Orienting Dice Tower

[TimeSpaceLightForce]

Using 1x1x1 dice, Marco makes the tallest dice tower (one die on top of another) 

where the front,back,left and right side do not have same dice face.

How many std dice shall he need?  (a) 3      (b) 4     ( c) 5     (d) 6     (e) 7

Edited September 26, 2014 by TimeSpaceLightForce

[TimeSpaceLightForce]

6?

 

 

The answer is six.

Consider the four numbers of each die on the side faces of the tower

taken in ccw order starting with the front.

A tower can be built for each orientation of the top die,

namely 1 2 6 5, 1 5 6 2, 1 4 6 3 and 1 3 6 4:

1 2 6 5 - 1 5 6 2 - 1 4 6 3 - 1 3 6 4 <- top die

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

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

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

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

6 4 1 3 - 6 3 1 4 - 6 2 1 5 - 6 5 1 2 <- bottom die

 

You found them 

[Barcallica]

6?

[bonanova]

It's at least four - let the 1-spot be the top face and just rotate the die.
And it's at most six - all six numbers appearing on each side.
But can that be done? Not if we begin with 1-spot four times on top.
Each number has to appear once in each position (top, bottom, four sides.)

[TimeSpaceLightForce]

..dont mind the bottom ,you cant see it and the top is covered with coin.

 

 the no.of pips on opposing faces of a std dice adds to 7.

[bonanova]

The answer is six.

Consider the four numbers of each die on the side faces of the tower

taken in ccw order starting with the front.

A tower can be built for each orientation of the top die,

namely 1 2 6 5, 1 5 6 2, 1 4 6 3 and 1 3 6 4:

1 2 6 5 - 1 5 6 2 - 1 4 6 3 - 1 3 6 4 <- top die

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

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

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

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

6 4 1 3 - 6 3 1 4 - 6 2 1 5 - 6 5 1 2 <- bottom die