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

Revision as of 19:28, 7 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 $40-14=\boxed{26}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 First rewrite <math>\frac{1}{20^{20}}</math> as <math>\frac{5^{20}}{10^{40}}</math>. 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 <math>{5^{20}}</math>.
-<math>\log{5^{20}} = 20\log{5}</math> and memming <math>\log{5}\approx0.69</math> (alternatively use the fact that <math>\log{5} = 1 - \log{2}</math>), <math>\floor{20\log{5}}+1=\floor{20\cdot0.69}+1=13+1=14</math> digits.
+<math>\log{5^{20}} = 20\log{5}</math> and memming <math>\log{5}\approx0.69</math> (alternatively use the fact that <math>\log{5} = 1 - \log{2}</math>), <math>\lfloor{20\log{5}}\rfloor+1=\lfloor{20\cdot0.69}\rfloor+1=13+1=14</math> digits.
 Our answer is <math>40-14=\boxed{26}</math>.

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