If you take a scientific calculator, type in any number, and then alternate applying the operations ATAN and COS, then eventually you will reach a point where you run into a cycle: after every COS step, you will see a result of approximately 0.786 (and after every ATAN step, a result of approximately 0.666 radians, or 38.2 degrees).
Susan and Eric are playing a dice game, with standard six-sided dice which have an equal probability of rolling each integer from 1 through 6. Each player takes turns scoring points as follows.
On Susan's turn, she rolls a die. If it's a 1, she rolls two dice in its place, continuing to replace any 1s by two new dice. When finally no 1s are showing, Susan scores points equal to the sum of the remaining dice, and then her turn ends. (For example, suppose she rolls a 1. Then it gets replaced by two dice; if these are a 3 and a 1, the 3 remains and the 1 gets replaced by two more; if these are a 2 and a 4, then Susan stops and scores 3 + 2 + 4 = 9 points.)
On Eric's turn, he rolls a die. If it's even, he replaces it with a roll of two dice; if the new total is even, he replaces it with a roll of three dice, etc. He continues to roll one more die than previously as long as his total is even. When he finally rolls an odd total, Eric scores that many points, and then his turn ends.
On average, how many points do Susan and Eric each score on a turn?
In the game of Flip Flop, players stand in a circle and take turns saying the numbers 1, 2, 3, etc., going clockwise.
When they get to a multiple of 7, the player whose turn it is says FLIP instead, and the direction switches: if they were going clockwise before, they now go counterclockwise, or vice versa.
When they get to a multiple if 8, the player whose turn it is says FLOP instead. The direction stays the same, but the next person is skipped over.
For a number n that is a multiple of both 7 and 8, the player says FLIP FLOP. Direction reverses and you skip over the next person. This means n + 1 is said by the person who said n - 2.
There is a puzzle with N^{2} pieces, each one shaped like a wedge that is 1/N of a disk. On the front of each piece, each of the two sides of the wedge is labeled with an integer between 1 and N, so that each of the N^{2} possible ordered pairs occurs exactly once. The goal of the puzzle is to form N full disks in such a way that the edges come together to have the same label. Below is an example of some pieces fitting together to form a full disk when N = 5.
Every day at noon exactly, people submit orders of positive integer numbers of paninis at your panini store. It takes you 1 minute to make 1 panini.
When Alice orders 2 paninis and Bob orders 100 paninis, you decide to make Alice's paninis first, because it results in less total customer wait time (104 vs. 202 minutes).
When the town gets wind of your scheme, they hatch devious plans. Instead of ordering 100 paninis, Bob decides to submit 100 orders of 1 panini each under different names, so he gets processed first. Oh no.
Call a panini-making scheme strategy-free if no customer, in any scenario, can strictly decrease their average wait time just by splitting their order up into several smaller (integer-size) orders.
Assume your customers are aware of your scheme and will not split up their orders if the scheme is strategy-free.
Jennifer and Serena are playing a game called Candy Split. They start with two piles of candy, one of size M and one of size K. On a player's turn, she eats one of the piles and splits the other pile into two (non-empty) piles any way she wants. Whoever can't make a legal move (i.e. is presented with two piles of one candy each) loses. Note that eating as much candy as possible is not the object of the game, though it may be a pleasant side effect.
The reason we're not asking you to submit an answer for the warm-up exercise is that it's a famous problem that can be found in many places on the internet. (Feel free to look it up if you can't figure out the answer.) What we are really interested is a harder problem:
Suppose Jennifer and Serena have three different types of candy. The game is now played with six piles of candy, two piles of each type. On a player's turn, she chooses a type of candy, eats one of the piles of that type, and splits the other. Whoever can't move loses. If the numbers are (M_{1}, K_{1}), (M_{2}, K_{2}) and (M_{3}, K_{3}), is it better to go first or second, and what is the winning strategy?
In order to solve this problem, you will need to learn a little bit of combinatorial game theory at this link: www.mathcamp.org/2019/cgt.pdf.
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All problems were written by the Mathcamp staff.