# Difference between revisions of "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)

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

Since $log_{a}bc = log_{a}b + log_{a}c$:

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

Take $log_{10}$ on both sides:

$$log_{10}x = \dfrac{1}{log_{10}10^{20} + log_{10}2^{20}}$$

$$= \dfrac{1}{20 \cdot log_{10}10 + 20 \cdot log_{10}2}$$

Remember that $log_{10}10 = 1$ and that, approximately, $log_{10}2 = 0.3010$

This makes our equation

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

Taking anti-logs: $x = 10^{-26.02}$

Let $[m]$ denote Greatest Integer Function of $m$.

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

~IceMatrix