Difference between revisions of "2020 AMC 10B Problems/Problem 12"
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==Solution 1== | ==Solution 1== | ||
+ | We have | ||
<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> | ||
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Just as in Solutions <math>2</math> and <math>3,</math> we rewrite <math>\dfrac{1}{20^{20}}</math> as <math>\dfrac{5^{20}}{10^{40}}.</math> We can then look at the number of digits in powers of <math>5</math>. <math>5^1=5</math>, <math>5^2=25</math>, <math>5^3=125</math>, <math>5^4=625</math>, <math>5^5=3125</math>, <math>5^6=15625</math>, <math>5^7=78125</math> and so on. We notice after a few iterations that every power of five with an exponent of <math>1 \mod 3</math>, the number of digits doesn't increase. This means <math>5^{20}</math> should have <math>20 - 6</math> digits since there are <math>6</math> numbers which are <math>1 \mod 3</math> from <math>0</math> to <math>20</math>, or <math>14</math> digits total. This means our expression can be written as <math>\dfrac{k\cdot10^{14}}{10^{40}}</math>, where <math>k</math> is in the range <math>[1,10)</math>. Canceling gives <math>\dfrac{k}{10^{26}}</math>, or <math>26</math> zeroes before the <math>k</math> since the number <math>k</math> should start on where the one would be in <math>10^{26}</math>. ~aop2014 | Just as in Solutions <math>2</math> and <math>3,</math> we rewrite <math>\dfrac{1}{20^{20}}</math> as <math>\dfrac{5^{20}}{10^{40}}.</math> We can then look at the number of digits in powers of <math>5</math>. <math>5^1=5</math>, <math>5^2=25</math>, <math>5^3=125</math>, <math>5^4=625</math>, <math>5^5=3125</math>, <math>5^6=15625</math>, <math>5^7=78125</math> and so on. We notice after a few iterations that every power of five with an exponent of <math>1 \mod 3</math>, the number of digits doesn't increase. This means <math>5^{20}</math> should have <math>20 - 6</math> digits since there are <math>6</math> numbers which are <math>1 \mod 3</math> from <math>0</math> to <math>20</math>, or <math>14</math> digits total. This means our expression can be written as <math>\dfrac{k\cdot10^{14}}{10^{40}}</math>, where <math>k</math> is in the range <math>[1,10)</math>. Canceling gives <math>\dfrac{k}{10^{26}}</math>, or <math>26</math> zeroes before the <math>k</math> since the number <math>k</math> should start on where the one would be in <math>10^{26}</math>. ~aop2014 | ||
− | == | + | ==Solution 5 (Scientific Notation)== |
We see that <math>\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}</math>. We see that this has <math>27-1=26</math> zeros after the decimal point before coming to <math>9</math>. | We see that <math>\frac{1}{20^{20}} = 9.5367432 \cdot \cdot \cdot \times 10^{-27}</math>. We see that this has <math>27-1=26</math> zeros after the decimal point before coming to <math>9</math>. | ||
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==Solution 6 (Logarithms Without Words)== | ==Solution 6 (Logarithms Without Words)== | ||
− | <cmath>|\lceil \log \dfrac{1}{20^{20}} \rceil| | + | <cmath>\begin{align*}|\lceil \log \dfrac{1}{20^{20}} \rceil| |
− | = |\lceil \log 20^{-20} \rceil| | + | &= |\lceil \log 20^{-20} \rceil| \\ |
− | = |\lceil -20 \log(20) \rceil| | + | &= |\lceil -20 \log(20) \rceil| \\ |
− | = |\lceil -20(\log 10 + \log 2) \rceil| | + | &= |\lceil -20(\log 10 + \log 2) \rceil| \\ |
− | = |\lceil -20(1 + 0.301) \rceil| | + | &= |\lceil -20(1 + 0.301) \rceil| \\ |
− | = |\lceil -26.02 \rceil| | + | &= |\lceil -26.02 \rceil| \\ |
− | = |-26| | + | &= |-26| \\ |
− | = \boxed{\textbf{(D) } \text{26}}</cmath> | + | &= \boxed{\textbf{(D) } \text{26}} \end{align*}</cmath> |
~phoenixfire | ~phoenixfire | ||
+ | |||
+ | ==Video Solution (HOW TO CRITICALLY THINK) | ||
+ | https://youtu.be/qnznImCDSW4 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Solution 7 (Algebra)== | ||
+ | Define <math>n</math> as the number of consecutive <math>0</math>s after the decimal point in the decimal representation of <math>\frac{1}{20^{20}}.</math> Thus, <math>\frac{1}{10^n}</math> has <math>n-1</math> zeroes followed by a <math>1</math> when expressed as a decimal, and is greater than <math>\frac{1}{20^{20}}.</math> So, we have that | ||
+ | <cmath>\frac{1}{10^n} > \frac{1}{20^{20}}.</cmath> | ||
+ | Taking the reciprocal and switching signs grants | ||
+ | <cmath>20^{20}>10^n.</cmath> | ||
+ | We can express <math>20^{20}</math> as <math>10^{20} \cdot 1024^2.</math> We know that | ||
+ | <cmath>\begin{align*} | ||
+ | (1000 + 24)^2 &= 1024^2 \\ | ||
