I have never written anything on Statistics (coz I hate statistics) on my blog so far and I believe I am going to start writing something on it from now on. I picked Normal Distribution because it is important. By Central limit theorem, in simple terms if you have many independent random variables that are generated by all kinds of crazy distribution, the sum of those random variables will tends toward the Normal distribution.

**Normal Distribution**

where .

Then the MGF (Moment Generating Function) would be the following:

Now using completing the square,

hence,

Since the integral is the cumulative distribution function of the normal distribution which shift to the right by units, the integral does sum to one. Therefore we left with

**Expected Value**

**Variance**

Hence,

]]>**2 is the only even prime number**

Any even number that is greater than 2 can be written as the product of 2 and a natural number greater than one so all even numbers greater than 2 are composite.

**91 is NOT a prime number**

I think almost all students in elementary school or even some in highschool thought 91 is a prime number. I am not kidding, if you are a teacher or a math tutor, you know what I mean. Almost 9 out of 10 students I encountered with always think 91 is a prime number. It could be that 91 isn’t in the times table? and it is odd and doesn’t end with a 5? and maybe it is obviously not divisible by 3 that makes people think 91 is highly unlikely a composite number? Or maybe people are too lazy to try to divide 91 by 7 and see for themselves.

**1979 is the year I was born and it is a prime number**

Yeah I am kind of proud to be born on the year that is a prime

**2017 is the year which bitcoin increased 20 fold**

It’s not because of bitcoin price has been increased from $1000 to $20,000 in the year 2017. The reason I think 2017 is an interesting prime number is that if you add up all the odd prime numbers from 3 all the way up to 2017, you will get a prime number as the result. 3 + 5 + 7 + 11 + 13 + … + 2017 = 283079 which is a prime number. You can check it with wolfram alpha. There are also some fun facts about 2017 I found on the internet.

- 20170123456789 is prime
- 2017Ï€ (round to nearest integer) = 6337 is primeÂ
- 2017e (round to nearest integer) = 5483 is prime
- The sum of the cube of gap of primes up to 2017 is a prime number.

- These three prime numbers are consecutive

**prime numbers that you can find in Pi**

This is first 50 digits of Pi.

Couple years ago I was reading a book on recreational math and the author quoted that the first 12 digits of Pi (after rounding off) is a prime number. 314159265359 is a prime number. Of course you can always check it with wolfram alpha. So I found 3, 31, 314159, andÂ 31415926535897932384626433832795028841 are also primes as well. There maybe more. I may wrote a script on maple to find more later when I have the time.

**a prime that is unlucky with the devil**

1000000000000066600000000000001 is a prime number. 666 is the number of a beast in the bible (I didn’t read the whole bible, that’s just what everyone is saying). The number 666 is considered as a devil which sits right inside in the middle of this prime number. And there are “13” zeros on the left and on the right of the number 666. Why 13 is unlucky? I don’t have a clue just go along with the majority.

**12345678910987654321 is a prime number**

This is no doubt one of the nice prime numbers even a 3 year old can remember.

**a prime number that ends with “19” and “67”**

1234567891011121314151617181967 is a prime number. This is a prime number that is easy to remember if you are a HKer and speak Cantonese because it is “19” and “67”. This number is very nice all you need to do is to write down the number 1, 2, 3, all the way up to 19 and then ends it with 67.

**the largest known prime as of March 2018**

I can’t end this article without writing down the largest known prime to human. As of March 2018 according to Wikipedia, the largest known prime is

which is found by the GIMPS (Greatest Internet Mersenne Prime Search) in 2017. I am not gonna bother typing the number out because it is 23,249,425 digits long!

]]>**Q24.**

Denote as the *i-*th bucket that contains the puck.

