2016 AMC 10A Problems/Problem 17
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 by Pi Academy
https://youtu.be/iTkJExTiGkM?si=2dEicIJoLsaeWE-y
~ Pi Academy
Video Solution 2
~IceMatrix
Video Solution 3
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.