成语数学

中国的小学离校考试 (PSLE) 「神题」: 猜成语
1) 20 除 3
2）1 除100
3）9寸+1寸=1尺
4）12345609
5）1,3,5,7,9

答案：:
1) 20/3= 6.666 六六大顺
2）百中挑一
3）得寸進尺
4）七零八落
5）举世无双

Win Win Problem

Yoyo lost $3, Pierre found it and returned to Yoyo.
Yoyo wanted to give all $3 to Pierre as gift, but Pierre refused.

Joe is the judge. How to solve it ?

Solution:
Joe donates $1 to make total $4,
divide between Yoyo and Pierre，

=> Yoyo and Pierre each gets $2
Also Yoyo, Pierre and Joe all lose $1.

Note: "三方各损一两" Win-Win Problem

Chicken & Rabbits

This is an ancient Math Puzzle before Algebra was invented, so the ancient intelligent Chinese used this funny "Method of Elimination" below:

Puzzle: In a cage housing some chicken and rabbits. There are total of  35 heads, and 94 legs. Find how many chicken and rabbits ?

Solution:

Imagine letting each animal raises up 2 legs, then there would be 35x2 = 70 legs raised.

94 - 70 = 24 legs

Note that all chicken raising 2 legs would have fallen on ground (no leg how to stand ?), so the 24 remaining legs still standing on the ground MUST belong ONLY to the rabbits standing on 2 legs (the other 2 legs raised up)

24 / 2 = 12 rabbits (Ans)

35 - 12 = 23 chicken (Ans)

Hotel Bill Puzzle

A 3-bed hotel room costs $30 per nite. Three men each pays $10 to share the room. The hotel boss decides to refund $5 as discount. The hotel clerk pockets $2, return each man only $1.

Each man nett pays $10 - $1 =$9
The room costs the 3 men $9 x3 = $ 27
plus $ 2 pocketed by clerk
Total = $ 29
Where is the missing $ 1 (= $30 - $29 )?

Solution 3 (Modelling): Monkeys & Coconuts

Let x the min number of coconuts initially.

1st monkey took "a" coconuts away, 2nd monkey "b" coconuts....5th monkey took "e" coconuts.

[a][a][a][a][a] + 1 =x

Loan 4 coconuts to the initial pool of x coconuts to divide by 5 evenly at each monkey.

[a][a][a][a][a] + 1 + 4 = [a][a][a][a][a] + 5 = x+4 = X (inflated x by 4 )
1st Monkey: [a’][a’][a’][a’][a’] = X
a’ = $Latex \frac {1}{5} X$

... Left 4a’= $Latex \frac {4}{5} X$
[a’][a’][a’][a’] => [b][b][b][b][b]

b= $latex \frac{1}{5} $ .4a’ = $latex \frac{4}{25}X$

... Left 4b= $Latex \frac {16}{25}X$
[b][b][b][b] => [c][c][c][c][c]
c=$Latex \frac {1}{5}$ .4b= $Latex \frac {16}{125}X$
... Left 4c= $Latex \frac {64}{125}X$
[c][c][c][c] => [d][d][d]d][d]
d= $Latex \frac {1}{5}$.4c= $Latex \frac {64}{625}X$
... Left 4d= $Latex \frac {256}{625}X$
[d][d][d]d] => [e][e][e][e][e]
e= $Latex \frac {1}{5}$.4d= 256.(X/3125)
Since e is integer
=> X = 3125 or multiples of 3125
Minimum X=3125
x+4 = 3125
x= 3121 = minimum Coconut initially.

Note: This solution used the Singapore Modelling Math taught in all Primary Schools for 11-year-old pupils.

Solution 2 (Eigenvalue): Monkeys & Coconuts

Solution 2: Use Linear Algebra

Eigenvalue equation: A.X = λ.X

A =S(x)= $Latex \frac{4}{5}(x-1)$  where x = coconuts

S(x)=λx
Set λ = 1
$Latex \frac{4}{5}(x-1)=x$
x =-4
Also fifth monkey:
$Latex S^5 (x)$ = $Latex (\frac{4}{5})^5 (x-1)+(\frac{4}{5})^5+(\frac{4}{5})^4+\cdots+\frac{4}{5}$

$Latex 5^5$ divides (x-1)

Minimum positive x= -4 mod ($Latex 5^5$ )= $Latex 5^5$-4= 3121 [QED]

Note: λ=1 because "before" and "after" (transformation A)  is the same.

Solution 1 (Sequence): Monkeys & Coconuts

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)

Monkeys & Coconuts Problem

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）.