+ | 1000000 + 48000 + 576 &= 1048576. \end{align*}</cmath> | ||
+ | Multiplying this with <math>10^{20}</math> adds <math>20</math> more zeroes at the end, which means the left side of our inequality has <math>27</math> digits. Then, the <math>10</math> on the right side must have an exponent less than <math>27,</math> meaning that <math>n<27.</math> | ||
+ | |||
+ | We are also given that the initial string of zeroes in the decimal representation of <math>\frac{1}{20^{20}}</math> is followed by a <math>9.</math> If we consider just these <math>n+1</math> digits after the decimal point, we can express it as the fraction <math>\frac{9}{10^{n+1}}.</math> Since there are more digits after this <math>9</math> in the decimal representation of <math>\frac{1}{20^{20}},</math> it is true that | ||
+ | <cmath>\frac{1}{20^{20}} > \frac{9}{10^{n+1}}.</cmath> | ||
+ | Rinse and repeat and we have | ||
+ | <cmath>20^{20} < \frac{10^{n+1}}{9}.</cmath> | ||
+ | Multiplying both sides by <math>\frac{9}{10^{20}}</math> gives | ||
+ | <cmath>9 \cdot 2^{20} < 10^{n-19}.</cmath> | ||
+ | We previously found that <math>2^{20} = 1048576,</math> which has <math>7</math> digits. Multiplying this by <math>9</math> will keep the number of digits the same (make sure to see why). Thus, | ||
+ | <cmath>n-19>6 \qquad \text{or} \qquad n > 25.</cmath> | ||
+ | |||
+ | If <math>n<27</math> and <math>n>25,</math> then <math>n= \boxed{\textbf{(D)} \ 26}.</math> | ||
+ | |||
+ | ~happyhari, mathbrek | ||
+ | |||
+ | ==Solution 8 (Finding a Pattern)== | ||
+ | |||
+ | <math>\frac{1}{2^{20}}</math> equals <math>\frac{1}{10^{20}}</math> times <math>\frac{1}{2^{20}}</math>. <math>\frac{1}{10^{20}}</math> has 19 zeroes before the 1 after the decimal point. <math>\frac{1}{2}</math> <math>\frac{1}{4}</math> and <math>\frac{1}{8}</math> have no zeroes after the decimal point. <math>\frac{1}{16}</math>, <math>\frac{1}{32}</math>, and <math>\frac{1}{64}</math> have 1 zero after the decimal point. If you do division for all the other powers of 2, you find that every three powers of 2 an extra zero is added (128, 256, and 512 all have 3 digits, 1024, 2048, and 4096 have 4 digits, and so on.) Therefore, the number of zeroes after the decimal point in <math>\frac{1}{2^{20}}</math> is 7 because 20/3 is 6 remainder 2. 19+7= <math>\boxed{\textbf{(D)} \ 26}.</math> | ||
+ | |||
+ | ~unhappyfarmer | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/t6yjfKXpwDs | https://youtu.be/t6yjfKXpwDs | ||
− | |||
− | |||
==See Also== | ==See Also== |
Latest revision as of 05:43, 21 June 2024
Contents
Problem
The decimal representation of 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
We have
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
Solution 5 (Scientific Notation)
We see that . We see that this has
zeros after the decimal point before coming to
.
Therefore, the answer is
Solution 6 (Logarithms Without Words)
~phoenixfire
==Video Solution (HOW TO CRITICALLY THINK) https://youtu.be/qnznImCDSW4
~Education, the Study of Everything
Solution 7 (Algebra)
Define as the number of consecutive
s after the decimal point in the decimal representation of
Thus,
has
zeroes followed by a
when expressed as a decimal, and is greater than
So, we have that
Taking the reciprocal and switching signs grants
We can express
as
We know that
Multiplying this with
adds
more zeroes at the end, which means the left side of our inequality has
digits. Then, the
on the right side must have an exponent less than
meaning that
We are also given that the initial string of zeroes in the decimal representation of is followed by a
If we consider just these
digits after the decimal point, we can express it as the fraction
Since there are more digits after this
in the decimal representation of
it is true that
Rinse and repeat and we have
Multiplying both sides by
gives
We previously found that
which has
digits. Multiplying this by
will keep the number of digits the same (make sure to see why). Thus,
If and
then
~happyhari, mathbrek
Solution 8 (Finding a Pattern)
equals
times
.
has 19 zeroes before the 1 after the decimal point.
and
have no zeroes after the decimal point.
,
, and
have 1 zero after the decimal point. If you do division for all the other powers of 2, you find that every three powers of 2 an extra zero is added (128, 256, and 512 all have 3 digits, 1024, 2048, and 4096 have 4 digits, and so on.) Therefore, the number of zeroes after the decimal point in
is 7 because 20/3 is 6 remainder 2. 19+7=
~unhappyfarmer
Video Solution
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.