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

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==Problem==
 
==Problem==
  
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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?
+
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>
  
 
==Solution 1==
 
==Solution 1==
 +
We have
  
 
<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>
  
−
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
+
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==
Line 21: Line 22:
 
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
 
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
  
−
==Solution 4 (Smarter Brute Force)==
+
==Solution 4 (Alternate Brute Force)==
 
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
 
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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==Solution 5 (Logarithms)==
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==Solution 5 (Scientific Notation)==
  
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<cmath>\lceil log \dfrac{1}{20^{20}} \rceil
+
We see that <math>\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}</math>. We see that this has <math>27-1=26</math> zeros after the decimal point before coming to <math>9</math>.
−
= \lceil log 20^{-20} \rceil
+
 
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= \lceil -20 log(20) \rceil
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Therefore, the answer is <math>\boxed{\textbf{(D)} ~26}</math>
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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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==Solution 6 (Logarithms Without Words)==
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= \lceil -26.02 \rceil
+
 
−
= \boxed{\textbf{(D) } \text{26}}</cmath>
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<cmath>\begin{align*}|\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}} \end{align*}</cmath>
  
 
~phoenixfire
 
~phoenixfire
  
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==Video Solution==
+
==Video Solution (HOW TO CRITICALLY THINK)
 +
https://youtu.be/qnznImCDSW4
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Solution 7 (Algebra)==
 +
Define <math>n</math> as the number of consecutive <math>0</math>s after the decimal point in the decimal representation of <math>\frac{1}{20^{20}}.</math> Thus, <math>\frac{1}{10^n}</math> has <math>n-1</math> zeroes followed by a <math>1</math> when expressed as a decimal, and is greater than <math>\frac{1}{20^{20}}.</math> So, we have that
 +
<cmath>\frac{1}{10^n} > \frac{1}{20^{20}}.</cmath>
 +
Taking the reciprocal and switching signs grants
 +
<cmath>20^{20}>10^n.</cmath>
 +
We can express <math>20^{20}</math> as <math>10^{20} \cdot 1024^2.</math> We know that
 +
<cmath>\begin{align*}
 +
(1000 + 24)^2 &= 1024^2 \\
 +
1000000 + 48000 + 576 &= 1048576. \end{align*}</cmath>
 +
Multiplying this with <math>10^{20}</math> adds <math>20</math> more zeroes at the end, which means the left side of our inequality has <math>27</math> digits. Then, the <math>10</math> on the right side must have an exponent less than <math>27,</math> meaning that <math>n<27.</math>
 +
 
 +
We are also given that the initial string of zeroes in the decimal representation of <math>\frac{1}{20^{20}}</math> is followed by a <math>9.</math> If we consider just these <math>n+1</math> digits after the decimal point, we can express it as the fraction <math>\frac{9}{10^{n+1}}.</math> Since there are more digits after this <math>9</math> in the decimal representation of <math>\frac{1}{20^{20}},</math> it is true that
 +
<cmath>\frac{1}{20^{20}} > \frac{9}{10^{n+1}}.</cmath>
 +
Rinse and repeat and we have
 +
<cmath>20^{20} < \frac{10^{n+1}}{9}.</cmath>
 +
Multiplying both sides by <math>\frac{9}{10^{20}}</math> gives
 +
<cmath>9 \cdot 2^{20} < 10^{n-19}.</cmath>
 +
We previously found that <math>2^{20} = 1048576,</math> which has <math>7</math> digits. Multiplying this by <math>9</math> will keep the number of digits the same (make sure to see why). Thus,
 +
<cmath>n-19>6 \qquad \text{or} \qquad n > 25.</cmath>
 +
 
 +
If <math>n<27</math> and <math>n>25,</math> then <math>n= \boxed{\textbf{(D)} \ 26}.</math>
 +
 
