Paradise anyone?

[bonanova]

The faces of two cubes are inscribed with single-digit numbers.

The cubes (let's call them dice) are rolled repeatedly and the sums of their top faces are recorded.

It is found empirically that

p(2) = p(12) = 1/36

p(3) = p(11) = 2/36

p(4) = p(10) = 3/36

p(5) = p( 9) = 4/36

p(6) = p( 8) = 5/36

p(7) =         6/36

What can we say about the numbers inscribed on the dice?

[gavinksong]

they're standard dice?

[gavinksong]

One of the die has the numbers 0-5 and the other has the numbers 2-7.

[karthickgururaj]

Without solving the problem, we could also do the following..

 

Let a0, a1, ..., a5 and b0, b1, .., b5 be 12 single digit numbers, such that:

  (x^a0 + x^a1 + .. + x^a5).(x^b0 + x^b1 + .. + x^b5) = (x² + 2.x³ + 3.x⁴ + 4.x⁵ + 5.x⁶ + 6.x⁷ + 5.x⁸ + 4.x⁹ + 3.x¹⁰ + 2.x¹¹ + x¹²)

 

What are the numbers then?

[witzar]

There are five solutions:

a=[0, 1, 1, 2, 2, 3] b=[2, 4, 5, 6, 7, 9]
a=[0, 1, 2, 3, 4, 5] b=[2, 3, 4, 5, 6, 7]
a=[0, 2, 3, 4, 5, 7] b=[2, 3, 3, 4, 4, 5]
a=[1, 2, 2, 3, 3, 4] b=[1, 3, 4, 5, 6, 8]
a=[1, 2, 3, 4, 5, 6] b=[1, 2, 3, 4, 5, 6]

[karthickgururaj]

Without solving the problem, we could also do the following..

 

Let a0, a1, ..., a5 and b0, b1, .., b5 be 12 single digit numbers, such that:

  (x^a0 + x^a1 + .. + x^a5).(x^b0 + x^b1 + .. + x^b5) = (x² + 2.x³ + 3.x⁴ + 4.x⁵ + 5.x⁶ + 6.x⁷ + 5.x⁸ + 4.x⁹ + 3.x¹⁰ + 2.x¹¹ + x¹²)

 

What are the numbers then?

  

..can be written as:

  x².(1 + 2.x + 3.x² + 4.x³ + 5.x⁴ + 6.x⁵ + 5.x⁶ + 4.x⁷ + 3.x⁸ + 2.x⁹ + x¹⁰) 

 

which is nothing but: x².(1 + x + x² + x³ + x⁴ + x⁵)²

 

(It took me sometime to realize that. If I instead had, (1 + 2.x + 3.x² + 4.x³ + ... + 9.x¹⁰), I could have simply integrated the terms.).

 

I couldn't factorize 1 + x + x² + x³ + x⁴ + x⁵ further, cheated a bit here and asked wolfram alpha. So the final factorized expression is:

  x².(x+1)².(x²-x+1)².(x²+x+1)²

 

This has got four terms: A(x) = x, B(x) = (x+1), C(x) = (x²-x+1), D(x) = (x²+x+1) (of degree = 1, 1, 2 and 2 respectively).

 

We are interested in factorizing the RHS into two polynomials of degree 6.

 

Further, we must have 6 "terms" in each polynomial. For instance, (1+x⁶) is a sixth degree polynomial with only two terms - so that is not a solution we are interested in. (1+5.x⁶) on the other hand is ok, since it has 6 terms (5.x⁶ is counted as 5 terms). Each term would then correspond to one face of the dice. One way to check this is to evaluate the polynomial with x = 1. The result must be 6. The values of the various terms when x = 1 evaluate to: A(1) = 1, B(1) = 2, C(1) = 1, D(1) = 3.

 

So we need to chose a combination of terms A(x), B(x), C(x) and D(x) to make two polynomials, p(x) and q(x) such that:

a. The terms are multiplied. Each term must be used exactly twice.

b. The degree of each polynomial is 6

c. p(1) = q(1) = 6.

 

For satisfying ©, both B(x) and D(x) must appear exactly once in p(x) and q(x) - this narrows down the solution possibilities considerably. The degree of B(x).D(x) is 3.

 

 

I saw witzar has already posted a solution which looks complete, so I didn't workout the last step..

 

@witzar: what was your approach? Something similar?

[witzar]

@witzar: what was your approach?

Just brute force.

[karthickgururaj]

@witzar: what was your approach?

Just brute force.

Ah, ok. Then it may be of some value for me to continue with my approach to closure, if only to check whether I get the same results..

 

 

I found one mistake in #6: the degree of the polynomials need not be 6. It can be anything less than 9 (since the degree corresponds to the number on a face of the dice, which is a single digit number). So the only requirements are:

 

a. The terms are multiplied. Each term must be used exactly twice.

b. p(1) = q(1) = 6.

 

The possible combinations (A(x) is denoted as 'a', B(x) as 'b', etc) will be like: (something . B(x) . D(x)), where 'something' is made of A(x) and C(x) and degree of 'something' must be 3.

 

So we have,

p = b.d; q = a.a.c.c.b.d

p = (x+1)*(x²+x+1) = x³+x²+x²+x+x+1

q = x*x*(x²-x+1)*(x²-x+1)*(x+1)*(x²+x+1) = x⁹+x⁷+x⁶+x⁵+x⁴+x²

Corresponds to the solution: { {3, 2, 2, 1, 1, 0}, {9, 7, 6, 5, 4, 2} }

 

p = a.b.d; q = a.c.c.b.d

p = x*(x+1)*(x²+x+1) = x⁴+x³+x³+x²+x²+x

q = x*(x²-x+1)*(x²-x+1)*(x+1)*(x²+x+1) =x⁸+x⁶+x⁵+x⁴+x³+x 

Corresponds to the solution: { {4, 3, 3, 2, 2, 1}, {8, 6, 5, 4, 3, 1} }

 

p = c.b.d; q = a.a.c.b.d

p = (x²-x+1)*(x+1)*(x²+x+1) = x⁵+x⁴+x³+x²+x+1

q = x*x*(x²-x+1)*(x+1)*(x²+x+1) = x⁷+x⁶+x⁵+x⁴+x³+x²

Corresponds to the solution: { {5, 4, 3, 2, 1, 0}, {7, 6, 5, 4, 3, 2} }

 

p = a.a.b.d; q = c.c.b.d

p = x*x*(x+1)*(x²+x+1) = x⁵+x⁴+x⁴+x³+x³+x²

q = (x²-x+1)*(x²-x+1)*(x+1)*(x²+x+1) =x⁷+x⁵+x⁴+x³+x²+1

Corresponds to the solution: { {5, 4, 4, 3, 3, 2}, {7, 5, 4, 3, 2, 0} }

 

p = a.c.b.d; q = a.c.b.d

p = q = x*(x+1)*(x²-x+1)*(x²+x+1) = x⁶+x⁵+x⁴+x³+x²+x

Corresponds to the solution: { {6, 5, 4, 3, 2, 1}, {6, 5, 4, 3, 2, 1} }

 

 

Matches with witzar's post..