# 考考你: $10哪去了? Where is$10?

John borrows $500 from dad and$500 from mom. He uses $970 to buy a pair of shoes, balance$30.
He returns $10 to dad,$10 to mom, keeps $10 for himself. So he owes$490 to dad, $490 to mom,$490 + $490 =$980,
with his $10 =$990.

Where is the missing $10 (=$500 + $500 –$990) ?

# PUZZLE

SOLVE PUZZLE……

I am a 5 letter word.
I am normally below u
If u remove my 1st letter
u’ll find me above u
If u remove my 1st & 2nd letters, u cant see me

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# Car Parking Lot Number

Sometimes solving problem is not straight forward, what about see the problem in the other way ?

Do not scroll down until you get the answer…
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Ans: 87 (see upside down)

# 500 Doors

Five hundreds closed doors along a corridor are numbered from 1 to 500.
A person walks through the corridor and opens each door.
A 2nd person walks through the corridor and closes every alternate door.

Continuing in this manner, the i-th person comes and toggles (opened becomes closed, vice-versa) the position of every i-th door starting from door i.

Question: Which of the 500 doors are open after the 500-th person has walked through.

[HINT]: Solving such abstract problem, it helps to visualize on small sample (eg. 10 doors) to find the pattern.

Initial doors: (All closed)
■■■■■■■■■■
After 1st person (open all doors)
□□□□□□□□□□

After 2nd person
□■ □ ■□ ■ □■ □ ■

After 3rd person (toggles all 3n doors)
□■[■]■□[□]□■[■]■

After 4th person (toggles 4n)
□■ ■[□]□□□[□]■■

After 5th person (toggles 5n)
□■ ■ □[■]□□□■[□]

After 6th person (toggles 6n)
□■ ■ □ ■[■]□□■□

After 7th person (toggles 7n)
□■ ■ □ ■ ■[■]□■□

After 8th person (toggles 8n)
□■ ■ □ ■ ■ ■[■]■□

After 9th person (toggles 9n)
□■ ■ □ ■ ■ ■ ■[□]□

After 10th person (toggles 10n)
□■ ■ □ ■ ■ ■ ■ □[■]
Notice the 3 open doors are: {1, 4, 9}.

Do they give you any clue ?

They are all perfect square:
$1 = 1^{2}$
$4 = 2^{2}$
$9 = 3^{2}$

For 20 doors, we can get the 4 open doors : {1, 4, 9, 16}.
$16 = 4^{2}$

Therefore,
for 500 doors there will be N open doors:
$N = \sqrt {500} = 22.36$
$22^2 = 484$

{1, 4, 9, 16, 25, 36, 49, …, 484}

Proof:
Perfect square has odd number of divisors.
4 : {1, 2, 4} (odd) 3 divisors
9: {1, 3, 9} (odd) 3 divisors
16: {1, 2, 4, 8, 16} (odd) 5 divisors
but
8: {1, 2, 4, 8} (even) 4 divisors

For an open door, after even number of toggles will close it, but odd number of toggles will open it again, hence all doors of perfect square will be open.

# Phone Number Tells Your Age

（1）Take the last digit number of your telephone number

（2）Multiply it by 2

（4）Multiply by 50

[Note: Next year (2014) you add 1764, and so on …]

（6）Last step: subtract your year of birth

Now you should get a 3-digit number

The first digit is your number in (1), followed by next 2 digits give your age.

# Prof Su Buqing Problem

Prof Su 苏步青, the founding pioneer Math professor of the China’s top universities (Zhejiang 浙江大学 and Fudan 复旦大学), was one of the few mathematicians who had longevity above 100 years old (the other was French Mathematician Hadammard).

http://en.m.wikipedia.org/wiki/Su_Buqing

Two men A and B are 100 km apart, walking towards each other, A at speed 6 km/hour and B at 4 km/hour.
A brings a dog which runs at 10 km/hour between them,  starting from A towards B, upon reaching B it runs back to reach A, then back to B again, and so on…

Find total distance the dog has covered when A and B finally meet ?

# 成语数学

1) 20 除 3
2）1 除100
3）9寸+1寸=1尺
4）12345609
5）1,3,5,7,9

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’ = $\frac {1}{5} X$

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

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

… Left 4b= $\frac {16}{25}X$
[b][b][b][b] => [c][c][c][c][c]
c=$\frac {1}{5}$ .4b= $\frac {16}{125}X$
… Left 4c= $\frac {64}{125}X$
[c][c][c][c] => [d][d][d]d][d]
d= $\frac {1}{5}$.4c= $\frac {64}{625}X$
… Left 4d= $\frac {256}{625}X$
[d][d][d]d] => [e][e][e][e][e]
e= $\frac {1}{5}$.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)= $\frac{4}{5}(x-1)$  where x = coconuts

S(x)=λx

Since each iteration of the transformation caused the coconut status ‘unchanged’, which means λ = 1 (see remark below)

$\frac{4}{5}(x-1)=x$
We get
x = – 4

Also by recursive, after the fifth monkey: $S^5 (x)$ = $(\frac{4}{5})^5 (x-1)- (\frac{4}{5})^4-(\frac{4}{5})^3- (\frac{4}{5})^2- \frac{4}{5}$

$S^5 (x)$ = $(\frac{4}{5})^5 (x) - (\frac{4}{5})^5 - (\frac{4}{5})^4 - (\frac{4}{5})^3+(\frac{4}{5})^2 - \frac{4}{5}$

$5^5$ divides (x)

Minimum positive x= – 4 mod ($5^{5}$ )= $5^{5} - 4$= 3,121 [QED]

Note: The meaning of eigenvalue  λ in linear transformation is the change  by a scalar of λ factor (lengthening or shortening by λ) after the transformation. Here

λ = 1 because “before” and “after” (transformation A)  is the SAME status (“divide coconuts by 5 and left 1”).

# Solution 1 (Sequence): Monkeys & Coconuts

Monkeys & Coconuts Problem

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

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

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

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

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

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

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

Since
$U_5$ is integer  ,
$5^5 divides (k+4)$
k+4 ≡ 0 mod($5^5$)
k≡-4 mod($5^5$)
Minimum {k} = $5^5 -4$= 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）.