# 2004 AMC 10B Problems/Problem 4

## Problem

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$?

$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$

## Solution 1

The product of all six numbers is $6!=720$. The products of numbers that can be visible are $720/1$, $720/2$, ..., $720/6$. The answer to this problem is their greatest common divisor -- which is $720/L$, where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$. Clearly $L=60$ and the answer is $720/60=\boxed{\mathrm{(B)}\ 12}$.

## Solution 2

Clearly, $P$ cannot have a prime factor other than $2$, $3$ and $5$.

We can not guarantee that the product will be divisible by $5$, as the number $5$ can end on the bottom.

We can guarantee that the product will be divisible by $3$ (one of $3$ and $6$ will always be visible), but not by $3^2$.

Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by $2^2$. This is the most we can guarantee, as when the $4$ is on the bottom side, the two visible even numbers are $2$ and $6$, and their product is not divisible by $2^3$.

## Solution 3

The product P can be one of the following six numbers excluding the number that is hidden under, so we have: \begin{align*} 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\ 1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5 \end{align*}

The largest number that is certain to divide product P is basically GCD of all the above 6 products which is $2^2 \cdot 3$.

Hence $P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}$.