2024 AMC 12B Problems/Problem 6: Difference between revisions

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Problem 6

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, number of digits of a number $n$ in base $b$ : $\text{Number of digits} = \lfloor \log_b n \rfloor + 1$ In this question, it is $\lfloor \log_{5} 5\times 10^{13}\rfloor+1$ . 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

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

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

@@ Line 74: / Line 74: @@
 ==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}}

2024 AMC 12B (Problems • Answer Key • Resources)
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