Difference between revisions of "2016 AMC 10A Problems/Problem 17"
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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 by <math>AOPS12142015 | + | Solution by <math>AOPS12142015</math> |
==See Also== | ==See Also== |
Revision as of 13:20, 20 May 2016
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
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
.
Pattern Solution
Let
,
1 (
)
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 get
So the sum of the digits of =
.
Solution by
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.