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

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)

$\lceil log \dfrac{1}{20^{20}} \rceil = \lceil log 20^{-20} \rceil = \lceil -20 log(20) \rceil = \lceil -20(log 10 + log 2) \rceil = \lceil -20(1 + 0.301) \rceil = \lceil -26.02 \rceil = -26$ Number of zeroes = $\boxed{\textbf{(D) } \text{26}}$

~phoenixfire

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

@@ Line 32: / Line 32: @@
 = \lceil -20(1 + 0.301) \rceil
 = \lceil -26.02 \rceil
-= \boxed{\textbf{(D) } \text{26}}</cmath>
+= -26</cmath>
+Number of zeroes = <math>\boxed{\textbf{(D) } \text{26}}</math>
 ~phoenixfire

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