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Last night at Morty's - Alex's weird die

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Another quiet night at Morty's. Until Alex came in.

I bought a bunch of factory reject dice the other day, he said.

They're OK except the numbers are all wrong. Mostly they have

extra two's and three's, but this one [he held one up, far enough

away so the numbers could not be read] has all different numbers.

Still, the numbers aren't 1-6.

I have a wager for anyone here who thinks he's a genius.

I'll roll the thing three times against that wall over there.

The bottom and back die faces won't be visible, but I'll give

you the sum of the four faces that are visible.

As a bonus, I'll give you the sum of the top and front faces.

I'll buy a pint for anyone who can tell me all six numbers on the die.

If you try and can't figure it out, you'll buy me a pint.

He rolled the die three times and called out the numbers:

Roll #1 = 28 and 18

Roll #2 = 18 and 7

Roll #3 = 22 and 6

Jim thought for a while, then said, no way. There's too many possibilities.

So did Ian and Jamie.

Davey paused to scratch his beard and said, I'll try.

Would you have taken the bet?

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I take the bet and win. Here's why:

This configuration is pretty simple. We know that no number repeats and that two adjacent sides have to equal 18, 7, and 6 respectively. 7 & 6 are pretty easy. 7 is either 1,6 2,5 or 3,4 . 6 is either 1,5 or 2,4. It can't be 3,3 because there are no repeats. We also can do this for 18, but I won't here. Now we can list out what the opposing sides have to be for the top and front combinations. We can see that the opposing sides for 22 must be 22-6 or 16. Thus the highest opposing sides here has to be 15,1. We can eliminate the 17,1 and 16,2 combinations for front & top (FT) 18. Now let's look at all the opposing lists. If numbers have to be opposing, they can't be adjacent. Looking at the front and side combinations and the opposing lists, it looks like we need some 2's 4's and 5's. Let's put them together:

(dashes for spacing only, boxes represent faces of the die)

---[5]

[ ] [2] [4] [ ]

---[ ]

Now we know that if we have FT 2,5 and one opposing 4 the other must be a 7:

----[5]

[7] [2] [4] [ ]

----[ ]

Now let's look at FT 2,4 opposing must be a 5 so we go to the list and see it's match must be an 11.

---[5]

[7] [2] [4] [ ]

---[11]

So does this work for our FT18? Let's see. One combination we have is 11,7. We have that with a 2 opposing. Let's go to the list and see that the 2 must be opposite an 8.

---[5]

[7] [2] [4] [8]

---[11]

Do we meet all our original requirements?

Roll #1 = 28 and 18

Roll #2 = 18 and 7

Roll #3 = 22 and 6

FT 18 and total 28? [11+7]+2+8 = yep

FT 7 and total 18? [5+2]+7+4 = yep

FT 6 and total 22? [2+4]+11+5 = yep

Buy me a pint Alex! :)

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Sliding a frosty cold one down the bar to writersblock!

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Another quiet night at Morty's. Until Alex came in.

I bought a bunch of factory reject dice the other day, he said.

They're OK except the numbers are all wrong. Mostly they have

extra two's and three's, but this one [he held one up, far enough

away so the numbers could not be read] has all different numbers.

Still, the numbers aren't 1-6.

I have a wager for anyone here who thinks he's a genius.

I'll roll the thing three times against that wall over there.

The bottom and back die faces won't be visible, but I'll give

you the sum of the four faces that are visible.

As a bonus, I'll give you the sum of the top and front faces.

I'll buy a pint for anyone who can tell me all six numbers on the die.

If you try and can't figure it out, you'll buy me a pint.

He rolled the die three times and called out the numbers:

Roll #1 = 28 and 18

Roll #2 = 18 and 7

Roll #3 = 22 and 6

Jim thought for a while, then said, no way. There's too many possibilities.

So did Ian and Jamie.

Davey paused to scratch his beard and said, I'll try.

Would you have taken the bet?

no

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...

Still, the numbers aren't 1-6.

...

It seems that the problem cannot be solved with this assumption--unless negative numbers are allowed.

For example, what two numbers, excluding 1-6, add up to 6?

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It seems that the problem cannot be solved with this assumption--unless negative numbers are allowed.

For example, what two numbers, excluding 1-6, add up to 6?

I believe you have misread this, I believe he meant to say its not a standard die that has 1 through 6, but it doesn't mean that any number 1 through 6 is not on the die

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I believe you have misread this, I believe he meant to say its not a standard die that has 1 through 6, but it doesn't mean that any number 1 through 6 is not on the die

woops, after re-reading it that makes more sense.

carry on...

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I realize that this is an older puzzle, but I just found it and tried solving it. I am not disagreeing with the posted solution. However, the solution I came up with I think is also valid. Would someone check me on this.

