Each of the natural numbers (positive integers) can be put into one of three directories, according to how many factors /divisors it has.

Given a number N, a divisor is an integer number d such that d divides N evenly with zero remainder. For example, 5 divides 10, or 10 ÷ 2 = 5 with zero remainder. A divisor is also called a factor as we can write 10 = 2 × 5.

Number 1, which is the beginning of numbers, is itself a directory. It has only one divisor, which is 1.

An integer like 5 has only two divisors: one and itself. Any number in this directory is called a *prime number*.

All other numbers must have at least three divisors: one, itself and one proper divisor (which is neither 1 nor itself). Any number in this directory is called a *composite number *.

Examples:

Numbers 2, 5, 7 are prime numbers (also called primes).

Number 4, 6, 9, 12 are composite numbers.

Number 1 is neither prime nor composite numbers.

1a) Please remember that 1 is not a prime!

2. It is helpful to remember a list of smaller primes, like:

2, 3, 5, 7, 11, 13, 17, 19

(in total 8 primes that’s under 20)

We’d like to list some interesting facts and properties about primes, as below:

3. The last digit (units-place) of a prime CANNOT be any digit from 0, 2, 4, 6, 8 (except prime 2 itself); neither can it be digit 5 (except prime 5). Only numbers with last digit 1,3,7,9 are possible candidates of prime numbers 1, 3, 7, 9 (besides 2 and 5, which are exceptions).

4. About 1/4 of the numbers in the first hundred (1-100) are primes.

Among the first hundred,

there are 10 primes that ends with either digit 1 or digit 9;

there are 13 primes that ends either either 3 or 7; they are:

3, 13, 23, 43, 53, 73, 83 (ends in digit “3”)

7, 17, 37, 47, 67, 97 (ends in digit “7”)

And 2 and 5 are primes. To summarize, there are 25 primes in the first hundred.

To look a bit deeper, let’s look at the a few questions as follows.

- Can any composite number be written as the product of primes?
- Are there finitely many prime numbers or infinitely many primes? (“Infinite” means that we cannot count all primes until the end: no matter how many we’ve counted, there are always some more ..)
- Is there a prime-generating math machine? (like a function we input any number, the output is always a prime)

These more amazing stuff with prime numbers! (to be posted)

]]>Please answer the following questions too.

(a) Look at the sum that you find. What are the digits at odd places (units-digit, and hundreds-digit)? What is the sum of digits at odd places?

(b) Look at the sum that you find. What are the digits at even places (tens-digit, and thousands-digit)?

What is the sum of digits at even places?

__Question 2.__ Knowing how to look at the clock.

(a) Draw a clock for 9:10.

What is the time 2 hours later?

(b) Draw a clock for 11:35.

What is the time 1/2 hour later? (or 30 minutes later)

__Question 3.__ Complete the Pattern by filling in the blanks.

Let us consider some lists of natural numbers (integer sequences). In these lists, we can find any number by adding or subtracting a fixed number (except 1st number). We will describe of adding /subtracting a fixed number as **the rule **. An example is provided below.

48, 45, 42, _39_, _36_, _33_

**Rule:** -3

-3 indicates that we always subtract 3 to get the next number.

To describe adding a fixed number, using “+” sign followed by the fixed number to add.

Please follow the example above to complete:

(a) 59, 63, 67, ___, ___, ___

Rule: ____

(b) 49, 43, 37, ___, ___, ___

Rule: ____

π = 3 ⁄ √̅5̅ (1+ (3 ⁄ √̅5̅ ) )— (**)

By standard notation, **√̅5̅ **is the square root of 5. Grab a calculator to verify how close it is to the genuine value π = 3.1415 .. And in neat form too!

It is still puzzling today how ** came to this value. We have guessed that he came to this particular form of π from one of the following equations:

(1/x) (1+ (1/x)) = π — Eq.(1)

or

x = π x^{2}– 1 — Eq. (2)

Both share a positive root that is pretty close to √̅5̅ /3.

From the second equation, if we set out to solve π in terms of x, then we will find

π = 1/x (1 + 1/x) = (1+x) ⁄ (x

^{2})

that’s exactly the first equation!

We can get the π -value expression (which appears at the very beginning of the article), by simple replacing x with the root √̅5̅ /3.

Now take another look at the expression π , and rewrite it in the form below:

π = (√̅5̅ /3 + 1) (3^{2} ⁄ 5)

If picking out all digits at the right hand side! but ignore operations and square /square root, then What do we get? We get 53135: a palindrome. That’s double neat!!

]]>A is approximately

(86 a) ⁄ {(b/2) + c}(note the result is in degree) — (**)

Let’s look at a few examples to verify that the formula above does give us the correct result.

For 1-1-√̅2 triangle, both of the acute angles shall be 45 degrees. While using formula (**), we get:

A is approximately (in degrees):: (86 x 1) ⁄ {(1/2) + √̅2} = 44.9

For 1- √̅3 -2 triangle, the smaller acute angle is 30 degree. While using formula (**), we get:

A is approximately (in degrees):: (86 x 1) ⁄ {√̅3/2 + 2} = 30.01

— So that’s pretty close.

Finally, let us look at the famous 3-4-5 right triangle: with the given formula,

A is approximately (in degrees) (86 x 3) ⁄ {(4/2) + 5} = 36.86

You can verify the smaller acute angle of such a triangle is roughly 36.87 degree.

Wondering how do we get the constant 86? You can find this constant from the formula:

(3/2) x (180 ⁄ π) = 86

An alternative formula that gives you roughly the same result is:

A is approximately

(3 a) ⁄ {b + 2 c} x (180 degree /π)— (*)

Question for thought:

1) With this formula, can we also find the larger acute angle of the right triangle?

