Difference between revisions of "2020 AMC 10B Problems/Problem 12"
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==Solution 2== | ==Solution 2== | ||
+ | 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>. | ||
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+ | <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. | ||
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+ | Our answer is <math>40-14=\boxed{26}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 19:26, 7 February 2020
Contents
[hide]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 ), $\floor{20\log{5}}+1=\floor{20\cdot0.69}+1=13+1=14$ (Error compiling LaTeX. Unknown error_msg) 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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