If all 4 pucks are distributed into the same Green bucket, then we don’t need to worry about the Red, Blue, and Yellow buckets since they won’t have 4 pucks in them. So,

Now, 3 pucks are distributed into the same Green bucket, then we only need to exclude the case where 3 pucks are put into the same Red bucket. There are 4 ways to permute *aaab* and 3 numbers to pick for *a* and *b* which is 6 ways, so

If 2 pucks are distributed into the same Green bucket, then we only need to consider the case where all 3 pucks are distributed evenly into three different Red buckets and all 2 pucks are distributed evenly into two different Blue buckets. The most difficult part of the problem is in this case here. We need to find the number of ways to permute *n* repeated objects, *aabc*, which is , and since the permutation of *b* and *c* has already counted, we need to only consider how many numbers go into the choice of *a* which is 3.

Final step, add the three results and we get

**Q25.**

For and

When *k* = 1,

. The 8 quadruples are easy to find, they are

When *k* = 2,

When *k* = 3,

or

Therefore, *R* can only be 3, 6, or 8.

When *R* = 3,

Therefore 2017 quadruples.

When *R* = 6,

Therefore 2017 quadruples.

When *R* = 8,

Now, obviously has no solution.

If *k* = 2,

is a solution.

In conclusion, *N* = 8 + 2017 + 2017 + 1 = 4043.

Therefore the sum of digits of *N* is 11.

No matter how many digits of you write down on a piece of paper, it is still just an approximation. The tail of never terminates. It is an irrational number. Most mathematicians don’t really like approximation, so you rarely see one write 3.14 as . It’s just not correct and not cool to write (other than PIE=314). So today I am gonna show you a cool way to write .

The symbol “!” means factorial. By definition,

for all positive integers n. So

,

,

and so on.

What about ? The above definition didn’t include non-positive integer factorial. So we need to find a way to extend the factorials to a larger set of number that includes decimal numbers. One way to do it is by using Gamma function.

First of all, we need to verify that this Gamma function indeed satisfies

and by using integration by parts, we get

And therefore the Gamma function does indeed extend the factorials to the real numbers, and in fact also the complex numbers as well.

Hence, we see that

Now, let’s compute .

By using substitution of , we get

Now let ,

then by using polar coordinates,

By substituting we get

Therefore,

or

or

if you accept the factorial notation

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Actually the method is very simple, even an eight year old can do it I promise. So here it is:

- Draw a quadrilateral (4-sided shape), any quadrilateral will do even the non-convex one.
- Find the midpoint of all four sides with your ruler.
- Connect those four midpoints together and you are done.

Here are some examples.

The proof is very simple.

Let the four points ,Â ,Â ,Â be the vertices of a quadrilateral.

Then the four midpoints are

Find the slopes of the sides and show that the opposite sides are parallel:

Therefore *PQRS* is a parallelogram.

**Sum of First Powers**

**Sum of Second Powers**

**Sum of Third Powers**

**Sum of Fourth Powers**

**Sum of Fifth Powers**

**Sum of Sixth Powers**

**Sum of Seventh Powers**

**Sum of Eighth Powers**

There are also some interesting facts about the number 2187. If you write the number in backward, you get 7812. When you add the two numbers, you get 9999. When you split the number 2187 into two 2-digit numbers 21 and 87 and then multiple them, you get 1827. And also notice that .

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(1) Â Suppose .

Case

Case

Case

Case

(2) Â Since there are more 2’s than 5’s in 2017! and , all we need to do is to count the number of 5’s in the prime factorization of 2017!.

(3) Â Suppose that .

, for some

By completing the square,

Case

Case

Case

Case

If we define zero being a perfect square, then . Otherwise .

(4) Â

constant term:

In particular,

In general, for any integer *n,*

(5) Â Let be the radius of the smallest semi-circle.

Note that the perimeter of semi-circle is ,

(6) Â Let the side length of B, C, D, E, F, G, H, and I be b, c, d, e, f, g, h, and i respectively.

(7) Â By Fermat’s little theorem,

(8) Â By Pythagoras theorem,

(9) Â By Euclidean Algorithm,

hence

(10) Â Divide the numbers from 1 to 100 into congruence classes mod 11.

[1] : 1, 12, … , 100; There are 10 members so we can choose at most 5 members, since we don’t want any pair of numbers with difference of 11. (Any adjacent members in the class have difference of 11)

[2] : 2, 13, … , 90; There are 9 members so we can choose at most 5 members.

