Monday, 9 March 2015

每周数学 (十二): Consecutive Sums 21/03/2015

*Consecutive Sum*

Some numbers can be expressed as the sum of a string of consecutive  positive numbers,

Exactly which numbers have this property?

1. What are the numbers have no consecutive sum? Odd or even integers? Is there anything to  do  with  average?

2. Exactly How many solutions will it be? 

3.How to determine the number of solutions? What is the Methodology?

4.Fn= ?

5. 1=, 2= 3=, 4=, 5=, 6=, 7=, 8=, 9=,10=,…

For example, observing that:

5=2+3 
9=2+3+4 =4+5  
11=5+6  
18=3+4+5+6 =5+6+7


What are the consecutive numbers that sum to 30?       105=?        

*连续总和*

有些数字可以表示为一串连续的正数之和,

究竟哪些数字具有此属性?

1. 没有连续和的数字有哪些? 奇数还是整数? 与平均值有什么关系吗?

2. 究竟有多少个解决方案?

3.如何确定解决方案的数量? 什么是方法论?

4.Fn= ?

5. 1=, 2= 3=, 4=, 5=, 6=, 7=, 8=, 9=,10=,...

例如,观察到:

5=2+3
9=2+3+4 =4+5
11=5+6
18=3+4+5+6 =5+6+7


总和为 30 的连续数字是什么? 105=?

Solutions
*Consecutive Sum*?

1. A number consists of at least one odd factor could be expressed as sum of consecutive numbers.

2. Therefore numbers 2 and all the power of 2 will not have consecutive Sum.

3. examples:
 
3=1+2
5= 2+3
6=1+2+3
7=3+4
9=4+5=2+3+4

......

*15=15x1=5x3=3x5*
15x1: ave=1
=-6-5-4-3-2-1+0+1+2+3+4+5+6+7+8
*=7+8*
5x3
=1+2+3+4+5

3x5
=4+5+6

30=3x10=5x6=15x2
=9+10+11
=4+5+6+7+8

*15x2*
=-5-4-3-2-1+0+1+2+3+4+5+6+7+8+9
=6+7+8+9



🙂🙂🙏🙏

 

Wednesday, 4 March 2015

每周数学(十一): Traffic Jam (also know as Fishing Boat or Leap-Frog) 14/03/2015

*Fishing boat*

Ten Men are fishing from a boat, five in the front, five in the back, and there is one empty seat in the middle.

The five in front are catching all the fish, so the five at the back want to change seats.

To avoid capsizing the boat, they agree to do so using the following rules:

1.A man may move from his seat to and empty seat next to him.

2.A man may step over only one man to an empty seat.

3.No other move are allowed.

What is the minimum number of moves necessary for the men to switch places?

If there are n men from each side, how many moves is needed for the swap?

*渔船* 

十个人在船上钓鱼,五个在前面,五个在后面,中间有一个空座位。 

前面的五个人都有鱼获,所以后面的五个人想换座位。 


为避免翻t船,他们同意使用以下规则: 

1.一个人可以从他的座位移动到他旁边的空座位。 

2.个人只能跨过另一人到一个空位。 

3.不允许其他动作。 

交换位置所需的最少移动次数是多少呢? 

*如果双方各有 n 个人,交换需要多少个步骤?*



Leapfrog


no of   no of    no of   total no

pegs   slides  Jumps  Moves

1.          2.           1.         3

2.          4.           4.         8

3.          6.           9.        15

4.          8.           16.       24


n.         2n.         n^2.    n(n+2)


Strategic:

1. Simplify the problems

2. Critical concepts, critical moves?

3. Total shifts : 2n(n+1)

              Jumps: n^2

1 jump =2 shifts

Total Slides

= Totalshifts-2x total jumps

= 2n(n+1)-2n^2

=2n

Tuesday, 3 March 2015

每周数学(九) :The Tower of Hanoi 河内之塔 28/02/2015

*Tower of Hanoi*

consists of three pegs or towers with n disks placed one over the other. The objective of the puzzle is to move the stack to another peg following these simple rules. Only one disk can be moved at a time. No disk can be placed on top of the smaller disk.

