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

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<cmath>\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}</cmath>
 
<cmath>\dfrac{1}{20^{20}} = \dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}</cmath>
  
Now we do some estimation. Notice that <math>2^{20} \approx 1024^2</math>, which means that <math>2^{20}</math> is a little more than <math>1000^2=1,000,000</math>. Multiplying it with <math>10^{20}</math>, we get that the denominator is about <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>. Notice that when we divide <math>1</math> by an <math>n</math> digit number, there are <math>n-1</math> zeros before the first nonzero digit. This means that when we divide <math>1</math> by the <math>27</math> digit integer <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>, there are <math>\boxed{\textbf{(D) } \text{26}}</math> zeros in the initial string after the decimal point. -PCChess
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Now we do some estimation. Notice that <math>2^{20} = 1024^2</math>, which means that <math>2^{20}</math> is a little more than <math>1000^2=1,000,000</math>. Multiplying it with <math>10^{20}</math>, we get that the denominator is about <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>. Notice that when we divide <math>1</math> by an <math>n</math> digit number, there are <math>n-1</math> zeros before the first nonzero digit. This means that when we divide <math>1</math> by the <math>27</math> digit integer <math>1\underbrace{00\dots0}_{26 \text{ zeros}}</math>, there are <math>\boxed{\textbf{(D) } \text{26}}</math> zeros in the initial string after the decimal point. -PCChess
  
 
==Solution 2==
 
==Solution 2==

Revision as of 20:29, 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

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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