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Difference between revisions of "2024 AMC 12B Problems/Problem 6"

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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