Difference between revisions of "2020 AMC 10B Problems/Problem 12"

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==Problem==
 
==Problem==
  
The decimal representation of<cmath>\dfrac{1}{20^{20}}</cmath>consists of a string of zeros after the decimal point, followed by a <math>9</math> and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
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The decimal representation of <cmath>\dfrac{1}{20^{20}}</cmath> consists of a string of zeros after the decimal point, followed by a <math>9</math> and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
  
 
<math>\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}</math>
 
<math>\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}</math>
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==Solution 1==
 
==Solution 1==
  
<cmath>\dfrac{1}{20^{20}} \approx \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}</cmath>
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<cmath>\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}</cmath>
  
Now we do some estimation. Notice that <math>2^{20}=1024^2</math>, which means that <math>2^{20}</math> is a little more than <math>1000^2=1,000,000</math>. Multiplying it with <math>10^{20}</math>, we get that the denominator is about <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>. Notice that when we divide <math>1</math> by an <math>n</math> digit number, there are <math>n-1</math> zeros before the first nonzero digit. This means that when we divide <math>1</math> by the <math>27</math> digit integer <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>, there are <math>\boxed{\textbf{(D) } \text{26}}</math> zeros in the initial string after the decimal point. -PCChess
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Now we do some estimation. Notice that <math>2^{20} = 1024^2</math>, which means that <math>2^{20}</math> is a little more than <math>1000^2=1,000,000</math>. Multiplying it with <math>10^{20}</math>, we get that the denominator is about <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>. Notice that when we divide <math>1</math> by an <math>n</math> digit number as long as n is not a power of 10, there are <math>n-1</math> zeros before the first nonzero digit. This means that when we divide <math>1</math> by the <math>27</math> digit integer <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>, there are <math>\boxed{\textbf{(D) } \text{26}}</math> zeros in the initial string after the decimal point. -PCChess
  
 
==Solution 2==
 
==Solution 2==
 
First rewrite <math>\frac{1}{20^{20}}</math> as <math>\frac{5^{20}}{10^{40}}</math>. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in <math>{5^{20}}</math>.
 
First rewrite <math>\frac{1}{20^{20}}</math> as <math>\frac{5^{20}}{10^{40}}</math>. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in <math>{5^{20}}</math>.
  
<math>\log{5^{20}} = 20\log{5}</math> and memming <math>\log{5}\approx0.69</math> (alternatively use the fact that <math>\log{5} = 1 - \log{2}</math>), <math>\floor{20\log{5}}+1=\floor{20\cdot0.69}+1=13+1=14</math> digits.  
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<math>\log{5^{20}} = 20\log{5}</math> and memming <math>\log{5}\approx0.69</math> (alternatively use the fact that <math>\log{5} = 1 - \log{2}</math>), <math>\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14</math> digits.  
  
Our answer is <math>40-14=\boxed{26}</math>.
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Our answer is <math>\boxed{\textbf{(D) } \text{26}}</math>.
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==Solution 3 (Brute Force)==
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Just as in Solution <math>2,</math> we rewrite <math>\dfrac{1}{20^{20}}</math> as <math>\dfrac{5^{20}}{10^{40}}.</math> We then calculate <math>5^{20}</math> entirely by hand, first doing <math>5^5 \cdot 5^5,</math> then multiplying that product by itself, resulting in <math>95,367,431,640,625.</math> Because this is <math>14</math> digits, after dividing this number by <math>10</math> fourteen times, the decimal point is before the <math>9.</math> Dividing the number again by <math>10</math> twenty-six more times allows a string of<math>\boxed{\textbf{(D) } \text{26}}</math> zeroes to be formed. -OreoChocolate
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==Solution 4 (Alternate Brute Force)==
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Just as in Solutions <math>2</math> and <math>3,</math> we rewrite <math>\dfrac{1}{20^{20}}</math> as <math>\dfrac{5^{20}}{10^{40}}.</math> We can then look at the number of digits in powers of <math>5</math>. <math>5^1=5</math>, <math>5^2=25</math>, <math>5^3=125</math>, <math>5^4=625</math>, <math>5^5=3125</math>, <math>5^6=15625</math>, <math>5^7=78125</math> and so on. We notice after a few iterations that every power of five with an exponent of <math>1 \mod 3</math>, the number of digits doesn't increase. This means <math>5^{20}</math> should have <math>20 - 6</math> digits since there are <math>6</math> numbers which are <math>1 \mod 3</math> from <math>0</math> to <math>20</math>, or <math>14</math> digits total. This means our expression can be written as <math>\dfrac{k\cdot10^{14}}{10^{40}}</math>, where <math>k</math> is in the range <math>[1,10)</math>. Canceling gives <math>\dfrac{k}{10^{26}}</math>, or <math>26</math> zeroes before the <math>k</math> since the number <math>k</math> should start on where the one would be in <math>10^{26}</math>. ~aop2014
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changed from 'Smarter Brute Force' to 'Alternate Brute Force'
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==Solution 5 (Logarithms Without Words)==
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<cmath>|\lceil \log \dfrac{1}{20^{20}} \rceil|
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= |\lceil \log 20^{-20} \rceil|
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= |\lceil -20 \log(20) \rceil|
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= |\lceil -20(\log 10 + \log 2) \rceil|
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= |\lceil -20(1 + 0.301) \rceil|
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= |\lceil -26.02 \rceil|
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= |-26|
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= \boxed{\textbf{(D) } \text{26}}</cmath>
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~phoenixfire
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 14:22, 1 May 2021

