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 Perfect Shuffle (Posted on 2004-05-19)
You have a deck of 52 cards - for convenience, number them 1 through 52. You cut the cards into two equal halves and shuffle them perfectly. That is, the cards were in the order
1,2,3,...,52
and now they are
1,27,2,28,...,26,52. Let's call this a perfect in-shuffle.

If you repeat this in-shuffling process, how many in-shuffles will it take for the deck to return to its initial ordering (taking for granted that the cards will eventually do so)?
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How does the solution change if you have a deck of 64 cards, or 10, or in general, n cards? For odd integer values of n, in-shuffling will take 1,2,3,...,n to 1,(n+3)/2,2,(n+5)/2,...,n,(n+1)/2. For example, when n=5, the first in-shuffle yields 1,4,2,5,3.

 No Solution Yet Submitted by SilverKnight Rating: 4.2500 (4 votes)

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 for powers of 6 | Comment 14 of 20 |
(In reply to re: there's a pattern there. I just know it! by Charlie)

Ok, so the left side is the exponent of 6.

`1 42 123 284 365 6206 4207 1119728 6489 3452410 930011 3154756`
`6^2 = 3 * 6^16^4 = 3 * 6^2    = 9 * 6^16^8 = 18 * 6^4    = 54 * 6^2   = 162 * 6^16^3 = 7 * 6^16^5 = 155 * 6^16^6 = 15 * 6^3    = 35 * 6^2   = 105 * 6^16^7 = 3999 * 6^3    = 9331 * 6^2   = 27993 * 6^16^9 = 959 * 6^4   = 1233 * 6^3   = 2877 * 6^2   = 8631 * 6^16^10 = 15 * 6^5    = 775 * 6^2   = 2325 * 6^16^11 = 788689 * 6^1`

All the sqare numbers (even exponents) follow some kinda pattern. and the ones with prime exponents kinda follow a pattern too. Except for 7, which was the same in the original problem. So, is there a pattern? My answer is "kinda" If there is a pattern then it's either incomplete (ie. too many exceptions) or it's too complicated for me.

Edited on May 25, 2004, 2:18 pm
 Posted by Danny on 2004-05-25 14:08:48

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