Difference between revisions of "2024 AMC 12B Problems/Problem 6"
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==Problem 6== | ==Problem 6== | ||
− | The national debt of the United States is on track to reach <math>5\times10^{13}</math> dollars by <math> | + | 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 5? (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, number of digits of a number <math>n</math> in base <math>b</math>: | |
− | + | <cmath> | |
+ | \text{Number of digits} = \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>\log_{5} 5\times 10^{13}=1+\frac{13}{\log_{10} 5}</cmath> | ||
Line 15: | Line 18: | ||
Hence, our answer is <math>\fbox{\textbf{(B) } 20}</math> | Hence, our answer is <math>\fbox{\textbf{(B) } 20}</math> | ||
− | ~tsun26 | + | ~tsun26 (small modification by notknowanything) |
==Solution 2== | ==Solution 2== | ||
Line 22: | Line 25: | ||
~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] | ||
+ | |||
+ | ==Video Solution 1 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=FUsMSwb-JUc | ||
+ | ==See also== | ||
+ | {{AMC12 box|year=2024|ab=B|num-b=5|num-a=7}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:54, 21 November 2024
Problem 6
The national debt of the United States is on track to reach dollars by . How many digits does this number of dollars have when written as a numeral in base 5? (The approximation of as is sufficient for this problem)
Solution 1
Generally, number of digits of a number in base : In this question, it is . Note that
Hence, our answer is
~tsun26 (small modification by notknowanything)
Solution 2
We see that and . Converting this to base gives us (trust me it doesn't take that long). So the final number in base is with zeroes at the end, which gives us digits. So the answer is .
~sidkris
Note - Base Conversion Step
To convert the number 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.
Now, reading the remainders from bottom to top:.
Thus, in base 5 is:
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=FUsMSwb-JUc
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.