Difference between revisions of "2020 AMC 10B Problems/Problem 12"
Icematrix2 (talk | contribs) |
(→Solution 1) |
||
Line 9: | Line 9: | ||
<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 as long as n is not a power of 10, 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== |
Latest revision as of 22:43, 9 January 2021
Contents
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 as long as n is not a power of 10, 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 .
Solution 3 (Brute Force)
Just as in Solution we rewrite as We then calculate entirely by hand, first doing then multiplying that product by itself, resulting in Because this is digits, after dividing this number by fourteen times, the decimal point is before the Dividing the number again by twenty-six more times allows a string of zeroes to be formed. -OreoChocolate
Solution 4 (Alternate Brute Force)
Just as in Solutions and we rewrite as We can then look at the number of digits in powers of . , , , , , , and so on. We notice after a few iterations that every power of five with an exponent of , the number of digits doesn't increase. This means should have digits since there are numbers which are from to , or digits total. This means our expression can be written as , where is in the range . Canceling gives , or zeroes before the since the number should start on where the one would be in . ~aop2014
changed from 'Smarter Brute Force' to 'Alternate Brute Force'
Solution 5 (Logarithms Without Words)
~phoenixfire
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.