Difference between revisions of "2016 AMC 10A Problems/Problem 17"
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<math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math> | <math>\textbf{(A) } 12 \qquad \textbf{(B) } 14 \qquad \textbf{(C) }16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 20</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Let <math>n = \frac{N}{5}</math>. Then, consider <math>5</math> blocks of <math>n</math> green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the <math>N + 1</math> positions between the green balls to insert the red ball. Less than <math>\frac{3}{5}</math> of the green balls will be on the same side of the red ball if the red ball is inserted | + | Let <math>n = \frac{N}{5}</math>. Then, consider <math>5</math> blocks of <math>n</math> green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the <math>N + 1</math> positions between the green balls to insert the red ball. Less than <math>\frac{3}{5}</math> of the green balls will be on the same side of the red ball if the red ball is inserted inside the middle block of <math>n</math> balls, and there are <math>n - 1</math> positions where this happens. Thus, <math>P(N) = 1 - \frac{n - 1}{N + 1} = \frac{4n + 2}{5n + 1}</math>, so |
<cmath>P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.</cmath> | <cmath>P(N) = \frac{4n + 2}{5n + 1} < \frac{321}{400}.</cmath> | ||
Line 24: | Line 24: | ||
Therefore, <math>N=5n>479</math>, so the least possible value of <math>N</math> is <math>480</math>. The sum of the digits of <math>480</math> is <math>\boxed{\textbf{(A) } 12}</math>. | Therefore, <math>N=5n>479</math>, so the least possible value of <math>N</math> is <math>480</math>. The sum of the digits of <math>480</math> is <math>\boxed{\textbf{(A) } 12}</math>. | ||
− | ==Pattern Solution== | + | ==Solution 2 (Pattern Solution)== |
− | Let <math>N | + | Let <math>N=5</math>, <math>P(N)=1</math> (Given) |
− | Let <math>N | + | Let <math>N=10</math>, <math>P(N)=\frac{10}{11}</math> |
− | Let <math>N | + | Let <math>N=15</math>, <math>P(N)=\frac{14}{16}</math> |
− | Notice that the fraction can be written as <math>1 | + | Notice that the fraction can be written as <math>1-\frac{\frac{N}{5}-1}{N+1}</math> |
− | Now it's quite simple to write the inequality as <math>1 | + | Now it's quite simple to write the inequality as <math>1-\frac{\frac{N}{5}-1}{N+1}<\frac{321}{400}</math> |
− | We can subtract <math>1</math> on both sides to obtain <math>- | + | We can subtract <math>1</math> on both sides to obtain <math>-\frac{\frac{N}{5}-1}{N+1}<-\frac{79}{400}</math> |
− | Dividing both sides by <math>-1</math>, we derive <math>\frac{\frac{N}{5}-1}{N+1} | + | Dividing both sides by <math>-1</math>, we derive <math>\frac{\frac{N}{5}-1}{N+1}>\frac{79}{400}</math>. (Switch the inequality sign when dividing by <math>-1</math>) |
We then cross multiply to get <math>80N - 400 > 79N + 79</math> | We then cross multiply to get <math>80N - 400 > 79N + 79</math> | ||
Line 46: | Line 46: | ||
Finally we get <math>N > 479</math> | Finally we get <math>N > 479</math> | ||
− | To | + | To achieve <math>N = 480</math> |
− | So the sum of the digits of <math>N</math> = <math>\boxed{\textbf{(A) } 12}</math> | + | So the sum of the digits of <math>N</math> = <math>\boxed{\textbf{(A) } 12}</math> |
− | + | ==Solution 3== | |
+ | We are trying to find the number of places to put the red ball, such that <math>\frac{3}{5}</math> of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with <math>N</math>: Trying a few values, we see that the ball "works" in places <math>1</math> to <math>\frac{2}{5}N + 1</math> and spaces <math>\frac{3}{5}N+1 </math> to <math>N+1</math>. This is a total of <math>\frac{4}{5}N + 2</math> spaces, over a total possible <math>N + 1</math> places to put the ball. So: | ||
+ | |||
+ | <math>\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.</math> And we know that the next value is what we are looking for, so <math>N+1 = 480</math>, and the sum of its digits is <math>\boxed{\textbf{(A) } 12}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | Draw <math>N</math> balls as described in the problem. First consider the left side arrangements. Notice that for at least <math>\frac{3}{5}N</math> balls must be to the right of the valid left side arrangements. Since we have <math>N+1</math> balls, we know that there are <math>\frac{2}{5}N+1</math> valid left side arrangements. This also goes for the right side arrangements, for a total of | ||
+ | <math>\frac{4}{5}N + 2</math> spaces, over a total possible <math>N + 1</math> places to put the ball. So: | ||
+ | |||
+ | <math>\frac{\frac{4}{5}N + 2}{N+1} = \frac{321}{400} \rightarrow N = 479.</math> And we know that the next value is what we are looking for, so <math>N+1 = 480</math>, and the sum of its digits is <math>\boxed{\textbf{(A) } 12}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/YsIPcwuycF0 | ||
+ | |||
+ | ~IceMatrix | ||
+ | ==Video Solution 2== | ||
+ | |||
+ | https://www.youtube.com/watch?v=iFSTuWCrosY | ||
==See Also== | ==See Also== |
Latest revision as of 03:26, 29 April 2024
Contents
Problem
Let be a positive multiple of
. One red ball and
green balls are arranged in a line in random order. Let
be the probability that at least
of the green balls are on the same side of the red ball. Observe that
and that
approaches
as
grows large. What is the sum of the digits of the least value of
such that
?
Solution 1
Let . Then, consider
blocks of
green balls in a line, along with the red ball. Shuffling the line is equivalent to choosing one of the
positions between the green balls to insert the red ball. Less than
of the green balls will be on the same side of the red ball if the red ball is inserted inside the middle block of
balls, and there are
positions where this happens. Thus,
, so
Multiplying both sides of the inequality by , we have
and by the distributive property,
Subtracting on both sides of the inequality gives us
Therefore, , so the least possible value of
is
. The sum of the digits of
is
.
Solution 2 (Pattern Solution)
Let ,
(Given)
Let ,
Let ,
Notice that the fraction can be written as
Now it's quite simple to write the inequality as
We can subtract on both sides to obtain
Dividing both sides by , we derive
. (Switch the inequality sign when dividing by
)
We then cross multiply to get
Finally we get
To achieve
So the sum of the digits of =
Solution 3
We are trying to find the number of places to put the red ball, such that of the green balls or more are on one side of it. Notice that we can put the ball in a number of spaces describable with
: Trying a few values, we see that the ball "works" in places
to
and spaces
to
. This is a total of
spaces, over a total possible
places to put the ball. So:
And we know that the next value is what we are looking for, so
, and the sum of its digits is
.
Solution 4
Draw balls as described in the problem. First consider the left side arrangements. Notice that for at least
balls must be to the right of the valid left side arrangements. Since we have
balls, we know that there are
valid left side arrangements. This also goes for the right side arrangements, for a total of
spaces, over a total possible
places to put the ball. So:
And we know that the next value is what we are looking for, so
, and the sum of its digits is
.
Video Solution
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=iFSTuWCrosY
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.