Mixing
How many fair dice of this kind exist?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I am not talking about the shape of the dice here, I am talking about another type. You will see what I mean soon.
For example, when there are 1 dice, a normal dice is a fair dice, because the probability of getting each number is the same by $ frac16 $
When 2 normal dice are thrown, you can get numbers from 2 to 12, but the probability of getting each of them are different. Think of Monopoly, the probability of getting 7 is $ frac636 $ while the probability of getting 2 is $ frac136 $ the goal is to make the probability of getting each number the same.
I got an example:
Dice 1: $ [1,2,3,4,5,6] $
Dice 2: $ [0,6,12,18,24,30] $
When these two dice are thrown and add the two number, the possible outcome can be every number from 1 to 36 and the probability of each is
So the question is, How many different pairs of dice of this kind are possible so that the possible outcome can be every number from 1 to 36 and the probability of each is $ frac136 $
The number on the dice can be negative too.
Example:
Dice 1: $ [-1,1,11,13,23,25] $
Dice 2: $ [2,3,6,7,10,11] $
Same number, different order is considered the same.
$$ [-1,1,11,13,23,25] [2,3,6,7,10,11] $$
$$ [1,11,13,23,25,-1] [2,3,6,7,10,11] $$
$$ [2,3,6,7,10,11] [-1,1,11,13,23,25] $$
They are all the same
So how many are there?
Is there a way to find it without trying one by one?
combinations puzzle
asked Jul 24 at 12:22
Pizzaroot
1056
add a comment |Â
up vote
2
down vote
favorite
I am not talking about the shape of the dice here, I am talking about another type. You will see what I mean soon.
For example, when there are 1 dice, a normal dice is a fair dice, because the probability of getting each number is the same by $ frac16 $
When 2 normal dice are thrown, you can get numbers from 2 to 12, but the probability of getting each of them are different. Think of Monopoly, the probability of getting 7 is $ frac636 $ while the probability of getting 2 is $ frac136 $ the goal is to make the probability of getting each number the same.
I got an example:
Dice 1: $ [1,2,3,4,5,6] $
Dice 2: $ [0,6,12,18,24,30] $
When these two dice are thrown and add the two number, the possible outcome can be every number from 1 to 36 and the probability of each is
So the question is, How many different pairs of dice of this kind are possible so that the possible outcome can be every number from 1 to 36 and the probability of each is $ frac136 $
The number on the dice can be negative too.
Example:
Dice 1: $ [-1,1,11,13,23,25] $
Dice 2: $ [2,3,6,7,10,11] $
Same number, different order is considered the same.
$$ [-1,1,11,13,23,25] [2,3,6,7,10,11] $$
$$ [1,11,13,23,25,-1] [2,3,6,7,10,11] $$
$$ [2,3,6,7,10,11] [-1,1,11,13,23,25] $$
They are all the same
So how many are there?
Is there a way to find it without trying one by one?
combinations puzzle
asked Jul 24 at 12:22
Pizzaroot
1056
4
Take your trivial solution Dice 1: [1,2,3,4,5,6] Dice 2: [0,6,12,18,24,30] and translate the first numbers by $n$ and the second numbers by $-n$, this yields infinitely many solutions.
– Arnaud Mortier
Jul 24 at 12:26
I think we can better define the problem by stating: "Dice 2 must include the number 0 and all other numbers on Dice 2 must be positive integers"
– Malkin
Jul 24 at 12:32
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am not talking about the shape of the dice here, I am talking about another type. You will see what I mean soon.
For example, when there are 1 dice, a normal dice is a fair dice, because the probability of getting each number is the same by $ frac16 $
When 2 normal dice are thrown, you can get numbers from 2 to 12, but the probability of getting each of them are different. Think of Monopoly, the probability of getting 7 is $ frac636 $ while the probability of getting 2 is $ frac136 $ the goal is to make the probability of getting each number the same.
I got an example:
Dice 1: $ [1,2,3,4,5,6] $
Dice 2: $ [0,6,12,18,24,30] $
When these two dice are thrown and add the two number, the possible outcome can be every number from 1 to 36 and the probability of each is
So the question is, How many different pairs of dice of this kind are possible so that the possible outcome can be every number from 1 to 36 and the probability of each is $ frac136 $
The number on the dice can be negative too.
Example:
Dice 1: $ [-1,1,11,13,23,25] $
Dice 2: $ [2,3,6,7,10,11] $
Same number, different order is considered the same.
$$ [-1,1,11,13,23,25] [2,3,6,7,10,11] $$
$$ [1,11,13,23,25,-1] [2,3,6,7,10,11] $$
$$ [2,3,6,7,10,11] [-1,1,11,13,23,25] $$
They are all the same
So how many are there?
Is there a way to find it without trying one by one?
combinations puzzle
asked Jul 24 at 12:22
Pizzaroot
1056
I am not talking about the shape of the dice here, I am talking about another type. You will see what I mean soon.
For example, when there are 1 dice, a normal dice is a fair dice, because the probability of getting each number is the same by $ frac16 $
When 2 normal dice are thrown, you can get numbers from 2 to 12, but the probability of getting each of them are different. Think of Monopoly, the probability of getting 7 is $ frac636 $ while the probability of getting 2 is $ frac136 $ the goal is to make the probability of getting each number the same.
I got an example:
Dice 1: $ [1,2,3,4,5,6] $
Dice 2: $ [0,6,12,18,24,30] $
When these two dice are thrown and add the two number, the possible outcome can be every number from 1 to 36 and the probability of each is
So the question is, How many different pairs of dice of this kind are possible so that the possible outcome can be every number from 1 to 36 and the probability of each is $ frac136 $
The number on the dice can be negative too.
