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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: 2⋅3⋅4⋅5⋅6=24⋅32⋅51⋅3⋅4⋅5⋅6=23⋅32⋅51⋅2⋅4⋅5⋅6=24⋅3⋅51⋅2⋅3⋅5⋅6=22⋅32⋅51⋅2⋅3⋅4⋅6=24⋅321⋅2⋅3⋅4⋅5=23⋅3⋅5

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

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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