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

Revision as of 11:14, 11 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 Solution $2$ 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$, %5^7=78125$ (Error compiling LaTeX. Unknown error_msg) 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 $20$ digits minus $6$ 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 $10^26$ .

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

@@ Line 21: / Line 21: @@
 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)==
+Just as in Solution <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>, %5^3=125<math>, </math>5^4=625<math>, </math>5^5=3125<math>, </math>5^6=15625<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</math> digits minus <math>6</math> 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 26 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>.
 ==Video Solution==
 https://youtu.be/t6yjfKXpwDs

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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