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Complete Numbers in Fraction Equations

The formula on our face page of “amazing numbers” is rather interesting:
1 – (1 ⁄ 28) = (1 ⁄ 2) + (1 ⁄ 4) + (1 ⁄ 7) + (1 ⁄ 14)

The point of interest is that: if you look at all divisors of 28: they are 1,2,4,7,14,28; with the exception of 28 which is itself, all divisors have appeared in this formula, and they appear in the form of so-called “unit fraction”, where numerator is 1. So (1 ⁄ 2), (1 ⁄ 4), etc. are all unit fractions.

Indeed, we present a fraction equation to make it a bit unusual, but there is a low-pitch but straightforward ways to present number 28. We have that:
28 = 1 + 2 + 4 + 7 + 14
To get to the earlier fraction form, just divide every term by the number 28.

The smallest complete number is 6 (=1+2+3), 28 is the 2nd complete number, and after that, you will not see a complete number until 496. So complete numbers are rare among all positive whole numbers.

Complete numbers 6 also has a nice fraction form, as:
1 – (1⁄6) = (1⁄2) + (1⁄3)

Prime numbers

Prime numbers are those that have 1 (one) and itself as the only two divisors. Examples of primes are 2, 3, 5, 7, 11. None of 4, 6, 9 is a prime since 4 = 2 × 2, 6 = 2 × 3, and 9 = 3 × 3.

If a number greater than one is not a prime, then it is a composite number, and can be factored into the product of primes — called prime factorization. We have given the prime factorization of 4, 6, 9 as above. For a couple of more examples:

12 = 2 × 2 × 3

36 = 2 × 3 × 3 × 3

28 = 2 × 2 × 7

So all natural numbers are divided into three classes: the number 1, the prime numbers, and the composite numbers.

A bonus point: π, besides representing in a circle, the ratio of circumference to diameter, also stands for a special function related to prime numbers. Function π(x) — for every integer x, represent the number of primes less than or equal (i.e. not exceeding) x. For example, we have:

π(2) = 1, π(3) = 2, π(10) = 4, π(20) = 8 etc.
[To find why π(10) = 4, recall the 4 prime numbers not exceeding 10: they are 2,3,5, and 7.]

On the Patterns (2)

★ On the Patterns

(2) Number Patterns * Numbers, Colours and Stars

Start by taking a look at the following chart.

1	2	★3	4	5	6	7	8	9	10
11	12	13	★14	15	16	17	18	19	20
21	22	23	24	★25	26	27	28	29	30
31	32	33	34	35	★36	37	38	39	40
41	42	43	44	45	46	★47	48	49	50
51	52	53	54	55	56	57	★58	59	60

First, let us look up to the stars:

What pattern is followed for the placement of stars ?

Each number besides the star is increased by ____. (Fill in the blank)

Following this pattern, the three more numbers that are besides and that

comes after 58 are: __ , __ , and __.

Now go back to the chart, and let us look at the coloured cell (those coloured by yellow)

What can you do to follow the yellowed-coloured cell? Please colour the numbers on the chart by continue the pattern that you discovered.

What are the common features of these yellow cells? Colour some new cells, and explain how, by colouring these cells, you have followed and extended the pattern which is already in the chart.

Take a moment to think. You can answer these questions!

3D objects with 3 views from top, front and side

For a 3D objects, given three views to you: one from top, one from front, and one from side, can you imagine what the original 3D objects looks like?

The question is not posed to a mechanic engineer, it would be trivial in that case. The question is raised to get a junior middle student to think a bit.

For a cylinder one of the three views is a circle, and the other two views are rectangles. For a cone one of the views is a circle, and the other two views are triangles. What if the three views given are a circle, a rectangle and a triangle? Can you figure out the original shape?

Find the Center of a Circle — Do you know how to do it?

Using a compass, you can draw a circle at any place, with any radius.

Now let’s reverse the problem. Given a circle, do you know how to find its center? (Of course, once the circle is found, there shall be no problem at all to tell its diameter, or radius.) You only see the circle itself, there is no explicit indication on where the center is.

With two set of restrictions on what kind of tools you can use, there are actually two questions. In general, you can find the center using any convenient method, including copy-and-paste the circle onto a paper, and then fold it. In particular (from classical Euclidean geometry), where it’s required to do so with a ruler (with which you are allowed to draw lines and line segments only) and a compass (with which you are allowed to draw circles only).

See the following article on how to do it in general.

How to find the center of a given circle?

If you attempt to solve this problem with a ruler and a compass, then you are required to know how to make a perpendicular bisector. This will be discussed in another posting.

Right Triangle: Find Hypotenuse Given Two Legs — Area-based Approach!

How Long Does the Hypotenuse of a Right Triangle Measure
— Find it Using Area-based Approach!

We will have an exciting journey of discovery, following the footstep of early pioneers in math.

Let the problem be raised as:
Problem Given two legs of a right triangle, what is the length of its hypotenuse?

After a bit thought, we decide to reduce the problem to the following form:
Standardized Problem Given a right triangle with one leg being 1, the other leg being x, what
length of the hypotenuse?

With familiarity with concepts of similar triangles and proportion, you will find that solution to
standardized problem leads immediately to a solution for the original one. Even without that,
the connection of the two problems can be intuitively understood. Suppose a triangle with two legs:
1 and x, and a hypotenuse of y, then we know a triangle with two legs 2, 2x will have a hypotenuse of 2y.

For this reason, below we will focus only on the standardized problem. The goal is to fill out a form
where leg one is always 1, leg two is any integer numbers: 1, 2, 3, .. etc.
This provides us with the length of the hypotenuse.
[table]
Leg-1(a),Leg-2(b),Hypotenuse(c)
1,1,?
1,2,?
1,3,?
1,4,?
[/table]

Let’s set out to work!

For row 1 – Leg two equals Leg one Equals 1
This is the case for a right isosceles triangle.

The right isosceles triangle is shown. Reflect it twice, to the horizontal leg and respectively
to vertical leg, as line symmetry. Then rotate the original triangle
around the right-angle corner for 180 degrees. In such way we obtain three new
triangles. The new three and the original one together form one square.

Since the original triangle has an area of (1/2) × 1 × 1 = (1/2), the four triangles have a total area of
4 × (1/2) = 2. While they together form one square with a side to be decided; let us suppose it to be y.
Then the area of square must equal 2. So y² = 2.

We will simply write for this case as Leg one = 1, Leg two = 1, and the Hypotenuse is sqrt 2.

Continue to part 2