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Monkeys & Coconuts Problem

Solution 1 : iteration problem => Use sequence
$Latex U_{j} =\frac {4}{5} U_{j- 1} -1 $

(initial coconuts)
$Latex U_0 =k$
Let
$Latex f(x)=\frac{4}{5}(x-1)=\frac{4}{5}(x+4)-4$
$Latex U_1 =f(U_0)=f(k)= \frac{4}{5}(k+4)-4$

$Latex U_2 =f(U_1)=f(\frac{4}{5}(k+4)-4)= \frac{4}{5}((\frac{4}{5}(k+4)-4+4)-4$

$Latex U_2=(\frac{4}{5})^2 (k+4)-4$

$Latex U_3=(\frac{4}{5})^3 (k+4)-4$

$Latex U_4=(\frac{4}{5})^4 (k+4)-4$

$Latex U_5=(\frac{4}{5})^5 (k+4)-4$

Since
$Latex U_5$ is integer  ,
$Latex 5^5 divides (k+4)$
k+4 ≡ 0 mod($Latex 5^5$)
k≡-4 mod($Latex 5^5$)
Minimum {k} = $Latex 5^5 -4$= 3121 [QED]

Note: The solution was given by Paul Richard Halmos (March 3, 1916 – October 2, 2006)

5 monkeys found some coconuts at the beach.

1st monkey came, divided the coconuts into 5 groups, left 1 coconut which it threw to the sea, and took away 1 group of coconuts.
2nd monkey came, divided the remaining coconuts into 5 groups, left 1 coconut again thrown to the sea, and took away 1 group.
Same for 3rd , 4th and 5th monkeys.

Find: how many coconuts are there initially?

Note: This problem was created by Nobel Physicist Prof Paul Dirac (8 August 1902 – 20 October 1984). Prof Tsung-Dao Lee (李政道) (1926 ~) , Nobel Physicist, set it as a test for the young gifted students in the Chinese university of Science and Technology （中国科技大学－天才儿童班）.

The French method of drawing curves is very systematic:

“Pratique de l’etude d’une fonction”

Let f be the function represented by the curve C

Steps:

1. Simplify f(x). Determine the Domain of definition (D) of f;
2. Determine the sub-domain E of D, taking into account of the periodicity (eg. cos, sin, etc) and symmetry of f;
3. Study the Continuity of f;
4. Study the derivative of fand determine f'(x);
5. Find the limits of fwithin the boundary of the intervals in E;
6. Construct the Table of Variation;
7. Study the infinite branches;
8. Study the remarkable points: point of inflection, intersection points with the X and Y axes;
9. Draw the representative curve C.

Example:

$latex \displaystyle\text{f: } x \mapsto \frac{2x^{3}+27}{2x^2}$
Step 1: Determine the Domain of Definition D
D = R* = R –…

Visually cut a cake 1/5 portions of equal size:

1) divide into half:

2) divide 1/5 of the right half:

3) divide half, obtain 1/5 = right of (3)

$latex \frac{1}{5}= \frac{1}{2} (\frac{1}{2}(1- \frac{1}{5}))= \frac{1}{2} (\frac{1}{2} (\frac{4}{5}))=\frac{1}{2}(\frac{2}{5})$

4) By symmetry another 1/5 at (2)=(4)

5) divide left into 3 portions, each 1/5

$latex \frac{1}{5}= \frac{1}{3}(\frac{1}{2}+ \frac{1}{2}.\frac{1}{5}) = \frac{1}{3}.\frac{6}{10}$

Generalized Analytic Geometry

Find the equation of the circle which cuts the tangent 2x-y=0 at M(1,4), passing thru point A(4,-1).

Solution:

1st generalization:
Let the point circle be:
(x-1)² + (y-4)² =0

2nd generalization:
It cuts the tangent 2x-y=0
(x-1)² + (y-4)² +k(2x-y) =0 …(C)

Pass thru A(4,-1)
x=4, y= -1
=> k= -2
(C): (x-3)² + (y-1)² =…
[QED]

Math Chants make learning Math formulas or Math properties fun and easy for memory . Some of them we learned in secondary school stay in the brain for whole life, even after leaving schools for decades.

Math chant is particularly easy in Chinese language because of its single syllable sound with 4 musical tones (like do-rei-mi-fa) – which may explain why Chinese students are good in Math, as shown in the International Math Olympiad championships frequently won by China and Singapore school students.

1. A crude example is the quadratic formula which people may remember as a little chant:
“ex equals minus bee plus or minus the square root of bee squared minus four ay see all over two ay.”

$latex \boxed{
x = \frac{-b \pm \sqrt{b^{2}-4ac}}
{2a}
}$

2. $latex \mathbb{NZQRC}$
Nine Zulu Queens Rule China

3. $latex \boxed {\cos 3A = 4\cos^{3}…

What is “sin A” concretely ?

1. Draw a circle (diameter 1)
2. Connect any 3 points on the circle to form a triangle of angles A, B, C.
3. The length of sides opposite A, B, C are sin A, sin B, sin C, respectively.

Proof:
By Sine Rule:

$latex \frac{a}{sin A} = \frac{b}{sin B} =\frac{c}{sin C} = 2R = 1$
where sides a,b,c opposite angles A, B, C respectively.
a = sin A
b = sin B
c = sin C