Example:
Dice 1: $ [-1,1,11,13,23,25] $
Dice 2: $ [2,3,6,7,10,11] $
Same number, different order is considered the same.
$$ [-1,1,11,13,23,25] [2,3,6,7,10,11] $$
$$ [1,11,13,23,25,-1] [2,3,6,7,10,11] $$
$$ [2,3,6,7,10,11] [-1,1,11,13,23,25] $$
They are all the same
So how many are there?
Is there a way to find it without trying one by one?
combinations puzzle
asked Jul 24 at 12:22
Pizzaroot
1056
asked Jul 24 at 12:22
Pizzaroot
1056
asked Jul 24 at 12:22
Pizzaroot
1056
asked Jul 24 at 12:22
Pizzaroot
1056
1056
4
Take your trivial solution Dice 1: [1,2,3,4,5,6] Dice 2: [0,6,12,18,24,30] and translate the first numbers by $n$ and the second numbers by $-n$, this yields infinitely many solutions.
– Arnaud Mortier
Jul 24 at 12:26
I think we can better define the problem by stating: "Dice 2 must include the number 0 and all other numbers on Dice 2 must be positive integers"
– Malkin
Jul 24 at 12:32
add a comment |Â
4
Take your trivial solution Dice 1: [1,2,3,4,5,6] Dice 2: [0,6,12,18,24,30] and translate the first numbers by $n$ and the second numbers by $-n$, this yields infinitely many solutions.
– Arnaud Mortier
Jul 24 at 12:26
I think we can better define the problem by stating: "Dice 2 must include the number 0 and all other numbers on Dice 2 must be positive integers"
– Malkin
Jul 24 at 12:32
4
4
Take your trivial solution Dice 1: [1,2,3,4,5,6] Dice 2: [0,6,12,18,24,30] and translate the first numbers by $n$ and the second numbers by $-n$, this yields infinitely many solutions.
– Arnaud Mortier
Jul 24 at 12:26
Take your trivial solution Dice 1: [1,2,3,4,5,6] Dice 2: [0,6,12,18,24,30] and translate the first numbers by $n$ and the second numbers by $-n$, this yields infinitely many solutions.
– Arnaud Mortier
Jul 24 at 12:26
I think we can better define the problem by stating: "Dice 2 must include the number 0 and all other numbers on Dice 2 must be positive integers"
– Malkin
Jul 24 at 12:32
I think we can better define the problem by stating: "Dice 2 must include the number 0 and all other numbers on Dice 2 must be positive integers"
– Malkin
Jul 24 at 12:32
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
5
down vote
As stated in MigMit's answer, we need two polynomials, each having the coefficient $1$ for exactly six terms coefficient $0$ for all other terms,
whose product is $1 + x + x^2 + cdots + x^35.$
Note that
$$ x^36 - 1 = (x - 1)(x^35 + x^34 + cdots + x^3 + x^2 + x + 1),$$
and since the roots of $x^36 - 1$ over complex numbers are all the $36$th roots of unity, the roots of $x^35 + x^34 + cdots + x + 1$
are all the $36$th roots of unity except $1$ itself.
That is,
$$ x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= prod_n=1^17(x - e^ipi n/18).
$$
Multiplying together the various factors of $(x - e^ipi n/18)$
to obtain polynomials with integer coefficients,
beginmultline
x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= \
(x + 1)(x^2 + 1) (x^2 - x + 1)(x^2 + x + 1) (x^4 - x^2 + 1)\
(x^6 - x^3 + 1) (x^6 + x^3 + 1) (x^12 - x^6 + 1).
endmultline
To eliminate the negative coefficients we can multiply as follows:
beginalign
(x + 1)(x^2 - x + 1) &= x^3 + 1,\
(x^2 + 1)(x^4 - x^2 + 1) &= x^6 + 1,\
(x^2 - x + 1)(x^2 + x + 1) &= x^4 + x^2 + 1,\
(x^4 - x^2 + 1)(x^4 + x^2 + 1) &= x^8 + x^4 + 1,\
(x^6 - x^3 + 1)(x^6 + x^3 + 1) &= x^12 + x^6 + 1,\
(x^12 - x^6 + 1)(x^12 + x^6 + 1) &= x^24 + x^12 + 1,\
(x + 1)(x^2 - x + 1)(x^6 - x^3 + 1) &= x^9 + 1,\
(x^2 + 1)(x^4 - x^2 + 1)(x^12 - x^6 + 1) &= x^18 + 1.
endalign
From this we get the following factorizations in which all coefficients are $1$ or $0$:
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^2 + x + 1) (x^3 + 1) (x^6 + 1) (x^24 + x^12 + 1)\
&= (x + 1)(x^2 + 1) (x^8 + x^4 + 1) (x^24 + x^12 + 1)\
&= (x^2 + x + 1) (x^6 + x^3 + 1) (x^9 + 1)(x^18 + 1) \
&= (x + 1) (x^4 + x^2 + 1) (x^12 + x^6 + 1) (x^18 + 1)\
&= (x + 1)(x^4 + x^2 + 1) (x^6 + 1) (x^24 + x^12 + 1).
endalign
In each case we can multiply either polynomial with two non-zero terms by either polynomial with three non-zero terms to get a polynomial with six non-zero terms.
