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} | + | 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 |
==Solution 2== | ==Solution 2== |
Revision as of 20:29, 7 February 2020
Problem
The decimal representation ofconsists 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
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
First rewrite as . 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 .
and memming (alternatively use the fact that ), digits.
Our answer is .
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 |
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