# Happy Halloween From  The Muse Garden!

In the tradition of my Holiday Math Puzzles, I’m here with an appropriately themed puzzle for this time of year.

## Candy Distribution

It’s that time of year all right. You’re out and about, trick or treating with your friends or family and when you come home, you decide to dump all the candy out on the floor to sort through it. But then, as siblings often do, you begin to bicker about who has “more” than the other. In fact, there are some candies that you really like, and some you don’t like. You’d rather have a bunch of chocolate than a bunch of peppermints, for instance.  But wait a minute! Of course you can’t like the same thing! Your sibling actually likes peppermints!

Here’s a table of the different candies you have. Each candy has a “value” to it; that is, how much you “want” it. Try to split up the candies such that you and your sibling both have as equal value as possible at the end. And no fighting!

 Candy Quantity Your Value (per piece) Sibling’s Value (per piece) Candy Corn 150 25 50 Peppermints 50 5 50 Peanut Butter Cups 10 100 75 Hershey Bars 25 50 10 Kit Kat 20 75 30

# Chopsticks Game – A Combinatorial Challenge

So I don’t know if anyone else is familiar with this game, but I just remember playing it with friends in middle school and it occurred to me the other day that it would be an interesting game to analyze combinatorially, and perhaps write a game playing algorithm for. This game can be found in more detail here: http://en.wikipedia.org/wiki/Chopsticks_(hand_game)

Players: 2+.

Rules of Play: Players each begin with two “piles” of points, and each pile has 1 point to begin with. We used fingers to represent this, one finger on each hand.

On each turn, a player can choose to do one of two things:

1. Send points from one of the player’s pile to one of the opponents piles. So if Player 1 wanted to send 1 point to Player 2’s left pile, then Player 2 would have 2 points in the left pile and Player 1 would still have 1 point in his left pile. Player 1 does not lose points, they are simply “cloned” over to the opponents pile.
2. If the player has an even number of points in one pile and zero points in the other pile, the player may elect to split his points evenly between the two piles. This consumes the player’s turn. Example: if Player 1 has (0     4) then he can use his turn to split his points, giving him (2     2).

If a player gets exactly 5 points in either pile, that pile loses all of its points and reverts to 0. However, if points applied goes over 5 (such as adding 2 points to a 4 point pile), then the remainder of points are added. (meaning that 4 + 2 = 1). The opponent simply gets points mod 5.

If a player gets to 0 points in both their piles, then they lose. The last person that has points remaining wins.
$=========================$
Okay, so let’s break this down. Here’s an example game for those of you that are more visually oriented (follow the turns by reading left to right, moves are marked with red arrows):

• On turn 4, player 2 adds 3 points to player 1’s 2 points, making 5. The rules state that any pile with exactly 5 points reverts to zero.
• On turn 7, player 2 adds 3 points to 3 points. $3+3=6$ as we all know, but $6 \equiv 1 \mod 5$, so player 1 now has one point in his pile.
• On turn 13, player 2 decides to split his points, turning his one pile of 4 into two piles of 2. This consumes his turn.
• On turn 15, player 2 adds 2 points to player 1’s 3, thus reverting his pile to zero. Player 1 now has no more points to play with, so player 2 wins.

We can think about this game as a combinatorial problem. What are the optimal positions to play? How would one program a computer to play this game? I plan to create an interactive web game where players can try this for themselves.

# Weighing The Gold Coins (Brain Teaser)

You have 10 bags with 10 gold coins each. Nine bags have genuine coins that weigh 10g each, but one bag has fake coins that are only 9g each. Using a digital scale and only one weighing, how do you figure out which bag has the fake coins?

# 4 Is The Magic Number – Riddle

So fishes56 alerted me to this riddle in a comment here, and if you didn’t see it I wanted to share it here in a separate post cause its a little fun thing to think about.

12 is 6, 6 is 3, 3 is 5, 5 is 4, 4 is 4.

4 is the magic number, why?

68 is 10, 10 is 3, 3 is 5, 5 is 4, 4 is 4.

4 is the magic number, why?

26 is 9, 9 is 4, 4 is 4. 4 is the magic number, why?

Give it a try. :)

# The Collatz Conjecture and Hailstone Sequences: Deceptively Simple, Devilishly Difficult

Here is a very simple math problem:

If a number is even, divide the number by two. If a number is odd, multiply the number by three, and then add one. Do this over and over with your results. Will you always get down to the value 1 no matter what number you choose?

Go ahead and test this out for yourself. Plug a few numbers in. Try it out. I’ll wait.

Back? Good. Here’s my example: I start with the number $12$.

$12$ is even, so I divide it by $2$.

$12/2=6$.

$6/2=3$.

$3$ is odd, so we multiply it by $3$ then add $1$.

$(3*3)+1=10$.

$10/2=5$.

$(5*3)+1=16$.

$16/2=8$.

$8/2=4$.

$4/2=2$.

$2/2=1$.

And we have arrived at one, just like we thought we would. But is this ALWAYS the case, for ANY number I could possibly dream of?

This may sound easy, but in fact it is an unsolved mathematics problem, with some of the best minds in the field doing research on it. It’s easy enough to check low values, but numbers are infinite, and can get very, very, very large. How do we KNOW that it works for every single number that exists?

That is what is so difficult about this problem. While every single number we have checked (which is a LOT) ends up at 1 sooner or later (some numbers can bounce around for a very long time, hence why they are called “hailstone sequences”), we still have no method to prove that it works for every number. Mathematicians have been looking for some method to predict this, but despite getting into some pretty heavy mathematics in an attempt to attack this problem, we still do not know for sure.

This problem is known as the Collatz Conjecture and is very interesting to mathematicians young and old because it is so easy to explain and play with, yet so tough to exhaustively prove. What do you think? Will this problem ever be solved? And what would be the implications if it was?

Here is some simple Python code that can display the cycles for any number you type in:

#!/usr/bin/python
num = int(input("Enter a number: "))
while num!=1:
if num%2 == 0:
num = num/2
else:
num = (num*3)+1
print num


# Periodic Last Digits of Fibonacci Numbers

Here’s something I was playing around with the other day.

Consider the Fibonacci sequence: $0, 1, 1, 2,3,5,8,13,21, \dots$. $F(n) = F(n-1)+F(n-2)$. I wanted to see if there was a pattern regarding the values of the ones digit in the fibonacci sequence. This is what I found:

The ones digits of the Fibonacci numbers are periodic, with period 60. This means that if you continue the sequence and look only at the ones digits of the numbers, they begin to repeat the same pattern after the 60th number. In addition, every 15th number in the sequence has a ones digit of zero. I wonder why this is. I can’t seem to find any other patterns at the moment. It probably has something to do with mod.

Here is the list of the 60 digit period (ones digits only):

1
1
2
3
5
8
3
1
4
5
9
4
3
7
0
7
7
4
1
5
6
1
7
8
5
3
8
1
9
0
9
9
8
7
5
2
7
9
6
5
1
6
7
3
0
3
3
6
9
5
4
9
3
2
5
7
2
9
1
0
---this is where the cycle starts over
1
1
2
3
5
8
...

Can you find any other interesting properties?