*How many moves is required to move the eight dices as shown?*


*河内之塔*

由三个柱子组成,
有 n 个大小不一的圆盘堆叠成塔,大的放在小的之上。 

目标是按照这些简单的规则将圆盘移动到另一个柱子上。
 一次只能移动一个圆盘。 
任何圆盘都不能放置在较小圆盘的上面。

如图所示
*移动八个圆盘需要多少个步骤呢?*

*Tower of Hanoi*
*Mathematical Recursion*

no of 
disks.     no of moves
1.     1.                   1
2.     3.     1+1+1.  1+2
3.     7.     3+1+3.   1+2+2^2
4.   15.  7+1+7. 1+2+2^2+2^3

n.   Fn.     2F(n-1)+1
Fn=1+2+2^2+2^3+..  +2^(n-1)
2Fn=2+2^2+2^3+....  +2^n

Fn=2^n-1

*请问 若是有64个圆盘;需要多长的时间完成?*😀😀😀

Weekly Maths 8: Sum to 20 21/02/2015


Select all the cards with 1 to 5

You are now having a pool of cards with 4 sets of cards from 1 to 5, all cards are open, facing up.

Play between 2 players (0r 2 teams of players)

The players take turns to choose a card from the pool, and sum up the numbers of all the cards selected from both players

Whoever gets the last card that the total sum reaches 20 win the game.
Who will win? How?

Weekly Maths 7 : The Three Mathematicians 14/02/2015

3 Mathematics Tutors are seated one behind another.




Another person showed them 3 grey hats and 2 white hats, blindfolded them, put one hat on each head, and threw the rest away.  When the blindfolds  were off, they all looked in front of them.




Each was asked in turn what colour hat she or he had.  No one could answer. After a long thoughtful silence, the person in front who could see no one’s hat was able to respond correctly about his own hat. 

N


How was this done ?

👍😄😄👏👏👏✅
排列组合

1。W - W✅ -G✅
2。W - G✅  - W/G?
3。G✅ - W/G?  - W/G?

7 combinations

好!

Person at the back:
If he saw *2 white hats* in front, his hat *must be grey*

If he cannot guess, it means the first 2 people have *at least 1 grey hat*

Person in the middle:
If he sees *white hat in front*, it means his hat *must be grey*

If he cannot guess too, it means the front guy's hat is *grey*. 

Therefore, when they both do not answer, the front guy can deduce *his hat is grey!*

Weekly Maths 6 : Fibonacci Magic 07/02/2015



Get two participants as Volunteers

Each of them suggests a number, any number between 1 to 20.


The third number is the sum of the first two numbers, the forth number will be the sum of second & third number, so on and so forth,

The subsequence number will be the sum of the previous two numbers, until you have all the 10 numbers

Now, ask the volunteers to add up all the 10 numbers.

( Someone will be able to tell you the SUM well before they have completed the calculation. Why?)

Weekly Maths 5 :The Singapore Polytechnic Lockers 31/01/2015


At Singapore Polytechnic, there were 1,000 students and 1,000 lockers
(numbered 1-1000).

At the beginning of our story, all the lockers were closed.

The first student come by and opens every locker.

Following the first students, the second student goes along and closes every second locker.

The third student changes the state, ( if the locker is open, he closes it; if the locker is closed, he opens it) of every third locker.

The fourth student changes the state of every fourth locker, and so forth.

Finally, the thousandth student changes the state of the thousandth locker.



When the last student changes the state of the last locker,

Which lockers are open?

每周数学(四):韩信点兵 24/01/2015 Remainder Theorem

每周数学(四):韩信点兵
24/01/2015

淮安民间传说着一则故事——“韩信点兵”,其次有成语“韩信点兵,多多益善”。韩信带1500名兵士打仗,战死四五百人,站3人一排,多出2人;站5人一排,多出4人;站7人一排,多出6人。韩信马上说出人数:1049。
在一千多年前的《孙子算经》中,有这样一道算术题:“今有物不知其数,三三数之剩二,五五数之剩三,七七数之剩二,问物几何?”按照今天的话来说:一个数除以3余2,除以5余3,除以7余4,求这个数。这样的问题,也有人称为“韩信点兵”。它形成了一类问题,也就是初等数论中的解同余式。
①有一个数,除以3余2,除以4余1,问这个数除以12余几?
②一个数除以3余2,除以5余3,除以7余2,求符合条件的最小数。




1。《孙子算经》下巻26题
“今有物不知其数,三三数之剩二,五五数之剩三,七七数之剩二,问物几何?
2。韩信带1500名兵士打仗,战死四五百人,站3人一排,多出2人;站5人一排,多出4人;站7人一排,多出6人。韩信马上说出人数:.......。


解法:余数定理
参考资讯;余数定理
http://youtu.be/1LZ1Hqaw8BY
http://youtu.be/IRHuYTS6r68

http://episte.math.ntu.edu.tw/articles/sm/sm_01_01_2/page6.html

言兵莫过孙武,用兵莫过韓信

西汉·司马迁《史记·淮阴侯列传》:上问 曰:“如我能将几何?”信曰:“陛下不过能 将十万。”上曰:“子有何如?”曰:“臣多 多而益善耳。”

典故:
胯下之辱
成也萧何,败也萧何!
韓信点兵;多多益善!
明修栈道,暗渡陈仓;
背水一战;十面埋伏.....

