### Brent DeJong

commented on a video on YouTube.We know where we are now, which is tragic.

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Brent DeJong

Works at Shoreline Containers

Attends Grand Valley State Unviversity

71 followers|14,391 views

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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.

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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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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All twisted up.... never seen the dude like this :-P

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In his circles

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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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Brent DeJong

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Actually, now I see it's not quite as simple as I had thought. Pay attention to the interaction between the green and blue faces, and you'll see that while one green face turns it does nothing to the blue face on one cube, but something to the blue on the other. Then, if g is a ccw turn of the green face on one cube, and b a ccw turn of the blue face on the same cube, gbgbgbgb will result in one cube being solved but not the other.

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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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n solo esta definido para los enteros, pensé que era obvio. Hablar de pares e impares considerando los racionales tales que estos no estén contenidos en los enteros no tiene sentido(lo digo por tu sustitución de n= 1/2), lo siento falto colocarlo xD, con esto espero que le veas el sentido.... todos los números de la forma 2n tales que n pertenezca a los enteros son pares, y los de la forma 2n+1 tal que n pertenezca a los enteros son impares.

Yo estoy de acuerdo contigo, pero ¿ No te parece curioso que todos los otros primos puedan escribirse así 2n+1 y justo el primo 2 no ?

Yo estoy de acuerdo contigo, pero ¿ No te parece curioso que todos los otros primos puedan escribirse así 2n+1 y justo el primo 2 no ?

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

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People

In his circles

147 people

Work

Occupation

Student

Skills

Bilingual - Spanish. Good with mental math. can think critically and abstractly.

Employment

- Shoreline ContainersSummer help, 2012 - presentPart time when school's in and full time when school's out, this job is supposed to remind me why I'm in school. It doesn't do its job that well except at paycheck time.
- Student, 1997 - present
- Employment SolutionsTemp, 2010 - 2013

Basic Information

Gender

Male

Other names

De Jong

Apps with Google+ Sign-in

Story

Tagline

Twitter @dejongbrent

Introduction

6'3 redhead.

Education

- Grand Valley State UnviversitySpanish for secondary education, 2010 - present
- CCHS2007 - 2010
- HCS1997 - 2007

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