Thread: Math Dice rolls

The number on a standard six faced die are arranged such that numbers on opposite faces always add to 7.The product of the number appearing on the four lateral faces of a rolled die is calculated(ignoring the numbers on the top and bottom) What is the maximum possible value of this product?

Originally Posted by Dragon

The number on a standard six faced die are arranged such that numbers on opposite faces always add to 7.The product of the number appearing on the four lateral faces of a rolled die is calculated(ignoring the numbers on the top and bottom) What is the maximum possible value of this product?

here is the rolled out version of the cube:

Code:

            :-----:
            :  1  :
:-----:-----:-----:-----:
:  2  :  3  :  5  :  4  :
:-----:-----:-----:-----:
            :  6  :
            :-----:

another one:

Code:

            :-----:
            :  4  :
:-----:-----:-----:-----:
:  1  :  2  :  6  :  5  :
:-----:-----:-----:-----:
            :  3  :
            :-----:

and the last one:

Code:

            :-----:
            :  2  :
:-----:-----:-----:-----:
:  1  :  3  :  6  :  4  :
:-----:-----:-----:-----:
            :  5  :
            :-----:

Your job is to find the product of all those middle rows.

So for the first one: $\displaystyle 2\times3\times5\times4=120$

Second One: $\displaystyle 1\times2\times6\times5=60$

Last One: $\displaystyle 1\times3\times6\times4=74$

So the maximum possible value is: $\displaystyle \boxed{120}$

Originally Posted by Dragon

The number on a standard six faced die are arranged such that numbers on opposite faces always add to 7.The product of the number appearing on the four lateral faces of a rolled die is calculated(ignoring the numbers on the top and bottom) What is the maximum possible value of this product?

This product is $\displaystyle \frac{6!}{A\times (7-A)}$, where $\displaystyle A$ is the smaller of the numbers on the oposite faces of the die, so is
one of 1, 2 , 3. In these cases A=1, 2, 3 these products are 120, 72 and 60 respectivly, so the maximum posible value is 120.

RonL

Hello, Dragon!

A re-hash of what Quick said . . .

The number on a standard six faced die are arranged
such that numbers on opposite faces always add to 7.
The product of the number appearing on the four lateral faces of a rolled die
is calculated, ignoring the numbers on the top and bottom.
What is the maximum possible value of this product?

There is basically one possible arrangement for the six numbers.

Code:

          * - - - *
        /   2   / |
      * - - - *   |
      |       | 3 |
      |   1   |   *
      |       | /
      * - - - *

At the vertex where $\displaystyle 1,\,2,\,3$ share common edges,
. . the numbers can be arranged clockwise or counterclockwise.
The two resulting dice are mirror-images of each other.

Then: $\displaystyle 6$ is on the back, $\displaystyle 4$ is on the left, and 5 is on the bottom.


There are three sets of four "lateral numbers":

. . Front-right-back-left: $\displaystyle = 1 \times 3 \times 6 \times 4\:=\:72$

. . Front-top-back-bottom: $\displaystyle 1 \times 2 \times 6 \times 5 \:=\:60$

. . Top-right-bottom-left: $\displaystyle 2 \times 3 \times 5 \times 4 \:=\:\boxed{120}\;\;\Leftarrow$ maximum product