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## Question

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?

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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]

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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:

(xa0 + xa1 + .. + xa5).(xb0 + xb1 + .. + xb5) = (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?

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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:

(xa0 + xa1 + .. + xa5).(xb0 + xb1 + .. + xb5) = (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.xis 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?

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Just brute force.
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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..

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