# 2020 AMC 10B Problems/Problem 12

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## 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