That is,
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^5 + x^4 + x^3 + x^2 + x + 1)
(x^30 + x^24 + x^18 + x^12 + x^6 + 1) \
&= (x^8 + x^7 + x^6 + x^2 + x + 1)
(x^27 + x^24 + x^15 + x^12 + x^3 + 1) \
&= (x^9 + x^8 + x^5 + x^4 + x + 1)
(x^26 + x^24 + x^14 + x^12 + x^2 + 1) \
&= (x^10 + x^8 + x^6 + x^4 + x^2 + 1)
(x^25 + x^24 + x^13 + x^12 + x + 1) \
&= (x^11 + x^10 + x^9 + x^2 + x + 1)
(x^24 + x^21 + x^18 + x^6 + x^3 + 1) \
&= (x^13 + x^12 + x^7 + x^6 + x + 1)
(x^22 + x^20 + x^18 + x^4 + x^2 + 1)\
&= (x^15 + x^12 + x^9 + x^6 + x^3 + 1)
(x^20 + x^19 + x^18 + x^2 + x + 1).
endalign
Although each of the five factorizations into four polynomials is capable of producing two factorizations into polynomials of six non-zero terms,
some of them are duplicates of each other, and only seven are distinct.
answered Jul 25 at 3:39
David K
48.2k340107
add a comment |Â
up vote
3
down vote
If you accept dice with negative numbers, then the answer is, obviously, infinity. You can add $N$ to all numbers on one dice, and subtract $N$ from all numbers on another.
If you require them to be all non-negative, then it boils down to factorizing the polynomial $1+X+X^2+ldots+X^35$ into the product of two polynomials of degree $6$ with all their coefficients being $0$ or $1$. Which, I guess, could be done by hand, but probably not worth it.
Correction: not of degree $6$, but having exactly $6$ non-zero (and hence equal to $1$) coefficients.
answered Jul 24 at 12:28
MigMit
27113
I think there is only one solution if we require non-negativity (and sum $le 36$). Indeed, we must reach all values between $1$ and $36$ in order to produce a probability of $frac 1 36$ for each of them.
– nicomezi
Jul 24 at 12:32
1
No. Example [2,3,6,7,10,11][−1,1,11,13,23,25] from the question can be adjusted so that it fits: [1,2,5,6,9,10],[0,2,12,14,24,26].
– MigMit
Jul 24 at 12:44
Indeed. Not as simple as I thought.
– nicomezi
Jul 24 at 12:46
add a comment |Â
up vote
3
down vote
Add $1$ to each number in one of the dice:
$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$
$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$
$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$
$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$
$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$
$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$
$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$
36 is the product of four prime numbers: 2, 2, 3, 3. Arrange them in any of six ways, for example $a=2, b=3,c =3, d=2$.
$a=2$ so I let $A=0,1$.
( If $a$ were 3 I would let $A=0,1,2$. )
$b=3$ so $B=0,a,2a=0,2,4$.
$c=3$ so $C=0,ab,2ab=0,6,12$.
$d=2$ so $D=0,abc=0,18$
Form one die from $A+B$ and the other from $C+D$; or else $A+C$ and $B+D$.
Here, $A+C=0,1,6,7,12,13$ and $B+D=0,2,4,18,20,22$.
Finally add 1 to either die so they go from 1 to 36 instead of 0 to 35.
answered Jul 24 at 13:08
Empy2
31.8k12059
This is simpler and more reliable than the method I used. (I noticed errors in my method when I saw you had found dice I had not yet found.) So this seems to me the best complete answer so far.
– David K
Jul 25 at 3:40
These solutions are beautiful. Thank you. How do we know that this solution is complete? Might there exist sets $R,Ssubset0,...,35$ such that $R+S=0,...,35$ but which cannot be expressed as $k_1mathbbZ_2+k_2mathbbZ_3$?
– Malkin
Jul 28 at 16:35
add a comment |Â
up vote
2
down vote
Let $n$ be any integer.
Dice 1: $left[matrix1+n\ 2+n\ 3+n\ 4+n\ 5+n\ 6+nright]$
$qquad $Dice 2: $left[matrix0-n\ 6-n\ 12-n\ 18-n\ 24-n\ 30-nright]$
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
As stated in MigMit's answer, we need two polynomials, each having the coefficient $1$ for exactly six terms coefficient $0$ for all other terms,
whose product is $1 + x + x^2 + cdots + x^35.$
Note that
$$ x^36 - 1 = (x - 1)(x^35 + x^34 + cdots + x^3 + x^2 + x + 1),$$
and since the roots of $x^36 - 1$ over complex numbers are all the $36$th roots of unity, the roots of $x^35 + x^34 + cdots + x + 1$
are all the $36$th roots of unity except $1$ itself.
That is,
$$ x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= prod_n=1^17(x - e^ipi n/18).
$$
Multiplying together the various factors of $(x - e^ipi n/18)$
to obtain polynomials with integer coefficients,
beginmultline
x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= \
(x + 1)(x^2 + 1) (x^2 - x + 1)(x^2 + x + 1) (x^4 - x^2 + 1)\
(x^6 - x^3 + 1) (x^6 + x^3 + 1) (x^12 - x^6 + 1).
endmultline
To eliminate the negative coefficients we can multiply as follows:
beginalign
(x + 1)(x^2 - x + 1) &= x^3 + 1,\
(x^2 + 1)(x^4 - x^2 + 1) &= x^6 + 1,\
(x^2 - x + 1)(x^2 + x + 1) &= x^4 + x^2 + 1,\
(x^4 - x^2 + 1)(x^4 + x^2 + 1) &= x^8 + x^4 + 1,\
(x^6 - x^3 + 1)(x^6 + x^3 + 1) &= x^12 + x^6 + 1,\
(x^12 - x^6 + 1)(x^12 + x^6 + 1) &= x^24 + x^12 + 1,\
(x + 1)(x^2 - x + 1)(x^6 - x^3 + 1) &= x^9 + 1,\
(x^2 + 1)(x^4 - x^2 + 1)(x^12 - x^6 + 1) &= x^18 + 1.
endalign
From this we get the following factorizations in which all coefficients are $1$ or $0$:
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^2 + x + 1) (x^3 + 1) (x^6 + 1) (x^24 + x^12 + 1)\
&= (x + 1)(x^2 + 1) (x^8 + x^4 + 1) (x^24 + x^12 + 1)\
&= (x^2 + x + 1) (x^6 + x^3 + 1) (x^9 + 1)(x^18 + 1) \
&= (x + 1) (x^4 + x^2 + 1) (x^12 + x^6 + 1) (x^18 + 1)\
&= (x + 1)(x^4 + x^2 + 1) (x^6 + 1) (x^24 + x^12 + 1).
endalign
In each case we can multiply either polynomial with two non-zero terms by either polynomial with three non-zero terms to get a polynomial with six non-zero terms.
