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Brent DeJong
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Brent DeJong

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Here is the kit of uncolored originals for my Math for seven-year-olds project on Graph coloring, Chromatic numbers, and Eulerian paths and circuits.  Just print, copy each folio double-sided (with correct orientation, the cover is one-sided), fold each folio in half, assemble the folded folios side-by-side, and place into the cover to form your booklet, with a few staples for a binding.  See http://jdh.hamkins.org/math-for-seven-year-olds-graph-coloring-chromatic-numbers-eulerian-paths/

Andrej Bauer has assembled the images into a single pdf file: https://drive.google.com/file/d/0B7eG5PHUDcmZX1FnOHRhSU9ubUU/edit?usp=sharing, and filtered the color to black/white to improve printing.
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Brent DeJong

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For every state of one cube, there is exactly one state of the other cube in this puzzle, are there not? Then this puzzle has just as many states as the original rubik's cube: 12!*8!*3^12*2^8/12. Perhaps I have missed something; if so, enlighten me.

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There's no psychological challenge for the robot :(

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This is pretty cool :) Shows the physics of 60-Hz electricity pretty well :)

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triple negative = single negative lol

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My approach was: 136/4 = (100+36)/4 = 25+9 = 34. Average is 34. OK. 31, 33, 35, 37.

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We know where we are now, which is tragic.

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So, you're going to get a value that's a multiple of 4. The probability that the number of cards you're holding is the number you first calculate is 1/4, or 25%. 
So: if you're holding x cards, how many times do you have to add or subtract? Let's find out!
1 = 40 - 13*3......14 = 40 - 13*2.......27 = 40 - 13*1..........40 = 40
2 = 28 - 13*2......15 = 28 - 13*1.......28 = 28....................41 = 28 + 13*1
3 = 16 - 13*1......16 = 16................29 = 16 + 13*1..........42 = 16 + 13*2
4 = 4 ................17 = 4 + 13*1.......30 = 4 + 13*2 ...........43 = 4 + 13*3
5 = 44 - 13*3......18 = 44 - 13*2......31 = 44 - 13*1 .......... 44 = 44
6 = 32 - 13*2 .....19 = 32 - 13*1 .....32 = 32 ....................45 = 32 + 13*1
7 = 20 - 13*1 ......20 = 20...............33 = 20 + 13*1 .........46 = 20 + 13*2
8 = 8 .................21 = 8 + 13*1 .....34 = 8 + 13*2 ...........47 = 8 + 13*3
9 = 48 - 13*3 ..... 22 = 48 - 13*2 ....35 = 48 - 13*1 ...........48 = 48
10 = 36 - 13*2 ....23 = 36 - 13*1 ....36 = 36 .....................49 = 36 + 13*1
11 = 24 - 13*1 ....24 = 24 ..............37 = 24 + 13*1 ..........50 = 24 + 13*2
12 = 12 .............25 = 12 + 13*1 ....38 = 12 + 13*2 ..........51 = 12 + 13*3
13 = 0 + 13*1 ....26 = 0 + 13*2 ......39 = 0 + 13*3 ........... 52 = 0 + 13*4
So, if you're holding between 10 and 38 cards, you add or subtract 13 no more than 2 times. If we don't count adding straight thirteens to 0, the upper bound increases from 38 to 42. 

I conclude that optimal range is to leave at least 10 cards on the table and pick up at least 10 cards.

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This is recent? I saw this the first time I looked at Pascal's triangle... but then, I suppose the proof would take some time to develop.

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2 es el más raro porque es el único que es par? bueno, 3 es el único igual a 0 mod 3, 5 el único igual a 0 mod 5.... todos son iguales de raro pensando así :P
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Brent DeJong

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" Vai repetir a frase..." GANHOU!! kkkkkkkkkkkkkkkkk
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