>
 +
2^{13} = 2^{10} \times 2^3 \approx 10^3 \times 8 = 8000
 +
</cmath>
 +
<cmath>
 +
5 \times 10^{13} \approx 8000 \times 5^{14}
 +
</cmath>
 +
 
 +
converted <math>8000</math> to base 5, divide <math>8000</math> repeatedly by 5 and keep track of the remainders:
 +
 +
1. <math>8000 \div 5 = 1600</math>, remainder <math>0</math>
 +
 +
2. <math>1600 \div 5 = 320</math>, remainder <math>0</math>
 +
 +
3. <math>320 \div 5 = 64</math>, remainder <math>0</math>
 +
 +
4. <math>64 \div 5 = 12</math>, remainder <math>4</math>
 +
 +
5. <math>12 \div 5 = 2</math>, remainder <math>2</math>
 +
 +
6. <math>2 \div 5 = 0</math>, remainder <math>2</math>
 +
 +
Thus, <math>8000</math> in base 5 is <math>224000_5</math>, which has  6 digits
 +
When we multiply <math>224000_5</math> by <math>5^{14}</math>, the multiplication shifts the digits by 14 places to the left, adding 14 zeros.
 +
Thus, the total number of digits is:
 +
 +
6 + 14 = <math>\fbox{\textbf{(B)} 20}</math>.
 +
 +
 +
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]

Revision as of 02:32, 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 $\fbox{\textbf{(B)} 20}$.

Solution 3

\[5 \times 10^{13} = 5 \times (2^{13} \times 5^{13}) = 2^{13} \times 5^{14}\] \[2^{10} = 1024 \approx 10^3\] \[2^{13} = 2^{10} \times 2^3 \approx 10^3 \times 8 = 8000\] \[5 \times 10^{13} \approx 8000 \times 5^{14}\]

converted $8000$ to base 5, divide $8000$ repeatedly by 5 and keep track of the remainders:

1. $8000 \div 5 = 1600$, remainder $0$

2. $1600 \div 5 = 320$, remainder $0$

3. $320 \div 5 = 64$, remainder $0$

4. $64 \div 5 = 12$, remainder $4$

5. $12 \div 5 = 2$, remainder $2$

6. $2 \div 5 = 0$, remainder $2$

Thus, $8000$ in base 5 is $224000_5$, which has 6 digits When we multiply $224000_5$ by $5^{14}$, the multiplication shifts the digits by 14 places to the left, adding 14 zeros. Thus, the total number of digits is:

6 + 14 = $\fbox{\textbf{(B)} 20}$.


~luckuso

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_12B_Problems/Problem_6&oldid=233701"