[Guest]

Here's the game: You are manipulating three different characters, $, A, 1. You will be putting these characters into different sequences or groups, such as: $A1, $A$A11, etc.

Here are the rules:

If you have a group with "A" at the end, you may add "1" to it. So if you have $A$11A you can make it $A$11A1, if you have $AA, you can make it $AA1, etc.

If you have a group that is $x where x is any set of $,A or 1, you may add x. So if you have $A1, you can make it $A1A1, if you have $aa11 you can make it $aa11aa11, ect.

If you have "AAA" anywhere in your group, you can change that to "1". So if you have $AAA$ it becomes $1$. If you have $11AAAA it can become $111A or $11A1. etc.

Edit (because I'm silly and didn't copy and paste correctly first time):

Rule 4: If you ever get "11" anywhere in a group, you can drop it. So $A11 can become $A, $111 can become $1, and $A11A can become $AA. (Sorry about the initial omission - see later post)

Your goal is to get from $A to $1. Here's your start:

$A

(now you can do $A1, or $AA - the choice is yours)

If you solve it, show your progression. Shortest route to $1 wins!

[Guest]

No, there will not be any number in the series 2^n that will be divisible by 3

[Guest]

So, after reading the posts, i think the question can be translated as -- Is there an integer that is a multiple of 3 in the series 2,4,8,16,32,... (or 2^n series).

Solving this may provides the actual solution... Any comments?

I was actually thinking it was:

Is there an integer n such that

1. 2ⁿ mod 3 = 0 or

2. (2ⁿ + 1) mod 3 = 0 (I'm not sure why i was thinking this one...)

It is trivial to show there is no integer that makes case 1 true. I'm not sure about case 2. In addition, for all n where n mod 3 != 0, then (n-3) mod 3 != 0. Additionally, all the test cases that I ran through (admittedly small number but...) eventually reduced the count of A's to an even or prime number count..

I don't think this is solvable...

[Guest]

No, there will not be any number in the series 2^n that will be divisible by 3

I found the answer after some Googling, but wouldn't want to ruin anyones fun trying to prove my first idea wrong, which of course it is

[Guest]

I am leaving this topic after this conclusion... Bye..

None of the numbers in 2^n sequence is divisible by 3, and hence this problem is insolvable.

[Guest]

$A

$AA

$AAAA

$AAAAAAAA

$AAAAAAAAAAAAAAAA

$1A11AAAAA

$1AAAAAA

$111

$1

[Guest]

sorry my sollution doesnt work, i missed an A.

we need some value x, given that x is not dividable by 3, such that...

x*2^n -3*c is divisible by 3. yeah looks like there is no solution.