2020 AMC 10B Problems/Problem 12

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 $x = \dfrac{1}{10^{20}\cdot2^{20}}$

Remember: $log_{a}b + log_{a}c = log_{a}bc$ And $log_{a}bc = log_{a}b + log_{a}c$ (We will use the second property) Like this

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

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 $log_{10}2 = 0.3010$ Note that this value is an approximate. This makes our equation

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

Removing the log we get $x = 10^{-26.02}$

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

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

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