Difference between revisions of "2016 AMC 10A Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | 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 in 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{ | + | 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 in 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> | ||
+ | |||
+ | Multiplying both sides of the inequality by <math>400(5n+1)</math>, we have | ||
+ | |||
+ | <cmath>400(4n+2)<321(5n+1),</cmath> | ||
+ | |||
+ | and by the distributive property, | ||
+ | |||
+ | <cmath>1600n+800<1605n+321.</cmath> | ||
+ | |||
+ | Subtracting <math>1600n+321</math> on both sides of the inequality gives us | ||
+ | |||
+ | <cmath>479<5n.</cmath> | ||
+ | |||
+ | 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>. | ||
==See Also== | ==See Also== |
Revision as of 18:57, 17 February 2016
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
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 in 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
.
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 |
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