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)

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