**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

]]>

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 , , , 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 .

]]>

(1) Suppose .

Case

Case

Case

Case

(2) 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) Suppose that .

, for some

By completing the square,

Case

Case

Case

Case

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

(4)

constant term:

In particular,

In general, for any integer *n,*

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

Note that the perimeter of semi-circle is ,

(6) 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) By Fermat’s little theorem,

(8) By Pythagoras theorem,

(9) By Euclidean Algorithm,

hence

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

*Solution*

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

*Solution*

(3) Simplify . (3 marks)

*Solution*

(4) Factorize

(a) .

(b) .

(c) . (4 marks)

*Solution*

(a)

(b)

(c)

(5) 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. (4 marks)

*Solution*

Let be the number of female members.

(6) Consider the compound inequality

(a) Solve (*).

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

*Solution*

(a)

(b)

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

(a) Find .

(b) Find the perimeter of .

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

*Solution*

(a)

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

(c) 3

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

(a) Find .

(b) Solve the equation . (5 marks)

*Solution*

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

(2) – (1) gives

(b)

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

(a) Find , , and .

(b) 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)

(b)

(10) 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) Find the equation of . (2 marks)

(b) 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. (3 marks)

*Solution*

(a)

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

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

*Solution*

(a) Let be the original volume of milk.

(b)

Therefore, disagree.

(12) 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) Find and . (3 marks)

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

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

*Solution*

(a) Since the median age is 7.5, hence

Now, .

Since

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

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

Case ,

Case ,

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

(a) Prove that . (2 marks)

(b) Suppose that , and .

(i) Find .

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

*Solution*

(a)

(b) (i)

Hence, is isosceles.

(b) (ii)

By Pyth. thm,

Therefore, is right-angled.

(14) 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) Find , , and . (5 marks)

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

*Solution*

(a)

By matching the coefficient,

(b)

For no real root.

For two real roots.

Therefore, has two real roots.

(15) If 4 boys and 5 girls randomly form a queue, find the probability that no boys are next to each other in the queue. (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) 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. (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) The 1st term and the 38th term of an arithmetic sequence are 666 and 555 respectively. Find

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

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

*Solution*

(a)

(b)

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

Therefore, the greatest such is 444.

(18) Let .

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

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

(c) Under a transformation, is changed to $latex \dfrac{-1}{3}x^2 – 12x – 121$. Describe the geometric meaning of the transformation. (2 marks)

*Solution*

(a)

Therefore, the vertex is .

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

(c) Since

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

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

(a) Find and . (4 marks)

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

*Solution*

(a) By Sine Law,

By Cosine Law, .

(b) By Sine law,

Therefore, do not agree.

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

(a) Prove that . (3 marks)

(b) 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) Find the equation of .

(ii) 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. (9 marks)

*Solution*

(a) 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) Let . Since ,

Let circle

Substitute into circle C yields . Hence,

3(1)-19(2):

Therefore, the equation of circle C is

(b) (ii) 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.

]]>Actually 777 is quite an interesting number. In gambling, it is a lucky number to most people because “777” is a jackpot number. In computer science, “777” means you have been granted read, write, and executable permission on a file (in UNIX or UNIX-like operating system such as Linux, BSD, Mac OSX, etc). To most HKers, 777, 365, and 21 mean the number of vote of today’s HK chief executive election. To me, I have another interpretation.

Math is beautiful, isn’t it?

The joke has made into wikipedia LOL

]]>So, what is Golden Ratio?

In Euclid’s Elements, Book VI Definition 2, Euclid wrote (in Greek of course):

Ακρον καὶ μέσον λόγον εὐθεῖα τετμῆσθαι λέγεται, ὅταν ᾖ ὡς ἡ ὅλη πρὸς τὸ μεῖζον τμῆμα, οὕτως τὸ μεῖζον πρὸς τὸ ἔλαττὸν

Or in English,

A straight-line is said to have been cut in extreme and mean ratio when as the whole is to the greater segment so the greater (segment is) to the lesser.

Or in plain English, it basically says that when you divide a line segment into two uneven pieces, the ratio of the whole line segment to the longer piece equals the ratio of the longer piece to the shorter piece. And this ratio is called the Golden Ratio (often denote as ).

After solving the quadratic equation, you will get:

Now, let’s consider the Fibonacci sequence.

and examine some nth power of the golden ratio.

This suggests that

Let’s define , then

Assume , then

Hence, by math induction, .

Now, go back to the equation , the two roots are

Hence,

(1) – (2) yields

Or

Since

then

As

As

as n is large.

The last approximation comes in handy when you want to find the nth Fibonacci number, let’s say the 30th Fibonacci number.

So, you can compute the 30th Fibonacci number to be 832040 in seconds with any normal scientific calculator.

]]>There are five houses in five different colors in a row. In each house lives a person with a different nationality. The five owners drink a certain type of beverage, smoke a certain brand of cigar, and keep a certain pet. No owners have the same pet, smoke the same brand of cigar, or drink the same beverage.

The question is: *Who owns the fish?*

Hints:

- The Brit lives in the red house.
- The Swede keeps dogs as pets.
- The Dane drinks tea.
- The green house is on the left of the white house.
- The green house’s owner drinks coffee.
- The owner who smokes Pall Mall rears birds.
- The owner of the yellow house smokes Dunhill.
- The owner living in the center house drinks milk.
- The Norwegian lives in the first house.
- The owner who smokes blends lives next to the one who keeps cats.
- The owner who keeps horses lives next to the man who smokes Dunhill.
- The owner who smokes BlueMaster drinks beer.
- The German smokes Prince.
- The Norwegian lives next to the blue house.
- The man who smokes blend has a neighbor who drinks water.