2004 AMC 10B Problems/Problem 4

Problem

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

Solution 1

The product of all six numbers is . The products of numbers that can be visible are , , ..., . The answer to this problem is their greatest common divisor -- which is , where is the least common multiple of . Clearly and the answer is .

Solution 2

Clearly, cannot have a prime factor other than , and .

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

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

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

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$ (Error compiling LaTeX. Unknown error_msg)P=3\cdot2^2=\boxed{\mathrm{(B)}\ 12}$.

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.