[3] : 3, 14, … , 91; There are 9 members so we can choose at most 5 members.

… … , similarly for classes such as [4] to [11], each of them have 9 members. Therefore we can at most pick numbers without any pair in the set having difference of 11.

]]>

(1)Â Simplify and express your answer with positive indices.Â (3 marks)

*Solution*

(2)Â Make the subject of the formula .Â (3 marks)

*Solution*

(3)Â Simplify .Â (3 marks)

*Solution*

(4)Â Factorize

(a)Â .

(b)Â .

(c)Â .Â (4 marks)

*Solution*

(a)Â

(b)Â

(c)Â

(5)Â In a recreation club, there are 180 members and the number of male members is 40% more than the number of female members. Find the difference of the number of male members and the number of female members.Â (4 marks)

*Solution*

Let be the number of female members.

(6)Â Consider the compound inequality

(a)Â Solve (*).

(b)Â Write down the greatest negative integer satisfying (*).Â (4 marks)

*Solution*

(a)Â

(b)Â

(7)Â In a polar coordinate system, is the pole. The polar coordinate of the points and are and respectively.

(a)Â Find .

(b)Â Find the perimeter of .

(c)Â Write down the number of folds of rotational symmetry of .Â (4 marks)

*Solution*

(a)Â

(b)Â Obviously the triangle is isosceles and hence and hence is equilateral. Therefore the perimeter is 36.

(c)Â 3

(8)Â It is given that is the sum of two parts, one part varies as and the other part varies as . Suppose that and .

(a)Â Find .

(b)Â Solve the equation .Â (5 marks)

*Solution*

(a)Â Let for some non-zero real numbers and .

(2) – (1) gives

(b)Â

(9)Â The frequency distribution table and the cumulative frequency distribution table below show the distribution of the heights of the plants in a garden.

(a)Â Find , , and .

(b)Â If a plant is randomly selected from the garden, find the probability that the height of the selected plant is less than 1.25 m but not less than 0.65 m.

*Solution*

(a)Â

(b)Â

(10)Â The coordinates of the points and are and respectively. Let be a moving point in the rectangular coordinate plane such that is equidistant from and . Denote the locus of by .

(a)Â Find the equation of .Â (2 marks)

(b)Â intersects the *x*-axis and *y*-axis at and respectively. Denote the origin by . Let be the circle which passes through , , and . Someone claims that the circumference of exceeds 30. Is the claim correct? Explain your answer.Â (3 marks)

*Solution*

(a)Â

(b)Â Let , then , hence

Let , then , hence

Since is a right triangle where , hence is the diameter of the circle .

Therefore, the claim is correct.

(11)Â An inverted right circular conical vessel contains some milk. The vessel is held vertically. The depth of milk in the vessel is . Peter pours of milk into the vessel without overflowing. He now finds that the depth of milk in the vessel is .

(a)Â Express the final volume of milk in the vessel in terms of .Â (3 marks)

(b)Â Peter claims that the final area of the wet curved surface of the vessel is at least . Do you agree? Explain your answer.Â (3 marks)

*Solution*

(a)Â Let be the original volume of milk.

(b)Â

Therefore, disagree.

(12)Â The bar chart below shows the distribution of the ages of the children in a group, where and . The median of the ages of the children in the group is 7.5.

(a)Â Find and .Â (3 marks)

(b)Â Four more children now join the group. It is found that the ages of these four children are all different and the range of the ages of the children in the group remains unchanged. Find

(i)Â the greatest possible median of the ages of the children in the group.

(ii)Â the least possible mean of the ages of the children in the group.Â (4 marks)

*Solution*

(a)Â Since the median age is 7.5, hence

Now, .

Since

(b) (i)Â To maximize the median, we add age 7, 8, 9, 10. Therefore the new median will be 8.

(b) (ii)Â To minimize the mean, we add age 6, 7, 8, 9.

Case ,

Case ,

(13)Â In figure 1, is a triangle. , , and are points lying on such that , and .

(a)Â Prove that .Â (2 marks)

(b) Suppose that , and .

(i)Â Find .

