bonanova Posted November 2, 2014 Report Share Posted November 2, 2014 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? Quote Link to comment Share on other sites More sharing options...

0 witzar Posted November 2, 2014 Report Share Posted November 2, 2014 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] 1 Quote Link to comment Share on other sites More sharing options...

0 gavinksong Posted November 2, 2014 Report Share Posted November 2, 2014 they're standard dice? Quote Link to comment Share on other sites More sharing options...

0 gavinksong Posted November 2, 2014 Report Share Posted November 2, 2014 One of the die has the numbers 0-5 and the other has the numbers 2-7. Quote Link to comment Share on other sites More sharing options...

0 karthickgururaj Posted November 2, 2014 Report Share Posted November 2, 2014 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}) = (x^{2} + 2.x^{3} + 3.x^{4} + 4.x^{5} + 5.x^{6} + 6.x^{7} + 5.x^{8} + 4.x^{9} + 3.x^{10} + 2.x^{11} + x^{12}) What are the numbers then? Quote Link to comment Share on other sites More sharing options...

0 karthickgururaj Posted November 2, 2014 Report Share Posted November 2, 2014 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}) = (x^{2} + 2.x^{3} + 3.x^{4} + 4.x^{5} + 5.x^{6} + 6.x^{7} + 5.x^{8} + 4.x^{9} + 3.x^{10} + 2.x^{11} + x^{12}) What are the numbers then? ..can be written as: x^{2}.(1 + 2.x + 3.x^{2} + 4.x^{3} + 5.x^{4} + 6.x^{5} + 5.x^{6} + 4.x^{7} + 3.x^{8} + 2.x^{9} + x^{10}) which is nothing but: x^{2}.(1 + x + x^{2} + x^{3} + x^{4} + x^{5})^{2} (It took me sometime to realize that. If I instead had, (1 + 2.x + 3.x^{2} + 4.x^{3} + ... + 9.x^{10}), I could have simply integrated the terms.). I couldn't factorize 1 + x + x^{2} + x^{3} + x^{4} + x^{5} further, cheated a bit here and asked wolfram alpha. So the final factorized expression is: x^{2}.(x+1)^{2}.(x^{2}-x+1)^{2}.(x^{2}+x+1)^{2} This has got four terms: A(x) = x, B(x) = (x+1), C(x) = (x^{2}-x+1), D(x) = (x^{2}+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+x^{6}) is a sixth degree polynomial with only two terms - so that is not a solution we are interested in. (1+5.x^{6}) on the other hand is ok, since it has 6 terms (5.x^{6 }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? Quote Link to comment Share on other sites More sharing options...

0 witzar Posted November 2, 2014 Report Share Posted November 2, 2014 @witzar: what was your approach?Just brute force. Quote Link to comment Share on other sites More sharing options...

0 karthickgururaj Posted November 3, 2014 Report Share Posted November 3, 2014 @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.dp = (x+1)*(x^{2}+x+1) = x^{3}+x^{2}+x^{2}+x+x+1q = x*x*(x^{2}-x+1)*(x^{2}-x+1)*(x+1)*(x^{2}+x+1) = x^{9}+x^{7}+x^{6}+x^{5}+x^{4}+x^{2}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.dp = x*(x+1)*(x^{2}+x+1) = x^{4}+x^{3}+x^{3}+x^{2}+x^{2}+xq = x*(x^{2}-x+1)*(x^{2}-x+1)*(x+1)*(x^{2}+x+1) =x^{8}+x^{6}+x^{5}+x^{4}+x^{3}+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.dp = (x^{2}-x+1)*(x+1)*(x^{2}+x+1) = x^{5}+x^{4}+x^{3}+x^{2}+x+1q = x*x*(x^{2}-x+1)*(x+1)*(x^{2}+x+1) = x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}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.dp = x*x*(x+1)*(x^{2}+x+1) = x^{5}+x^{4}+x^{4}+x^{3}+x^{3}+x^{2}q = (x^{2}-x+1)*(x^{2}-x+1)*(x+1)*(x^{2}+x+1) =x^{7}+x^{5}+x^{4}+x^{3}+x^{2}+1Corresponds to the solution: { {5, 4, 4, 3, 3, 2}, {7, 5, 4, 3, 2, 0} } p = a.c.b.d; q = a.c.b.dp = q = x*(x+1)*(x^{2}-x+1)*(x^{2}+x+1) = x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+xCorresponds to the solution: { {6, 5, 4, 3, 2, 1}, {6, 5, 4, 3, 2, 1} } Matches with witzar's post.. 1 Quote Link to comment Share on other sites More sharing options...

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## 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

(2) =p(12) = 1/36p(3) =p(11) = 2/36p(4) =p(10) = 3/36p(5) =p( 9) = 4/36p(6) =p( 8) = 5/36p(7) = 6/36pWhat can we say about the numbers inscribed on the dice?

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