FT = 18 TOT = 28 SIDES = 10

FT = 7 TOT = 18 SIDES = 11

FT = 6 TOT = 22 SIDES = 16

I get 13,8,5,3,2,1

checking this

FT = 18 = 13 + 5

SIDES = 10 = 8 + 2

FT = 7 = 5 + 2

SIDES = 11 = 7 + 4

FT = 6 = 5 + 1

SIDES = 16 = 13 + 3

I was working on this while at work, so I may be wrong, but if so can someone explain to me how?

thanks

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I realize that this is an older puzzle, but I just found it and tried solving it. I am not disagreeing with the posted solution. However, the solution I came up with I think is also valid. Would someone check me on this.

FT = 18 TOT = 28 SIDES = 10

FT = 7 TOT = 18 SIDES = 11

FT = 6 TOT = 22 SIDES = 16

I get 13,8,5,3,2,1

checking this

FT = 18 = 13 + 5

SIDES = 10 = 8 + 2

FT = 7 = 5 + 2

SIDES = 11 = 7 + 4

FT = 6 = 5 + 1

SIDES = 16 = 13 + 3

I was working on this while at work, so I may be wrong, but if so can someone explain to me how?

thanks

You're using seven numbers. ;)

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I think there is more than one answer to this. Writersblock's works an here is another set

numbers 2, 3, 4, 7, 9, 14 will also work

4+14 = 18, +3+7 = 28

3+4 = 7, +9+2 = 18

2+4 = 6, +7+9 = 22

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I think there is more than one answer to this. Writersblock's works an here is another set

numbers 2, 3, 4, 7, 9, 14 will also work

4+14 = 18, +3+7 = 28

3+4 = 7, +9+2 = 18

2+4 = 6, +7+9 = 22

The numbers work, but try to place those in the correct orientation on a die...I think you may find a problem. ;)

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Got one more combination for this:

1st throw: 7+11=18, 4+6=10

2nd throw: 2+5=7, 4+7=11

3rd throw: 2+4=6, 5+11=16

So, the dice contains:

2,4,5,6,7,11

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Got one more combination for this:

1st throw: 7+11=18, 4+6=10

2nd throw: 2+5=7, 4+7=11

3rd throw: 2+4=6, 5+11=16

So, the dice contains:

2,4,5,6,7,11

As with the previous solution you have failed to see that the numbers that total 10, 11 and 16 are on opposing sides of the dice, therefore the 4 cannot be used twice. It is can either be opposite the 6 or the 7, but not both.

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I believe there are multiple answers for Morty's:

1-2-3-4-9-15

2-3-4-5-8-13

2-3-4-5-7-13

2-3-4-5-6-13

2-4-5-7-8-11

2-4-5-6-7-11

This is my first puzzle. Did I do ok?

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There is only one answer when you take into account that certain numbers need to be opposite each other

The way I worked it out was by observing that the dice has landed each time on a different axis,  If it had been the same axis at least 2 of the sides (total - front & top) would have been the same value.  I then did a brute force by assuming first that the third throws top and front was 1,5 and then 2,4.  This would mean that the sides for the first two throws would both need to include either 1 or 5, or 2 or 4. also means that we can eliminate (1,15 and 5,11) or (2,14 and 4,12) from the third throw.  From there it is reasonably easy to prove that the third throw cannot be (1,5) on the top & face.  (2,4) on the top & face. is a little more tricky and I would suggest drawing a cube with possible values on each respective face. In the end I found that only Writersblock's answer existed.

An interesting follow up question would be if Morty cheated by switching dice between 3 dice with the same numbers  (2,4,5,7,8,11) but different configurations could he ensure that there would be no possible solution no matter which way the dice landed on each throw?

NOTE I haven't personally figured this out yet

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On 1/21/2017 at 9:15 PM, Kt.Kpop said:

I believe there are multiple answers for Morty's:

1-2-3-4-9-15

2-3-4-5-8-13

2-3-4-5-7-13

2-3-4-5-6-13

2-4-5-7-8-11

2-4-5-6-7-11

This is my first puzzle. Did I do ok?

Hi Kt.Kpop.

Welcome to the Den, and congratulations on solving the puzzle.

Feel free to post some of your favorites here, too.

On 1/25/2017 at 4:27 PM, phaze said:

An interesting follow up question would be if Morty cheated by switching dice between 3 dice with the same numbers  (2,4,5,7,8,11) but different configurations could he ensure that there would be no possible solution no matter which way the dice landed on each throw?

NOTE I haven't personally figured this out yet

Interesting twist phaze. Is this an accurate paraphrase?

Arrange 2,4,5,7,8,11 (differently) on three dice in such a manner that throwing each of them once cannot disclose enough information to deduce the six numbers separately.

My intuition says that it is possible. But proving it seems to require a list of every combination of all possible throws, 24 possibilities for each throw, and the application of a program that is verified to find a solution if one exists for each set of three throws. Proving something does not exist is always harder. Still it's interesting.

 

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