 +
~happyhari, mathbrek
 +
 
 +
==Solution 8 (Finding a Pattern)==
 +
 
 +
<math>\frac{1}{2^{20}}</math> equals <math>\frac{1}{10^{20}}</math> times <math>\frac{1}{2^{20}}</math>. <math>\frac{1}{10^{20}}</math> has 19 zeroes before the 1 after the decimal point. <math>\frac{1}{2}</math> <math>\frac{1}{4}</math> and <math>\frac{1}{8}</math> have no zeroes after the decimal point. <math>\frac{1}{16}</math>, <math>\frac{1}{32}</math>, and <math>\frac{1}{64}</math> have 1 zero after the decimal point. If you do division for all the other powers of 2, you find that every three powers of 2 an extra zero is added (128, 256, and 512 all have 3 digits, 1024, 2048, and 4096 have 4 digits, and so on.) Therefore, the number of zeroes after the decimal point in <math>\frac{1}{2^{20}}</math> is 7 because 20/3 is 6 remainder 2. 19+7= <math>\boxed{\textbf{(D)} \ 26}.</math>
 +
 
 +
~unhappyfarmer
 +
 
 +
==Video Solution by TheBeautyofMath==
 
https://youtu.be/t6yjfKXpwDs
 
https://youtu.be/t6yjfKXpwDs
  

Revision as of 18:52, 20 August 2024

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

We have

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

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

Solution 5 (Scientific Notation)

We see that $\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}$. We see that this has $27-1=26$ zeros after the decimal point before coming to $9$.

Therefore, the answer is $\boxed{\textbf{(D)} ~26}$

Solution 6 (Logarithms Without Words)

\begin{align*}|\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}} \end{align*}

~phoenixfire

==Video Solution (HOW TO CRITICALLY THINK) https://youtu.be/qnznImCDSW4

~Education, the Study of Everything

Solution 7 (Algebra)

Define $n$ as the number of consecutive $0$s after the decimal point in the decimal representation of $\frac{1}{20^{20}}.$ Thus, $\frac{1}{10^n}$ has $n-1$ zeroes followed by a $1$ when expressed as a decimal, and is greater than $\frac{1}{20^{20}}.$ So, we have that \[\frac{1}{10^n} > \frac{1}{20^{20}}.\] Taking the reciprocal and switching signs grants \[20^{20}>10^n.\] We can express $20^{20}$ as $10^{20} \cdot 1024^2.$ We know that \begin{align*} (1000 + 24)^2 &= 1024^2 \\ 1000000 + 48000 + 576 &= 1048576. \end{align*} Multiplying this with $10^{20}$ adds $20$ more zeroes at the end, which means the left side of our inequality has $27$ digits. Then, the $10$ on the right side must have an exponent less than $27,$ meaning that $n<27.$

We are also given that the initial string of zeroes in the decimal representation of $\frac{1}{20^{20}}$ is followed by a $9.$ If we consider just these $n+1$ digits after the decimal point, we can express it as the fraction $\frac{9}{10^{n+1}}.$ Since there are more digits after this $9$ in the decimal representation of $\frac{1}{20^{20}},$ it is true that \[\frac{1}{20^{20}} > \frac{9}{10^{n+1}}.\] Rinse and repeat and we have \[20^{20} < \frac{10^{n+1}}{9}.\] Multiplying both sides by $\frac{9}{10^{20}}$ gives \[9 \cdot 2^{20} < 10^{n-19}.\] We previously found that $2^{20} = 1048576,$ which has $7$ digits. Multiplying this by $9$ will keep the number of digits the same (make sure to see why). Thus, \[n-19>6 \qquad \text{or} \qquad n > 25.\]

If $n<27$ and $n>25,$ then $n= \boxed{\textbf{(D)} \ 26}.$

~happyhari, mathbrek

Solution 8 (Finding a Pattern)

$\frac{1}{2^{20}}$ equals $\frac{1}{10^{20}}$ times $\frac{1}{2^{20}}$. $\frac{1}{10^{20}}$ has 19 zeroes before the 1 after the decimal point. $\frac{1}{2}$ $\frac{1}{4}$ and $\frac{1}{8}$ have no zeroes after the decimal point. $\frac{1}{16}$, $\frac{1}{32}$, and $\frac{1}{64}$ have 1 zero after the decimal point. If you do division for all the other powers of 2, you find that every three powers of 2 an extra zero is added (128, 256, and 512 all have 3 digits, 1024, 2048, and 4096 have 4 digits, and so on.) Therefore, the number of zeroes after the decimal point in $\frac{1}{2^{20}}$ is 7 because 20/3 is 6 remainder 2. 19+7= $\boxed{\textbf{(D)} \ 26}.$

~unhappyfarmer

Video Solution by TheBeautyofMath

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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