(ii)Â Is a right-angled triangle? Explain your answer.Â (5 marks)

*Solution*

(a)Â

(b) (i)Â

Hence, is isosceles.

(b) (ii)

By Pyth. thm,

Therefore, is right-angled.

(14)Â Let , where , , and are constants. When is divided by and when is divided by , the two remainders are equal. It is given that , where , , and are constants.

(a)Â Find , , and .Â (5 marks)

(b)Â How many real roots does the equation have? Explain your answer.Â (5 marks)

*Solution*

(a)Â

By matching the coefficient,

(b)Â

For no real root.

For two real roots.

Therefore, has two real roots.

(15)Â If 4 boys and 5 girls randomly form a queue, find the probability that no boys are next to each other in the queue.Â (3 marks)

*Solution*

Consider this : _ G _ G _ G _ G _ G _

There are 6 places to put 4 B’s so there are 6P4 ways to place the boys. Moreover, there are 5! ways to permute those 5 G’s. Therefore,

(16)Â In a test, the mean of the distribution of the scores of a class of students is 61 marks. The standard scores of Albert and Mary are -2.6 and 1.4 respectively. Albert gets 22 marks. A student claims that the range of the distribution is at most 59 marks. Is the claim correct? Explain your answer.Â (3 marks)

*Solution*

So, the standard deviation is 15. Now, let’s calculate what Mary gets on the test.

, which is greater than 59. Therefore the claim is false.

(17) Â The 1st term and the 38th term of an arithmetic sequence are 666 and 555 respectively. Find

(a) Â the common difference of the sequence. Â (2 marks)

(b) Â the greatest value of such that the sum of the first term of the sequence is positive. Â (3 marks)

*Solution*

(a) Â

(b) Â

Since is positive integer, we can divide both side by without switching the sign.

Therefore, the greatest such is 444.

(18) Â Let .

(a) Â Using the method of completing the square, find the coordinates of the vertex of the graph of . Â (2 marks)

(b) Â The graph of is obtained by translating the graph of vertically. If the graph of touches the *x*-axis, find . Â (2 marks)

(c) Â Under a transformation, is changed to . Describe the geometric meaning of the transformation. Â (2 marks)

*Solution*

(a) Â

Therefore, the vertex is .

(b) Â Basically translate the graph of 13 units upward.

(c) Â Since

Therefore, the transformation is reflecting the graph of f(x) horizontally along the *y*-axis.

(19) Â Figure 2 shows a geometric model in the form of tetrahedron. It is given that .

(a) Â Find and . Â (4 marks)

(b) Â A craftsman claims that the angle between and the face is . Do you agree? Explain your answer. Â (2 marks)

*Solution*

(a) Â By Sine Law,

By Cosine Law, .

(b) Â By Sine law,

Therefore, do not agree.

(20) Â is an obtuse-angled triangle. Denote the in-centre and the circumcentre of by and respectively. It is given that , , and are collinear.

(a) Â Prove that . Â (3 marks)

(b) Â A rectangular coordinate system is introduced so that the coordinates of and are and respectively while the *y*-coordinate of is 19. Let be the circle which passes through , , and .

(i) Â Find the equation of .

(ii) Â Let and be two tangents to such that the slope of each tangent is and the *y*-intercept of is greater than that of . cuts the *x*-axis and *y*-axis at and respectively while cuts the *x*-axis and the *y*-axis at and respectively. Someone claims that the area of the trapezium exceeds 17000. Is the claim correct? Explain your answer. Â (9 marks)

*Solution*

(a) Â Since , , and are collinear, we can draw a line passing through all three points and meet the line segment at . Since is incenter, is angle bisector of , hence . Since is circumcenter, then is perpendicular bisector of and so . obviously by common sides. Hence, by ASA. Therefore the corresponding sides of the two congruent triangles are congruent and so .

(b) (i) Â Let . Â Since ,

Let circle

Substitute into circle C yields . Hence,

3(1)-19(2):

Therefore, the equation of circle C is

(b) (ii) Â Since and have slope of , They both have the equation of the form:

They are both tangents to the circle hence:

Since they are tangents to the circle,

Let

Let

Therefore, the claim is correct.

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