• Apr 26th 2006, 12:21 PM
kimki
Ok here is the question:

For a charity event, some students assembled one million small plastic cubes to form a massive cube. How many small cube faces could be seen on one face of the finished massive cubes?

I thought it would be cube root of 1 000 000 = 100 squared.

Thats not the answer thats in the book.

• Apr 26th 2006, 12:27 PM
CaptainBlack
Quote:

Originally Posted by kimki
Ok here is the question:

For a charity event, some students assembled one million small plastic cubes to form a massive cube. How many small cube faces could be seen on one face of the finished massive cubes?

I thought it would be cube root of 1 000 000 = 100 squared.

Thats not the answer thats in the book.

If the book does not say 10000, what does it say?

RonL
• Apr 26th 2006, 01:37 PM
kimki
it says 60 000
• Apr 26th 2006, 02:10 PM
ThePerfectHacker
If you have 1,000,000 cubes and you use them to form a monster cube, then you need that,
$(\mbox{side})^3=1,000,000$
That happens when the side got 100 cubes.
Each face of a cube is 100x100 cubes thus they are
10,000 cubes visible on one side. But there are 6 side to a cube thus, 60,000 cubes visible.
• Apr 26th 2006, 08:29 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
If you have 1,000,000 cubes and you use them to form a monster cube, then you need that,
$(\mbox{side})^3=1,000,000$
That happens when the side got 100 cubes.
Each face of a cube is 100x100 cubes thus they are
10,000 cubes visible on one side. But there are 6 side to a cube thus, 60,000 cubes visible.

So that implies that the wording of the question that we have been given
is either wrong or misleading

RonL
• Apr 27th 2006, 03:30 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
If you have 1,000,000 cubes and you use them to form a monster cube, then you need that,
$(\mbox{side})^3=1,000,000$
That happens when the side got 100 cubes.
Each face of a cube is 100x100 cubes thus they are
10,000 cubes visible on one side. But there are 6 side to a cube thus, 60,000 cubes visible.

Actually, if the problem works like that there are only 50,000 because the bottom face is not visible.

-Dan
• Apr 27th 2006, 04:16 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
If you have 1,000,000 cubes and you use them to form a monster cube, then you need that,
$(\mbox{side})^3=1,000,000$
That happens when the side got 100 cubes.
Each face of a cube is 100x100 cubes thus they are
10,000 cubes visible on one side. But there are 6 side to a cube thus, 60,000 cubes visible.

Should we take account of triple counting the 8 corner cubes or double
counting the 12*98 edge cubes?

RonL
• Apr 27th 2006, 07:22 AM
kimki
Also, in the question it states how many cubes are visable in ONE FACE. Doesnt that mean one side of the cube? ie 10,000?! 1002
• Apr 27th 2006, 12:43 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Actually, if the problem works like that there are only 50,000 because the bottom face is not visible.

-Dan

That type of fact only matters in physics type problems.
• Apr 27th 2006, 03:37 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
That type of fact only matters in physics type problems.

Translation: REAL problems. ;)

-Dan