That is,
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^5 + x^4 + x^3 + x^2 + x + 1)
(x^30 + x^24 + x^18 + x^12 + x^6 + 1) \
&= (x^8 + x^7 + x^6 + x^2 + x + 1)
(x^27 + x^24 + x^15 + x^12 + x^3 + 1) \
&= (x^9 + x^8 + x^5 + x^4 + x + 1)
(x^26 + x^24 + x^14 + x^12 + x^2 + 1) \
&= (x^10 + x^8 + x^6 + x^4 + x^2 + 1)
(x^25 + x^24 + x^13 + x^12 + x + 1) \
&= (x^11 + x^10 + x^9 + x^2 + x + 1)
(x^24 + x^21 + x^18 + x^6 + x^3 + 1) \
&= (x^13 + x^12 + x^7 + x^6 + x + 1)
(x^22 + x^20 + x^18 + x^4 + x^2 + 1)\
&= (x^15 + x^12 + x^9 + x^6 + x^3 + 1)
(x^20 + x^19 + x^18 + x^2 + x + 1).
endalign
Although each of the five factorizations into four polynomials is capable of producing two factorizations into polynomials of six non-zero terms,
some of them are duplicates of each other, and only seven are distinct.
answered Jul 25 at 3:39
David K
48.2k340107
add a comment |Â
up vote
5
down vote
As stated in MigMit's answer, we need two polynomials, each having the coefficient $1$ for exactly six terms coefficient $0$ for all other terms,
whose product is $1 + x + x^2 + cdots + x^35.$
Note that
$$ x^36 - 1 = (x - 1)(x^35 + x^34 + cdots + x^3 + x^2 + x + 1),$$
and since the roots of $x^36 - 1$ over complex numbers are all the $36$th roots of unity, the roots of $x^35 + x^34 + cdots + x + 1$
are all the $36$th roots of unity except $1$ itself.
That is,
$$ x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= prod_n=1^17(x - e^ipi n/18).
$$
Multiplying together the various factors of $(x - e^ipi n/18)$
to obtain polynomials with integer coefficients,
beginmultline
x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= \
(x + 1)(x^2 + 1) (x^2 - x + 1)(x^2 + x + 1) (x^4 - x^2 + 1)\
(x^6 - x^3 + 1) (x^6 + x^3 + 1) (x^12 - x^6 + 1).
endmultline
To eliminate the negative coefficients we can multiply as follows:
beginalign
(x + 1)(x^2 - x + 1) &= x^3 + 1,\
(x^2 + 1)(x^4 - x^2 + 1) &= x^6 + 1,\
(x^2 - x + 1)(x^2 + x + 1) &= x^4 + x^2 + 1,\
(x^4 - x^2 + 1)(x^4 + x^2 + 1) &= x^8 + x^4 + 1,\
(x^6 - x^3 + 1)(x^6 + x^3 + 1) &= x^12 + x^6 + 1,\
(x^12 - x^6 + 1)(x^12 + x^6 + 1) &= x^24 + x^12 + 1,\
(x + 1)(x^2 - x + 1)(x^6 - x^3 + 1) &= x^9 + 1,\
(x^2 + 1)(x^4 - x^2 + 1)(x^12 - x^6 + 1) &= x^18 + 1.
endalign
From this we get the following factorizations in which all coefficients are $1$ or $0$:
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^2 + x + 1) (x^3 + 1) (x^6 + 1) (x^24 + x^12 + 1)\
&= (x + 1)(x^2 + 1) (x^8 + x^4 + 1) (x^24 + x^12 + 1)\
&= (x^2 + x + 1) (x^6 + x^3 + 1) (x^9 + 1)(x^18 + 1) \
&= (x + 1) (x^4 + x^2 + 1) (x^12 + x^6 + 1) (x^18 + 1)\
&= (x + 1)(x^4 + x^2 + 1) (x^6 + 1) (x^24 + x^12 + 1).
endalign
In each case we can multiply either polynomial with two non-zero terms by either polynomial with three non-zero terms to get a polynomial with six non-zero terms.
That is,
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^5 + x^4 + x^3 + x^2 + x + 1)
(x^30 + x^24 + x^18 + x^12 + x^6 + 1) \
&= (x^8 + x^7 + x^6 + x^2 + x + 1)
(x^27 + x^24 + x^15 + x^12 + x^3 + 1) \
&= (x^9 + x^8 + x^5 + x^4 + x + 1)
(x^26 + x^24 + x^14 + x^12 + x^2 + 1) \
&= (x^10 + x^8 + x^6 + x^4 + x^2 + 1)
(x^25 + x^24 + x^13 + x^12 + x + 1) \
&= (x^11 + x^10 + x^9 + x^2 + x + 1)
(x^24 + x^21 + x^18 + x^6 + x^3 + 1) \
&= (x^13 + x^12 + x^7 + x^6 + x + 1)
(x^22 + x^20 + x^18 + x^4 + x^2 + 1)\
&= (x^15 + x^12 + x^9 + x^6 + x^3 + 1)
(x^20 + x^19 + x^18 + x^2 + x + 1).
endalign
Although each of the five factorizations into four polynomials is capable of producing two factorizations into polynomials of six non-zero terms,
some of them are duplicates of each other, and only seven are distinct.
answered Jul 25 at 3:39
David K
48.2k340107
add a comment |Â
up vote
5
down vote
up vote
5
down vote
As stated in MigMit's answer, we need two polynomials, each having the coefficient $1$ for exactly six terms coefficient $0$ for all other terms,
whose product is $1 + x + x^2 + cdots + x^35.$
Note that
$$ x^36 - 1 = (x - 1)(x^35 + x^34 + cdots + x^3 + x^2 + x + 1),$$
and since the roots of $x^36 - 1$ over complex numbers are all the $36$th roots of unity, the roots of $x^35 + x^34 + cdots + x + 1$
are all the $36$th roots of unity except $1$ itself.