Problem

The decimal representation of \[\dfrac{1}{20^{20}}\] consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?

$\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}$

Solution 1

\[\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}\]

Now we do some estimation. Notice that $2^{20} = 1024^2$, which means that $2^{20}$ is a little more than $1000^2=1,000,000$. Multiplying it with $10^{20}$, we get that the denominator is about $1\underbrace{00\dots0}_{26 \text{ zeros}}$. Notice that when we divide $1$ by an $n$ digit number as long as n is not a power of 10, there are $n-1$ zeros before the first nonzero digit. This means that when we divide $1$ by the $27$ digit integer $1\underbrace{00\dots0}_{26 \text{ zeros}}$, there are $\boxed{\textbf{(D) } \text{26}}$ zeros in the initial string after the decimal point. -PCChess

Solution 2

First rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$. Then, we know that when we write this in decimal form, there will be 40 digits after the decimal point. Therefore, we just have to find the number of digits in ${5^{20}}$.

$\log{5^{20}} = 20\log{5}$ and memming $\log{5}\approx0.69$ (alternatively use the fact that $\log{5} = 1 - \log{2}$), $\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14$ digits.

Our answer is $\boxed{\textbf{(D) } \text{26}}$.

Solution 3 (Brute Force)

Just as in Solution $2,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We then calculate $5^{20}$ entirely by hand, first doing $5^5 \cdot 5^5,$ then multiplying that product by itself, resulting in $95,367,431,640,625.$ Because this is $14$ digits, after dividing this number by $10$ fourteen times, the decimal point is before the $9.$ Dividing the number again by $10$ twenty-six more times allows a string of$\boxed{\textbf{(D) } \text{26}}$ zeroes to be formed. -OreoChocolate

Solution 4 (Alternate Brute Force)

Just as in Solutions $2$ and $3,$ we rewrite $\dfrac{1}{20^{20}}$ as $\dfrac{5^{20}}{10^{40}}.$ We can then look at the number of digits in powers of $5$. $5^1=5$, $5^2=25$, $5^3=125$, $5^4=625$, $5^5=3125$, $5^6=15625$, $5^7=78125$ and so on. We notice after a few iterations that every power of five with an exponent of $1 \mod 3$, the number of digits doesn't increase. This means $5^{20}$ should have $20 - 6$ digits since there are $6$ numbers which are $1 \mod 3$ from $0$ to $20$, or $14$ digits total. This means our expression can be written as $\dfrac{k\cdot10^{14}}{10^{40}}$, where $k$ is in the range $[1,10)$. Canceling gives $\dfrac{k}{10^{26}}$, or $26$ zeroes before the $k$ since the number $k$ should start on where the one would be in $10^{26}$. ~aop2014

changed from 'Smarter Brute Force' to 'Alternate Brute Force'

Solution 5 (Logarithms Without Words)

\[|\lceil \log \dfrac{1}{20^{20}} \rceil| = |\lceil \log 20^{-20} \rceil| = |\lceil -20 \log(20) \rceil| = |\lceil -20(\log 10 + \log 2) \rceil| = |\lceil -20(1 + 0.301) \rceil| = |\lceil -26.02 \rceil| = |-26| = \boxed{\textbf{(D) } \text{26}}\]

~phoenixfire

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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 10 Problems and Solutions

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