Ways are there of picking out the name ELLE from the following table

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The discussion focuses on two mathematical problems involving cubes and letter arrangements. The first problem asks how to form the name "ELLE" by moving through a grid of letters, allowing adjacent movements without repeating letters. The second problem involves counting painted and unpainted smaller cubes within a larger painted cube, specifically determining when the number of unpainted cubes equals those with one painted face. Participants share their understanding of the formulas needed to solve these problems, emphasizing the importance of counting and setting equations equal to each other. The conversation highlights the collaborative nature of problem-solving in mathematics.
futb0l
I have trouble understanding the MEANING of each question...

How many ways are there of picking out the name ELLE from the following table by moving from each letter to an adjacent letter ( up or down, left or right, and diagonal steps are allowed, but the same must not be used twice in one ELLE)?

E E E E
E L L E
E L L E
E E E E

---

A cube with edge of length N is painted. The cube is then divided into N^3 cubes of the same size. Among these small cubes, there are some which do not have a painted face, some which have one, two or three painted faces. For what value of N is the number of small cubes which have do NOT have a painted face equal to the number of small cubes which have only one painted face?
 
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1st Question - Now, I understand but not the 2nd one.
 
I assume that you have to use the entire cube to create these smaller cubes. Then each smaller cube will have to an edge of length 1. Each face of the larger cube will look something like this:

http://w1.315.telia.com/~u31523890/cube.GIF

(I've used two shades of red for purely pedagogical reasons). Now, looking at only one face, which of the smaller cubes will be painted in only one color? Obviously the ones which are not out on the edge (the ones that are lighter red in my drawing). Thus, for each face, there are (n - 2)(n - 2) smaller cubes with only one color on them, and there are 6 faces altogether, so there are 6(n - 2)^2 1-colored cubes altogether.

The cubes which have no painted faces, are those cubes which are totally "inside" the larger cube. They can also be easily counted. Then you just put that number equal to 6(n - 2)^2 and solve the equation...
 
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futb0l said:
A cube with edge of length N is painted. The cube is then divided into N^3 cubes of the same size. Among these small cubes, there are some which do not have a painted face, some which have one, two or three painted faces. For what value of N is the number of small cubes which have do NOT have a painted face equal to the number of small cubes which have only one painted face?

You have N^3 small cubes, so take the cube root of that to find the number of cubes in length the large cube is. The large cube will be N cubes long, but it's also N units long which means each small cube is 1 unit long.
Simple assumption but it's probably important.

To find the number of cubes with 1 face showing, look at one side of the cube; just a square. Count all of the little squares except for the squares that make up the perimeter. If I'm not mistaken, the formula would be something like

S = (N-2)^2

S is the number of squares being counted, N is one square (or square length). It's -2 because the perimeter is not counted.
Remember to include the other sides of the big cube, so multiply S by 6. S * 6 should give the number of cubes showing 1 face (I think).

Now find the number of cubes with no faces showing. The cubes with no faces showing would make up a cube inside of the large cube that would have a formula like this, I believe

C = (N-2)^3


Then set those equal to each other

6(N-2)^2 = (N-2)^3

The interesting thing about the formula is that it gives 2 answers. Try to guess which one is real :wink:


Damn, Muzza beat me to it. Might have beaten you to posting if I didn't make so many errors when checking my work.
 
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It's not a competition you know ;)
 
lol you guyes are funny.
 
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