That is,
$$ x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= prod_n=1^17(x - e^ipi n/18).
$$
Multiplying together the various factors of $(x - e^ipi n/18)$
to obtain polynomials with integer coefficients,
beginmultline
x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= \
(x + 1)(x^2 + 1) (x^2 - x + 1)(x^2 + x + 1) (x^4 - x^2 + 1)\
(x^6 - x^3 + 1) (x^6 + x^3 + 1) (x^12 - x^6 + 1).
endmultline
To eliminate the negative coefficients we can multiply as follows:
beginalign
(x + 1)(x^2 - x + 1) &= x^3 + 1,\
(x^2 + 1)(x^4 - x^2 + 1) &= x^6 + 1,\
(x^2 - x + 1)(x^2 + x + 1) &= x^4 + x^2 + 1,\
(x^4 - x^2 + 1)(x^4 + x^2 + 1) &= x^8 + x^4 + 1,\
(x^6 - x^3 + 1)(x^6 + x^3 + 1) &= x^12 + x^6 + 1,\
(x^12 - x^6 + 1)(x^12 + x^6 + 1) &= x^24 + x^12 + 1,\
(x + 1)(x^2 - x + 1)(x^6 - x^3 + 1) &= x^9 + 1,\
(x^2 + 1)(x^4 - x^2 + 1)(x^12 - x^6 + 1) &= x^18 + 1.
endalign
From this we get the following factorizations in which all coefficients are $1$ or $0$:
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^2 + x + 1) (x^3 + 1) (x^6 + 1) (x^24 + x^12 + 1)\
&= (x + 1)(x^2 + 1) (x^8 + x^4 + 1) (x^24 + x^12 + 1)\
&= (x^2 + x + 1) (x^6 + x^3 + 1) (x^9 + 1)(x^18 + 1) \
&= (x + 1) (x^4 + x^2 + 1) (x^12 + x^6 + 1) (x^18 + 1)\
&= (x + 1)(x^4 + x^2 + 1) (x^6 + 1) (x^24 + x^12 + 1).
endalign
In each case we can multiply either polynomial with two non-zero terms by either polynomial with three non-zero terms to get a polynomial with six non-zero terms.
That is,
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^5 + x^4 + x^3 + x^2 + x + 1)
(x^30 + x^24 + x^18 + x^12 + x^6 + 1) \
&= (x^8 + x^7 + x^6 + x^2 + x + 1)
(x^27 + x^24 + x^15 + x^12 + x^3 + 1) \
&= (x^9 + x^8 + x^5 + x^4 + x + 1)
(x^26 + x^24 + x^14 + x^12 + x^2 + 1) \
&= (x^10 + x^8 + x^6 + x^4 + x^2 + 1)
(x^25 + x^24 + x^13 + x^12 + x + 1) \
&= (x^11 + x^10 + x^9 + x^2 + x + 1)
(x^24 + x^21 + x^18 + x^6 + x^3 + 1) \
&= (x^13 + x^12 + x^7 + x^6 + x + 1)
(x^22 + x^20 + x^18 + x^4 + x^2 + 1)\
&= (x^15 + x^12 + x^9 + x^6 + x^3 + 1)
(x^20 + x^19 + x^18 + x^2 + x + 1).
endalign
Although each of the five factorizations into four polynomials is capable of producing two factorizations into polynomials of six non-zero terms,
some of them are duplicates of each other, and only seven are distinct.
answered Jul 25 at 3:39
David K
48.2k340107
As stated in MigMit's answer, we need two polynomials, each having the coefficient $1$ for exactly six terms coefficient $0$ for all other terms,
whose product is $1 + x + x^2 + cdots + x^35.$
Note that
$$ x^36 - 1 = (x - 1)(x^35 + x^34 + cdots + x^3 + x^2 + x + 1),$$
and since the roots of $x^36 - 1$ over complex numbers are all the $36$th roots of unity, the roots of $x^35 + x^34 + cdots + x + 1$
are all the $36$th roots of unity except $1$ itself.
That is,
$$ x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= prod_n=1^17(x - e^ipi n/18).
$$
Multiplying together the various factors of $(x - e^ipi n/18)$
to obtain polynomials with integer coefficients,
beginmultline
x^35 + x^34 + cdots + x^3 + x^2 + x + 1
= \
(x + 1)(x^2 + 1) (x^2 - x + 1)(x^2 + x + 1) (x^4 - x^2 + 1)\
(x^6 - x^3 + 1) (x^6 + x^3 + 1) (x^12 - x^6 + 1).
endmultline
To eliminate the negative coefficients we can multiply as follows:
beginalign
(x + 1)(x^2 - x + 1) &= x^3 + 1,\
(x^2 + 1)(x^4 - x^2 + 1) &= x^6 + 1,\
(x^2 - x + 1)(x^2 + x + 1) &= x^4 + x^2 + 1,\
(x^4 - x^2 + 1)(x^4 + x^2 + 1) &= x^8 + x^4 + 1,\
(x^6 - x^3 + 1)(x^6 + x^3 + 1) &= x^12 + x^6 + 1,\
(x^12 - x^6 + 1)(x^12 + x^6 + 1) &= x^24 + x^12 + 1,\
(x + 1)(x^2 - x + 1)(x^6 - x^3 + 1) &= x^9 + 1,\
(x^2 + 1)(x^4 - x^2 + 1)(x^12 - x^6 + 1) &= x^18 + 1.
