Difference between revisions of "2020 AMC 10B Problems/Problem 12"
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<math>\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}</math> | <math>\textbf{(A)} \text{ 23} \qquad \textbf{(B)} \text{ 24} \qquad \textbf{(C)} \text{ 25} \qquad \textbf{(D)} \text{ 26} \qquad \textbf{(E)} \text{ 27}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
<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}=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 | 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== | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 19:20, 7 February 2020
Problem
The decimal representation of![\[\dfrac{1}{20^{20}}\]](http://latex.artofproblemsolving.com/6/c/e/6ce6559c22b38f698ff7b67ddc88f2d4d89c4084.png)
consists of a string of zeros after the decimal point, followed by a 
and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
Solution 1
![\[\dfrac{1}{20^{20}}=\dfrac{1}{(10\cdot2)^{20}}=\dfrac{1}{10^{20}\cdot2^{20}}\]](http://latex.artofproblemsolving.com/8/3/6/8360396f4f09706ce19e4f59018cfe10059c5cfb.png)
Now we do some estimation. Notice that 
, which means that 
is a little more than 
. Multiplying it with 
, we get that the denominator is about 
. Notice that when we divide 
by an 
digit number, there are 
zeros before the first nonzero digit. This means that when we divide by the 
digit integer , there are 
zeros in the initial string after the decimal point. -PCChess
Solution 2
Video Solution
~IceMatrix
See Also
| 2020 AMC 10B (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
