Thursday, June 28, 2012 11:07:23 PM
Imagine the top card of a deck as card 1 and the bottom as card 52. For example the 3rd card from the top is 3. This is very important.
In this case the cards are placed with 15 cards in between them,as long as protocol is followed, no matter what. In the position of the deck they are always cards 6,22,and 38 check me if I'm wrong.
Then using alternation we eliminate the odd numbered cards and reverse the order of remaining even cards so now starting with the top, the cards are 52,50,48...2.
Repeating the alternation method removes cards 52,48,44...4. not 38,22,or6 because of sequence. now the order is reversed again. the cards remaining in order are now 2,6,10...50.
Repeat alternation to remove cards 2,10,18...50. reversing the order the cards from top to bottom remaining are 46,38,30,22,14, and 6.
what is left is a simple alternation that removes 46,30,and 14 leaving the last three cards 38,22,and 6 yet to be unturned.
Saturday, June 23, 2012 9:52:28 PM
I'm about to turn 24...and my 6th grade math teacher showed my class this trick. Old trick is old. Still pretty cool though!
Saturday, June 23, 2012 8:54:13 PM
Move four cards to the bottom? I'm not sure why folks are having trouble with this trick, but I can tell you that it`s much easier to make the first pile fourteen instead of ten and skip the "four cards to the bottom" part.