endalign
From this we get the following factorizations in which all coefficients are $1$ or $0$:
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^2 + x + 1) (x^3 + 1) (x^6 + 1) (x^24 + x^12 + 1)\
&= (x + 1)(x^2 + 1) (x^8 + x^4 + 1) (x^24 + x^12 + 1)\
&= (x^2 + x + 1) (x^6 + x^3 + 1) (x^9 + 1)(x^18 + 1) \
&= (x + 1) (x^4 + x^2 + 1) (x^12 + x^6 + 1) (x^18 + 1)\
&= (x + 1)(x^4 + x^2 + 1) (x^6 + 1) (x^24 + x^12 + 1).
endalign
In each case we can multiply either polynomial with two non-zero terms by either polynomial with three non-zero terms to get a polynomial with six non-zero terms.
That is,
beginalign
x^35 + x^34 + cdots + x + 1
&= (x^5 + x^4 + x^3 + x^2 + x + 1)
(x^30 + x^24 + x^18 + x^12 + x^6 + 1) \
&= (x^8 + x^7 + x^6 + x^2 + x + 1)
(x^27 + x^24 + x^15 + x^12 + x^3 + 1) \
&= (x^9 + x^8 + x^5 + x^4 + x + 1)
(x^26 + x^24 + x^14 + x^12 + x^2 + 1) \
&= (x^10 + x^8 + x^6 + x^4 + x^2 + 1)
(x^25 + x^24 + x^13 + x^12 + x + 1) \
&= (x^11 + x^10 + x^9 + x^2 + x + 1)
(x^24 + x^21 + x^18 + x^6 + x^3 + 1) \
&= (x^13 + x^12 + x^7 + x^6 + x + 1)
(x^22 + x^20 + x^18 + x^4 + x^2 + 1)\
&= (x^15 + x^12 + x^9 + x^6 + x^3 + 1)
(x^20 + x^19 + x^18 + x^2 + x + 1).
endalign
Although each of the five factorizations into four polynomials is capable of producing two factorizations into polynomials of six non-zero terms,
some of them are duplicates of each other, and only seven are distinct.
answered Jul 25 at 3:39
David K
48.2k340107
answered Jul 25 at 3:39
David K
48.2k340107
answered Jul 25 at 3:39
David K
48.2k340107
answered Jul 25 at 3:39
David K
48.2k340107
48.2k340107
add a comment |Â
add a comment |Â
up vote
3
down vote
If you accept dice with negative numbers, then the answer is, obviously, infinity. You can add $N$ to all numbers on one dice, and subtract $N$ from all numbers on another.
If you require them to be all non-negative, then it boils down to factorizing the polynomial $1+X+X^2+ldots+X^35$ into the product of two polynomials of degree $6$ with all their coefficients being $0$ or $1$. Which, I guess, could be done by hand, but probably not worth it.
Correction: not of degree $6$, but having exactly $6$ non-zero (and hence equal to $1$) coefficients.
answered Jul 24 at 12:28
MigMit
27113
I think there is only one solution if we require non-negativity (and sum $le 36$). Indeed, we must reach all values between $1$ and $36$ in order to produce a probability of $frac 1 36$ for each of them.
– nicomezi
Jul 24 at 12:32
1
No. Example [2,3,6,7,10,11][−1,1,11,13,23,25] from the question can be adjusted so that it fits: [1,2,5,6,9,10],[0,2,12,14,24,26].
– MigMit
Jul 24 at 12:44
Indeed. Not as simple as I thought.
– nicomezi
Jul 24 at 12:46
add a comment |Â
up vote
3
down vote
If you accept dice with negative numbers, then the answer is, obviously, infinity. You can add $N$ to all numbers on one dice, and subtract $N$ from all numbers on another.
If you require them to be all non-negative, then it boils down to factorizing the polynomial $1+X+X^2+ldots+X^35$ into the product of two polynomials of degree $6$ with all their coefficients being $0$ or $1$. Which, I guess, could be done by hand, but probably not worth it.
Correction: not of degree $6$, but having exactly $6$ non-zero (and hence equal to $1$) coefficients.
answered Jul 24 at 12:28
MigMit
27113
I think there is only one solution if we require non-negativity (and sum $le 36$). Indeed, we must reach all values between $1$ and $36$ in order to produce a probability of $frac 1 36$ for each of them.
– nicomezi
Jul 24 at 12:32
1
No. Example [2,3,6,7,10,11][−1,1,11,13,23,25] from the question can be adjusted so that it fits: [1,2,5,6,9,10],[0,2,12,14,24,26].
– MigMit
Jul 24 at 12:44
Indeed. Not as simple as I thought.
– nicomezi
Jul 24 at 12:46
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If you accept dice with negative numbers, then the answer is, obviously, infinity. You can add $N$ to all numbers on one dice, and subtract $N$ from all numbers on another.
If you require them to be all non-negative, then it boils down to factorizing the polynomial $1+X+X^2+ldots+X^35$ into the product of two polynomials of degree $6$ with all their coefficients being $0$ or $1$. Which, I guess, could be done by hand, but probably not worth it.
Correction: not of degree $6$, but having exactly $6$ non-zero (and hence equal to $1$) coefficients.
answered Jul 24 at 12:28
MigMit
27113
If you accept dice with negative numbers, then the answer is, obviously, infinity. You can add $N$ to all numbers on one dice, and subtract $N$ from all numbers on another.
If you require them to be all non-negative, then it boils down to factorizing the polynomial $1+X+X^2+ldots+X^35$ into the product of two polynomials of degree $6$ with all their coefficients being $0$ or $1$. Which, I guess, could be done by hand, but probably not worth it.
