The Numbers on a Dice

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Q: A die is rolled and the numbers on all the visible faces is multiplied together \(=S\). How would you choose a highest number such that it is guaranteed to divide the product.

A: The puzzle rests on a simple fact of prime factorization. Every number can be expressed as a product of prime numbers. For the given example, assume the face that is not visible is 1. The product of the remainder of numbers is \(6 \times 5\times\ldots \times 2\) or equivalently \(6!\). If the face that is invisible is 2, the number \(S\) is \(\frac{6!}{2}\), if it is 3 the number \(S\) is \(\frac{6!}{3}\) and so on. The number \(6!\) can be expressed as prime numbers as \(2^{4}\times3^{2}\times 5\). Let us enumerate the cases for each of the faces that is invisible, the product \(S\) expressed in terms of prime numbers are as follows

1. \(2^{4}\times3^{2}\times5\)
2. \(2^{3}\times3^{2}\times5\)
3. \(2^{4}\times3^{1}\times5\)
4. \(2^{2}\times3^{2}\times5\)
5. \(2^{4}\times3^{2}\)
6. \(2^{3}\times3^{1}\times5\)

Notice, the solution to our problem is the lowest exponent of each of the prime factors. This yields \(2^{2}\times3^{1}\times5^{0} = 12\).

Three good books to own to learn number theory,

Number Theory: A Lively Introduction with Proofs, Applications, and Stories

An Introduction to the Theory of Numbers

Elementary Number Theory (Springer Undergraduate Mathematics Series)

A Perfect Mathematical Christmas Gift