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

Revision as of 00:30, 11 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 $\boxed{\textbf{(D) } \text{26}}$ .

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

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

@@ Line 19: / Line 19: @@
 ==Solution 3 (Brute Force)==
-Just as in Solution <math>2,</math> we rewrite <math>\dfrac{1}{20^{20}}</math> as <math>\dfrac{5^{20}}{10^{40}}.</math> We then calculate <math>5^{20}</math> entirely by hand, first doing <math>5^5 \cdot 5^5,</math> then multiplying that product by itself, resulting in <math>95,367,431,640,625.</math> Because this is <math>14</math> digits, after dividing this number by <math>10</math> fourteen times, the decimal point is before the <math>9.</math> Dividing the number again by <math>10</math> twenty-six more times allows a string of <math>\boxed{26~(D)}</math> zeroes to be formed. -OreoChocolate
+Just as in Solution <math>2,</math> we rewrite <math>\dfrac{1}{20^{20}}</math> as <math>\dfrac{5^{20}}{10^{40}}.</math> We then calculate <math>5^{20}</math> entirely by hand, first doing <math>5^5 \cdot 5^5,</math> then multiplying that product by itself, resulting in <math>95,367,431,640,625.</math> Because this is <math>14</math> digits, after dividing this number by <math>10</math> fourteen times, the decimal point is before the <math>9.</math> Dividing the number again by <math>10</math> twenty-six more times allows a string of<math>\boxed{\textbf{(D) } \text{26}}</math> zeroes to be formed. -OreoChocolate
 ==Video Solution==

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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