2) We applied the formula to find the smaller acute angle of a right triangle. Can we apply this formula to any triangle?

3) Do the calculation yourself – find the smaller acute angle respectively for the following two right triangles:

3a) the 1-2-√̅5 triangle

3b) the 8-15-17 triangle

Wow! With the approximate formula as given in this article, you can find the angle, on a calculator even without a trigonometry / inverse trigonometry functionality. Isn’t that cool?

]]>Yeh! let us focus within the border drawn with a crayon, that is, within triangle DAB. How many isosceles are there within it? (For the moment, ignore the smaller pentagram A’C’E’B’D’ and all the triangle formed by their intersections.)

Let’s count it! Recognizing that DA = DB (why?), we will know that DAB, DA’B’ are two isosceles. D’AB is the 3rd isosceles. When connecting points C’ and E’, then DC’E’ and D’C’E’ are also isosceles (having found 5 isos).

There are more though. Through the properties of pentagons, we know that AE’ = BC’ = AB. Therefore ABE’ (with AB = AE’), BAC’ (with BA = BC’) are isosceles. Note that AC’ = C’E’ = E’B, we know that AC’E’ and (with AC’ = C’E’) and C’E’B (with C’E’ = E’B) are isosceles. Finally, AC’ = AD’ and BD’ = BE’ so we find two other isosceles: triangles AC’D’ and BD’E’. Up till now, we have count 11 isosceles in total!

Another nice feature is that, if the original ABCDE is a regular pentagon, then the smaller pentagon A’B’C’D’E’ (as formed by intersection points of the diagonals of the pentagon ABCDE), must also be a regular pentagon. If we look into some of these points A’, B’, C’, D’, E’, we got to find many many more isosceles. But currently let’s ignore them.

We also find a trapezoid A’B’AB. By connecting points C’ and E’, there are other two trapezoids, ABE’C’ and A’B’C’E’. Yes! A’C’AE’ is a trapezoid, it is not an isosceles trapezoid, yet! the shorter base A’C’ equals one of the sides C’A, and is also equal to one diagonal C’E’.

For parallelograms? Either we allow to include some points outside the triangle DAB, or we make more connections (e.g. include the smaller pentagram A’B’C’D’E’); if we do one of the two, then there will be lots of them!

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Questions chosen from previous Gauss contests

Gauss contests are organized by the Centre of Education for

Math and Computing, University of Waterloo

In the addition shown, *P* and *Q* each represent single digits, and the sum is *1PP7*. What is *P + Q*?

(A) 9 (B) 12 (C) 14 (D) 15 (E) 13

**Problem 2**

In the right-angled triangle PQR, we have that PQ = QR. The three segments QS, TU and VW are perpendicular to PR, and the segments ST and UV are perpendicular to QR, as shown. What fraction of triangle PQR is shaded?

(A) 3 ⁄ 16 (B) 3 ⁄ 8 (C) 5 ⁄ 16 (D) 5 ⁄ 32 (E) 7 ⁄ 32

**Problem 3**

A box contains a total of 400 tickets that come in five colours: blue, green, red, yellow, and orange. The ratio of blue to green to red tickets is 1 : 2 : 4. The ratio of green to yellow to orange tickets is 1 : 3 : 6. What is the smallest number of tickets that must be drawn to ensure that at least 50 tickets of the same colour have been selected?

(A) 50 (B) 246 (C) 148 (D) 196 (E) 115

**Problem 4**

Greg, Charlize, and Azarah run at different but constant speeds. Each pair ran a race on a track that measured 100 m from start to finish. In the first race, when Azarah crossed the finish line, Charlize was 20 m behind. In the second race, when Charlize crossed the finish line, Greg was 10 m behind. In the third race, when Azarah crossed the finish line, how many metres was Greg behind?

(A) 20 (B) 25 (C) 28 (D) 32 (E) 40

**Problem 5**

In right-angled, isosceles triangle FGH, segment FH = √̅8. Arc FH is part of the circumference of a circle with centre G and radius GH. The area of the shaded region is

(A) π – 2; (B) 4 π – 2 (C) 4 π – (1 ⁄ 2) √̅8 ; (D) 4 π – 4 (E) π – √̅8

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Stories, Facts and Implications

Here is some facts found in somebody’s math venture:

He marks a point A on a circle. He then split the radius of the same circle into 4 equal segments – call the length of one segment as b.

He measured along the circle using b (explain the measurement). After 7 steps he stopped and label that point as B. — As he noticed, the arc from A to B is slightly greater than a quarter of the circle.

Now he returns to split the radius of this circle into 5 equal parts – call the length of one segment as d. Then he measured along the circle using d. After 7 steps he stopped and label that point as D. — As he noticed, the arc from A to D is slightly less than a quarter of the circle.

Having a second look, he has noticed as well that the line connecting B and D “almost” passes through the centre of the circle (or line BD is “almost” a diameter).

Pause for a minute. Try to draw a diagram according to this description. If not sure, look into the figure below for a check.

[As in this figure, N is the center of the circle, NA is the radius of the big circle, and the two small circles are of radius that is (1/4), resp. (1/5) of NA. From A, going through 7 steps to reach B; from A, going the other direction 7 steps to reach D. Pls. note the length of step in the two directions are not the same! ]

After some thought and calculation, he decided the approximate value of π. The value he’s found is 3.15 – which is pretty close to the true value. (For sure you know π is 3.14159, don’t you?)

Now is your chance to take part: how did he finds that value?

For a bonus question: he has decided that (7/4) radians is approximately 100 degrees. Do you agree? How did he come to that conclusion?

[Explain the concepts of radians if needed.]

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