# Uncategorized

# Cube Coloring

Suppose we have a 3X3X3 cube. Dip the cube in red paint and split it into 27 cubes each with size 1X1X1. How many of the 1X1X1 cubes are painted on exactly two faces?

A cube has 8 vertices, 12 edges and 6 faces. Since three faces meet at each of the 8 vertices in a cube, the eight 1X1X1 cubes at each vertex will have 3 faces painted. Cubes with two faces painted are the cubes corresponding to the 12 edges of the 3X3X3 cube. Six of the 1X1X1 cubes in each face of the 3X3X3 are painted one face.

The number of 1X1X1 cubes with:

3 faces painted = 8

2 faces painted = 12

1 face painted = 6

No faces painted=1

So there are **twelve** 1X1X1 cubes with two faces painted.

It is clear that the vertices, edges and the faces of the cube play a major role in finding the painted faces when splitting a 3X3X3 cube into 27 1X1X1 cubes.

Is it possible to generalize this? Yes. For any integer N>= 3, consider the NxNxN cube. Paint this cube and then split it into 1X1X1 cubes. Clearly we will get N^{3} 1X1X1 cubes.

The number of 1X1X1 cubes with:

3 faces painted = 8

2 faces painted =

1 face painted =

No faces painted=

We can see that

We will look at example now. Take N=5. Then the cube is 5X5x5. After painting, we will get 125 1X1X1 cubes.

The number of 1X1X1 cubes with:

3 faces painted = 8

2 faces painted = = 12(5-2)=36

1 face painted = = 6(5-2)^{2}=54

No faces painted= = (5-2)^{3} = 27.

We will look at some problems based on this concept.

- In a birthday party, an NxNxN cubical special cake frosted on all 6 faces was cut into 1X1X1 cakes and distributed to all the participants. The number of people who got the cakes with no frost in it was 8 times the number of people who received the cakes with 3 faces frosted. How many people were there in the party?

Solution: Since there are 8 cakes with 3 faces frosted(the vertices), the number of people who got cakes with no frost is 8X8 = 64.

So, = 64 = 4^{3}

So, N=6.

Therefore: NxNxN = 6X6X6 = 216

Number of people in the party is 216.

Try these problems:

2. A cube of cheese is 4 cm wide, 4 cm long and 4 cm high. Three faces of the cube that meet in a corner are covered with thin layer of wax. The cheese is then cut into 64 small cubes with sides of length 1 cm. How many of these small cubes have no wax on them?

3. A cubical box without a top is 4 cm on each edge. It contains 64 identical 1-cm cubes that exactly fill the box. How many of these small cubes actually touch the box?

# Amicable numbers

The unravelling of the hidden properties of numbers never fails to fascinate math enthusiasts. Pythagoreans, known for the famous Pythagoras Theorem, saw a property for the pair of numbers 220 and 284 which they called Amicable. It started the hunt for other pairs of amicable numbers. What is that make pair of numbers amicable or friendly?

Find the proper divisors (all divisors except the number itself) of 220 and add them. Guess what? You will get 284. If you do the same with 284, you will end up with 220.

The sum of proper divisors of 220:

1+2+4+5+10+11+20+22+44+55+110 = 284

The sum of the proper divisors of 284:

1+2+4+71+142 = 220.

Two positive integers are amicable if the sum of proper divisors of one integer equals the other.

After Greeks, Arab mathematicians found several amicable numbers. Euler has given a rule to produce amicable numbers. We will look at Euler’s rule. Take prime numbers p, q, r as given below with n and m as positive integers with n > m.

Then are an amicable pair of integers.

For example, take n=2 and m=1. Then , q= 11 and r = 71. It follows that 4*5*11= 220 and 4*71= 284, the same two numbers we explored at the beginning. Euler foundnearly 60 amicable pairs using this rule.

However, Euler’s rule doesn’t provide all possible amicable numbers. There are amicable numbers not satisfying this rule. The amicable pair (1184, 1210) found by a sixteen year old Italian boy Nicolo Paganini in 1866 doesn’t satisfy Euler’s rule.

Please look at this nice video presented by Dr. James Grime on Amicable numbers.

Also look at http://mathworld.wolfram.com/AmicablePair.html for further reading.