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

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==Solution 5 (Logarithms)==
 
==Solution 5 (Logarithms)==
  
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Note: You need a good understanding of how Logarithms work.
 
  
 
<cmath>\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}</cmath>
 
<cmath>\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}</cmath>
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Let <cmath>x = \dfrac{1}{10^{20}\cdot2^{20}}</cmath>
 
Let <cmath>x = \dfrac{1}{10^{20}\cdot2^{20}}</cmath>
  
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Remember:
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We make use of <math>log_{a}bc = log_{a}b + log_{a}c</math>
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<cmath>log_{a}b + log_{a}c = log_{a}bc</cmath>
 
−
And
 
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<cmath>log_{a}bc = log_{a}b + log_{a}c</cmath>  
 
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(We will use the second property)
 
−
Like this
 
  
 
<cmath>\dfrac{1}{log_{10}(10^{20}\cdot2^{20})} = \dfrac{1}{log_{10}10^{20} + log_{10}2^{20}}</cmath>
 
<cmath>\dfrac{1}{log_{10}(10^{20}\cdot2^{20})} = \dfrac{1}{log_{10}10^{20} + log_{10}2^{20}}</cmath>
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<cmath>= \dfrac{1}{20 \cdot log_{10}10 + 20 \cdot log_{10}2}</cmath>
 
<cmath>= \dfrac{1}{20 \cdot log_{10}10 + 20 \cdot log_{10}2}</cmath>
  
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Remember that <cmath>log_{10}10 = 1</cmath> and <cmath>log_{10}2 = 0.3010</cmath>
+
Remember that <math>log_{10}10 = 1</math> and that, approximately, <math>log_{10}2 = 0.3010</math>
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Note that this value is an approximate.
+
 
 
This makes our equation
 
This makes our equation
  
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<cmath>log_{10}x = \dfrac{1}{20 + 20 \cdot {0.3010}}</cmath>
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<cmath>log_{10}x = \dfrac{1}{20 + 20 \cdot {0.3010}} = \dfrac{1}{20 + 6.020} = \dfrac{1}{26.02}</cmath>
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<cmath>= \dfrac{1}{20 + 6.020}</cmath>
 
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<cmath>= \dfrac{1}{26.02}</cmath>
 
  
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Removing the log we get
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Taking anti-logs: <math>x = 10^{-26.02}</math>
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<cmath>x = 10^{-26.02}</cmath>
 
  
 
Let <math>[m]</math> denote Greatest Integer Function of <math>m</math>.
 
Let <math>[m]</math> denote Greatest Integer Function of <math>m</math>.

Revision as of 12:44, 20 February 2020

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, 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 (Smarter 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

Solution 5 (Logarithms)

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

Let \[x = \dfrac{1}{10^{20}\cdot2^{20}}\]

We make use of $log_{a}bc = log_{a}b + log_{a}c$

\[\dfrac{1}{log_{10}(10^{20}\cdot2^{20})} = \dfrac{1}{log_{10}10^{20} + log_{10}2^{20}}\]

Taking $log_{10}$ on both the sides results in

\[log_{10}x = \dfrac{1}{log_{10}10^{20} + log_{10}2^{20}}\]

\[= \dfrac{1}{20 \cdot log_{10}10 + 20 \cdot log_{10}2}\]

Remember that $log_{10}10 = 1$ and that, approximately, $log_{10}2 = 0.3010$

This makes our equation

\[log_{10}x = \dfrac{1}{20 + 20 \cdot {0.3010}} = \dfrac{1}{20 + 6.020} = \dfrac{1}{26.02}\]

Taking anti-logs: $x = 10^{-26.02}$

Let $[m]$ denote Greatest Integer Function of $m$.

Number of zero's after the decimal point is \[| [-26.02] + 1 |= | -26 | = 26\]

And thus our answer is $\boxed{\textbf{(D) } \text{26}}$.

~phoenixfire

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

See Also

2020 AMC 10B (Problems • Answer Key • Resources)
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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