2024 AMC 12B Problems/Problem 6

Revision as of 02:37, 14 November 2024 by Cyantist (talk | contribs) (Solution 2)

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

~sidkris

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