Difference between revisions of "2016 AMC 10A Problems/Problem 17"
Math101010 (talk | contribs) (Created page with "Let <math>N</math> be a positive multiple of <math>5</math>. One red ball and <math>N</math> green balls are arranged in a line in random order. Let <math>P(N)</math> be the p...") |
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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> | ||
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+ | ==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{0.8N + 2}{N + 1}</math>. | ||
+ | Solving the inequality <math>P(N) = \frac{0.8N + 2}{N + 1} < \frac{321}{400}</math> gives <math>N > 479</math>, so the least value of <math>N</math> is <math>480</math>. The sum of the digits of <math>480</math> is <math>\boxed{12}</math>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2016|ab=A|num-b=16|num-a=18}} | ||
+ | {{MAA Notice}} |
Revision as of 22:46, 3 February 2016
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,
.
Solving the inequality
gives
, so the least 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 |
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