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

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

Solution 8 (Finding a Pattern)

$\frac{1}{2^{20}}$ equals $\frac{1}{10^{20}}$ times $\frac{1}{2^{20}}$ . $\frac{1}{10^{20}}$ has 19 zeroes before the 1 after the decimal point. $\frac{1}{2}$ $\frac{1}{4}$ and $\frac{1}{8}$ have no zeroes after the decimal point. $\frac{1}{16}$ , $\frac{1}{32}$ , and $\frac{1}{64}$ have 1 zero after the decimal point. If you do division for all the other powers of 2, you find that every three powers of 2 an extra zero is added (128, 256, and 512 all have 3 digits, 1024, 2048, and 4096 have 4 digits, and so on.) Therefore, the number of zeroes after the decimal point in $\frac{1}{2^{20}}$ is 7 because 20/3 is 6 remainder 2. 19+7= $\boxed{\textbf{(D)} \ 26}.$

~unhappyfarmer

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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