Correction: not of degree $6$, but having exactly $6$ non-zero (and hence equal to $1$) coefficients.
answered Jul 24 at 12:28
MigMit
27113
edited Jul 24 at 13:04
answered Jul 24 at 12:28
MigMit
27113
answered Jul 24 at 12:28
MigMit
27113
answered Jul 24 at 12:28
MigMit
27113
27113
I think there is only one solution if we require non-negativity (and sum $le 36$). Indeed, we must reach all values between $1$ and $36$ in order to produce a probability of $frac 1 36$ for each of them.
– nicomezi
Jul 24 at 12:32
1
No. Example [2,3,6,7,10,11][−1,1,11,13,23,25] from the question can be adjusted so that it fits: [1,2,5,6,9,10],[0,2,12,14,24,26].
– MigMit
Jul 24 at 12:44
Indeed. Not as simple as I thought.
– nicomezi
Jul 24 at 12:46
add a comment |Â
I think there is only one solution if we require non-negativity (and sum $le 36$). Indeed, we must reach all values between $1$ and $36$ in order to produce a probability of $frac 1 36$ for each of them.
– nicomezi
Jul 24 at 12:32
1
No. Example [2,3,6,7,10,11][−1,1,11,13,23,25] from the question can be adjusted so that it fits: [1,2,5,6,9,10],[0,2,12,14,24,26].
– MigMit
Jul 24 at 12:44
Indeed. Not as simple as I thought.
– nicomezi
Jul 24 at 12:46
I think there is only one solution if we require non-negativity (and sum $le 36$). Indeed, we must reach all values between $1$ and $36$ in order to produce a probability of $frac 1 36$ for each of them.
– nicomezi
Jul 24 at 12:32
I think there is only one solution if we require non-negativity (and sum $le 36$). Indeed, we must reach all values between $1$ and $36$ in order to produce a probability of $frac 1 36$ for each of them.
– nicomezi
Jul 24 at 12:32
1
1
No. Example [2,3,6,7,10,11][−1,1,11,13,23,25] from the question can be adjusted so that it fits: [1,2,5,6,9,10],[0,2,12,14,24,26].
– MigMit
Jul 24 at 12:44
No. Example [2,3,6,7,10,11][−1,1,11,13,23,25] from the question can be adjusted so that it fits: [1,2,5,6,9,10],[0,2,12,14,24,26].
– MigMit
Jul 24 at 12:44
Indeed. Not as simple as I thought.
– nicomezi
Jul 24 at 12:46
Indeed. Not as simple as I thought.
– nicomezi
Jul 24 at 12:46
add a comment |Â
up vote
3
down vote
Add $1$ to each number in one of the dice:
$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$
$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$
$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$
$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$
$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$
$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$
$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$
36 is the product of four prime numbers: 2, 2, 3, 3. Arrange them in any of six ways, for example $a=2, b=3,c =3, d=2$.
$a=2$ so I let $A=0,1$.
( If $a$ were 3 I would let $A=0,1,2$. )
$b=3$ so $B=0,a,2a=0,2,4$.
$c=3$ so $C=0,ab,2ab=0,6,12$.
$d=2$ so $D=0,abc=0,18$
Form one die from $A+B$ and the other from $C+D$; or else $A+C$ and $B+D$.
Here, $A+C=0,1,6,7,12,13$ and $B+D=0,2,4,18,20,22$.
Finally add 1 to either die so they go from 1 to 36 instead of 0 to 35.
answered Jul 24 at 13:08
Empy2
31.8k12059
This is simpler and more reliable than the method I used. (I noticed errors in my method when I saw you had found dice I had not yet found.) So this seems to me the best complete answer so far.
– David K
Jul 25 at 3:40
These solutions are beautiful. Thank you. How do we know that this solution is complete? Might there exist sets $R,Ssubset0,...,35$ such that $R+S=0,...,35$ but which cannot be expressed as $k_1mathbbZ_2+k_2mathbbZ_3$?
– Malkin
Jul 28 at 16:35
add a comment |Â
up vote
3
down vote
Add $1$ to each number in one of the dice:
$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$
$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$
$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$
$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$
$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$
$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$
$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$
36 is the product of four prime numbers: 2, 2, 3, 3. Arrange them in any of six ways, for example $a=2, b=3,c =3, d=2$.
$a=2$ so I let $A=0,1$.
( If $a$ were 3 I would let $A=0,1,2$. )
$b=3$ so $B=0,a,2a=0,2,4$.
$c=3$ so $C=0,ab,2ab=0,6,12$.
$d=2$ so $D=0,abc=0,18$
Form one die from $A+B$ and the other from $C+D$; or else $A+C$ and $B+D$.
Here, $A+C=0,1,6,7,12,13$ and $B+D=0,2,4,18,20,22$.
Finally add 1 to either die so they go from 1 to 36 instead of 0 to 35.
answered Jul 24 at 13:08
Empy2
31.8k12059
This is simpler and more reliable than the method I used. (I noticed errors in my method when I saw you had found dice I had not yet found.) So this seems to me the best complete answer so far.
– David K
Jul 25 at 3:40
These solutions are beautiful. Thank you. How do we know that this solution is complete? Might there exist sets $R,Ssubset0,...,35$ such that $R+S=0,...,35$ but which cannot be expressed as $k_1mathbbZ_2+k_2mathbbZ_3$?
– Malkin
Jul 28 at 16:35
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Add $1$ to each number in one of the dice:
$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$
$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$
$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$
$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$
$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$
$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$
$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$
36 is the product of four prime numbers: 2, 2, 3, 3. Arrange them in any of six ways, for example $a=2, b=3,c =3, d=2$.
$a=2$ so I let $A=0,1$.
( If $a$ were 3 I would let $A=0,1,2$. )
$b=3$ so $B=0,a,2a=0,2,4$.
$c=3$ so $C=0,ab,2ab=0,6,12$.
