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 edit
Cyantist (talk | contribs)
editor
252 edits
Rushilyeole (talk | contribs)
editor
3 edits
 
(21 intermediate revisions by 9 users not shown)
Line 1: Line 1:
==Problem 6==
==Problem==


The national debt of the United States is on track to reach <math>5\times10^{13}</math> dollars by <math>2023</math>. How many digits does this number of dollars have when written as a numeral in base 5? (The approximation of <math>\log_{10} 5</math> as <math>0.7</math> is sufficient for this problem)
The national debt of the United States is on track to reach <math>5\times10^{13}</math> dollars by <math>2033</math>. How many digits does this number of dollars have when written as a numeral in base <math>5</math>? (The approximation of <math>\log_{10} 5</math> as <math>0.7</math> is sufficient for this problem)


<math>\textbf{(A) } 18 \qquad\textbf{(B) } 20 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 24 \qquad\textbf{(E) } 26</math>
<math>\textbf{(A) } 18 \qquad\textbf{(B) } 20 \qquad\textbf{(C) } 22 \qquad\textbf{(D) } 24 \qquad\textbf{(E) } 26</math>


==Solution==
==Solution 1==
 
Generally, the number of digits of number <math>n</math> in base <math>b</math> is
The number of digits is just <math>\lceil \log_{5} 5\times 10^{13} \rceil</math>.
<cmath>\lfloor \log_b n \rfloor + 1.</cmath>
In this question, it is <math>\lfloor \log_{5} (5\times 10^{13})\rfloor+1</math>.
Note that  
Note that  
<cmath>\log_{5} 5\times 10^{13}=1+\frac{13}{\log_{10} 5}</cmath>
<cmath>\begin{align*}
<cmath>\approx 1+\frac{13}{0.7}</cmath>
\log_{5} (5\times 10^{13}) &= 1+\frac{13}{\log_{10} 5} \\
<cmath>\approx 19.5</cmath>
&\approx 1+\frac{13}{0.7} \\
&\approx 19.5
\end{align*}</cmath>
Hence, our answer is <math>\fbox{\textbf{(B)} 20}</math>


Hence, our answer is <math>\fbox{\textbf{(B) } 20}</math>
~tsun26 (small modification by notknowanything)
 
~tsun26


==Solution 2==
==Solution 2==
Line 22: Line 24:


~sidkris
~sidkris
Note - Base Conversion Step
To convert the number <math>8192</math> 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:<math> 2, 3, 0, 2, 3, 2 </math>.
Thus, <math>8192</math> in base 5 is:
<cmath>
\boxed{230232_5}
</cmath>
~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
Note - an approximation you can use here to convert 8192 to base 5: since <math>5^5 = 3125</math> and <math>5^6 = 15625</math>, we automatically know that you need 6 digits to represent <math>2^{13}</math> in base 5.
~ RushilYeole
==Solution 3==
<math>5 \times 10^{13} = 5 \times (5^{13} \times 2^{13}) = 2^{13} \times 5^{14} = 8192 \times 5^{14}.</math>
<math>5^5 = 3125</math> and <math>5^6 = 15625</math> (or just notice that it must be <math>> 8192</math>) <math>\implies 5^5 < 8192 < 5^6 \implies 5^{19} < 5 \times 10^{13} < 5^{20}</math>.
Since an integer <math>x</math> has <math>n</math> base-<math>a</math> digits when it satisfies <math>a^{n-1} \le x < a^n</math>, it follows that <math>5 \times 10^{13}</math> requires <math>\fbox{\textbf{(B)} 20}</math> base-5 digits.
~drnez
==Video Solution 1 by SpreadTheMathLove==
https://www.youtube.com/watch?v=FUsMSwb-JUc
==Video Solution 2 by TheBeautyofMath==
https://youtu.be/AKLPjTRPF4Q
~IceMatrix
==See also==
{{AMC12 box|year=2024|ab=B|num-b=5|num-a=7}}
{{MAA Notice}}

Latest revision as of 23:56, 3 November 2025

Problem

The national debt of the United States is on track to reach $5\times10^{13}$ dollars by $2033$. 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

Generally, the number of digits of number $n$ in base $b$ is \[\lfloor \log_b n \rfloor + 1.\] In this question, it is $\lfloor \log_{5} (5\times 10^{13})\rfloor+1$. Note that \begin{align*} \log_{5} (5\times 10^{13}) &= 1+\frac{13}{\log_{10} 5} \\ &\approx 1+\frac{13}{0.7} \\ &\approx 19.5 \end{align*} 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

Note - Base Conversion Step

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.

\[8192 \div 5 = 1638 \text{ remainder } 2\] \[1638 \div 5 = 327 \text{ remainder } 3\] \[327 \div 5 = 65 \text{ remainder } 2\] \[65 \div 5 = 13 \text{ remainder } 0\] \[13 \div 5 = 2 \text{ remainder } 3\] \[2 \div 5 = 0 \text{ remainder } 2\]

Now, reading the remainders from bottom to top:$2, 3, 0, 2, 3, 2$.

Thus, $8192$ in base 5 is:

\[\boxed{230232_5}\] ~luckuso

Note - an approximation you can use here to convert 8192 to base 5: since $5^5 = 3125$ and $5^6 = 15625$, we automatically know that you need 6 digits to represent $2^{13}$ in base 5.

~ RushilYeole

Solution 3

$5 \times 10^{13} = 5 \times (5^{13} \times 2^{13}) = 2^{13} \times 5^{14} = 8192 \times 5^{14}.$

$5^5 = 3125$ and $5^6 = 15625$ (or just notice that it must be $> 8192$) $\implies 5^5 < 8192 < 5^6 \implies 5^{19} < 5 \times 10^{13} < 5^{20}$.

Since an integer $x$ has $n$ base-$a$ digits when it satisfies $a^{n-1} \le x < a^n$, it follows that $5 \times 10^{13}$ requires $\fbox{\textbf{(B)} 20}$ base-5 digits.

~drnez

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=FUsMSwb-JUc

Video Solution 2 by TheBeautyofMath

https://youtu.be/AKLPjTRPF4Q

~IceMatrix

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=264979"