$d=2$ so $D=0,abc=0,18$
Form one die from $A+B$ and the other from $C+D$; or else $A+C$ and $B+D$.
Here, $A+C=0,1,6,7,12,13$ and $B+D=0,2,4,18,20,22$.
Finally add 1 to either die so they go from 1 to 36 instead of 0 to 35.
answered Jul 24 at 13:08
Empy2
31.8k12059
Add $1$ to each number in one of the dice:
$A: (0,1,2,3,4,5),(0,6,12,18,24,30)$
$B: (0,1,2,6,7,8),(0,3,12,15,24,27)$
$C: (0,1;2,9,10,11),(0,3,6,18,21,24)$
$D: (0,1,2,18,19,20),(0,3,6,9,12,15)$
$E: (0,1,4,5,8,9),(0,2,12,14,24,26)$
$F: (0,1,6,7,12,13),(0,2,4,18,20,22)$
$G: (0,1,12,13,24,25),(0,2,4,6,8,10)$
36 is the product of four prime numbers: 2, 2, 3, 3. Arrange them in any of six ways, for example $a=2, b=3,c =3, d=2$.
$a=2$ so I let $A=0,1$.
( If $a$ were 3 I would let $A=0,1,2$. )
$b=3$ so $B=0,a,2a=0,2,4$.
$c=3$ so $C=0,ab,2ab=0,6,12$.
$d=2$ so $D=0,abc=0,18$
Form one die from $A+B$ and the other from $C+D$; or else $A+C$ and $B+D$.
Here, $A+C=0,1,6,7,12,13$ and $B+D=0,2,4,18,20,22$.
Finally add 1 to either die so they go from 1 to 36 instead of 0 to 35.
answered Jul 24 at 13:08
Empy2
31.8k12059
edited Jul 24 at 21:00
answered Jul 24 at 13:08
Empy2
31.8k12059
answered Jul 24 at 13:08
Empy2
31.8k12059
answered Jul 24 at 13:08
Empy2
31.8k12059
31.8k12059
This is simpler and more reliable than the method I used. (I noticed errors in my method when I saw you had found dice I had not yet found.) So this seems to me the best complete answer so far.
– David K
Jul 25 at 3:40
These solutions are beautiful. Thank you. How do we know that this solution is complete? Might there exist sets $R,Ssubset0,...,35$ such that $R+S=0,...,35$ but which cannot be expressed as $k_1mathbbZ_2+k_2mathbbZ_3$?
– Malkin
Jul 28 at 16:35
add a comment |Â
This is simpler and more reliable than the method I used. (I noticed errors in my method when I saw you had found dice I had not yet found.) So this seems to me the best complete answer so far.
– David K
Jul 25 at 3:40
These solutions are beautiful. Thank you. How do we know that this solution is complete? Might there exist sets $R,Ssubset0,...,35$ such that $R+S=0,...,35$ but which cannot be expressed as $k_1mathbbZ_2+k_2mathbbZ_3$?
– Malkin
Jul 28 at 16:35
This is simpler and more reliable than the method I used. (I noticed errors in my method when I saw you had found dice I had not yet found.) So this seems to me the best complete answer so far.
– David K
Jul 25 at 3:40
This is simpler and more reliable than the method I used. (I noticed errors in my method when I saw you had found dice I had not yet found.) So this seems to me the best complete answer so far.
– David K
Jul 25 at 3:40
These solutions are beautiful. Thank you. How do we know that this solution is complete? Might there exist sets $R,Ssubset0,...,35$ such that $R+S=0,...,35$ but which cannot be expressed as $k_1mathbbZ_2+k_2mathbbZ_3$?
– Malkin
Jul 28 at 16:35
These solutions are beautiful. Thank you. How do we know that this solution is complete? Might there exist sets $R,Ssubset0,...,35$ such that $R+S=0,...,35$ but which cannot be expressed as $k_1mathbbZ_2+k_2mathbbZ_3$?
– Malkin
Jul 28 at 16:35
add a comment |Â
up vote
2
down vote
Let $n$ be any integer.
Dice 1: $left[matrix1+n\ 2+n\ 3+n\ 4+n\ 5+n\ 6+nright]$
$qquad $Dice 2: $left[matrix0-n\ 6-n\ 12-n\ 18-n\ 24-n\ 30-nright]$
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
add a comment |Â
up vote
2
down vote
Let $n$ be any integer.
Dice 1: $left[matrix1+n\ 2+n\ 3+n\ 4+n\ 5+n\ 6+nright]$
$qquad $Dice 2: $left[matrix0-n\ 6-n\ 12-n\ 18-n\ 24-n\ 30-nright]$
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $n$ be any integer.
Dice 1: $left[matrix1+n\ 2+n\ 3+n\ 4+n\ 5+n\ 6+nright]$
$qquad $Dice 2: $left[matrix0-n\ 6-n\ 12-n\ 18-n\ 24-n\ 30-nright]$
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
Let $n$ be any integer.
Dice 1: $left[matrix1+n\ 2+n\ 3+n\ 4+n\ 5+n\ 6+nright]$
$qquad $Dice 2: $left[matrix0-n\ 6-n\ 12-n\ 18-n\ 24-n\ 30-nright]$
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
answered Jul 24 at 12:30
Arnaud Mortier
18.9k22159
18.9k22159
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861275%2fhow-many-fair-dice-of-this-kind-exist%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
4
Take your trivial solution Dice 1: [1,2,3,4,5,6] Dice 2: [0,6,12,18,24,30] and translate the first numbers by $n$ and the second numbers by $-n$, this yields infinitely many solutions.
– Arnaud Mortier
Jul 24 at 12:26
I think we can better define the problem by stating: "Dice 2 must include the number 0 and all other numbers on Dice 2 must be positive integers"
– Malkin
Jul 24 at 12:32