谢谢建平同学提供...

每周数学(三):百钱买百鸡🐔17/01/2015 Problem of the hundred Fowls

每周数学(三)17/01/2015
Maths Questions for your children
👍誏儿孙们思考的数学!

Q1.  百钱買百鸡
今有鸡翁一,值钱伍;鸡母一,值钱三;鸡鶵三,值钱一。凡百钱买鸡百只,问鸡翁、母、 鶵各几何?

我特别喜欢古代数学家张丘建在《算经》一书中提出的这数学问题。有好多解法!也有好几个答案;(非学校传统式的標准试题)(又含人生哲理)

Q2. 和尚食饅头
一百个和尚吃一百个饅头
大和尚一人食三个,小和尚三人吃一个饅头。问大小和尚各几人?

💝
请享受解题的美妙过程!无穷乐趣。
Please enjoy the thinking process, the approaches and concepts.  😄
祝福安康

“百钱买百鸡” 趣味数学的啟示和哲理

特别喜欢古代数学家张丘建在《算经》提出的这一道数学问题;从思考解析中, 可以悟出一些人生哲理.

中小学教导的数学;一般上都是一个解法;一个答案;也许会造成学生往往习惯性地以为;在处理人生的问题上;只有一个方法, 一个过程;一个答案;一个结剧;这和实际处理事务的人生经验就有很大的出入.
 
这题“百钱买百鸡”;有许多巧妙的解法;答案也绝对不止一个(四组不同答案);思维模式的训练就可以应用到人生实战中了!😃😃😃😃



原题:
鸡翁一值钱五,鸡母一值钱三,鸡雏三值钱一。百钱买百鸡,问鸡翁、鸡母、鸡雏各几何?
翻译成白话文则是:
公鸡一只5塊钱;母鸡一只3塊钱, 三只小鸡一塊钱; 问:一百塊钱能买公鸡母鸡小鸡各几只?

数学人人会解;巧妙各有不同

1. 把复杂的问题简化之, 先解决简单的问题; 然后回头面对现实生活中复杂的问题

简化题:
一百个和尚吃一百个饅头; 大和尚一人吃三个, 三个小和尚吃一个饅头. 问:有多少个大和尚?多少位小和尚?

公元二千年, 就读小学三年级的小儿子提出了他的解法
假设只有四个和尚,一个大和尚吃三个饅头,三个小和尚吃一个饅头;四个和尚吃四个饅头;
答案就一目了然了....!

2.  增减平衡法
从简化法总结出25个大和尚吃75个饅头, 75个小和尚吃25个饅头, 刚好一百个和尚吃一百个饅头

回归到原题的百钱买百鸡的问题;其中一组答案不就是
O只公鸡;25只母鸡;75只小鸡吗!
公鸡比母鸡贵二塊钱
三只小鸡比三只母鸡便宜八塊钱
四只公鸡比四只母鸡贵八塊钱
一百塊钱维持不变
少买7只母鸡就能多买四只公鸡和三只小鸡了!
另一组答案就显而易见了
0 25 75
4 18 78
以此类推
8 11 81
12 4 84
共四组答案

3. 三元一次方程式 (无解还是多解?)

假设
公鸡的数目为 x
母鸡为 y
小鸡为 z
 
只能夠例出二个独立方程式
x+y+z=100 只-----------1
5x+3y+z/3 =$100-------2

解得:
7x+4y=100 
结果並非无解,而是多解
x=0, y=25
x=4, y=18
x=8, y=11
x=12, y=4
Z 也就不言而喻了

总结
从解答这道数学题目中;悟出了一个至关重要的人生哲理;
处理事物往往有多种不同的方案;必须从各方面进行考虑;处理事情的结果也会有许多不同的可能性 .

祝福大家
安然无恙

陈宗兴
Teng Chong Heng
91066168
16 Phoenix Rise
Hua Mei Garden
S668218
tengchongheng@gmail.com
(退休理工学院讲师 工程师)
😃😃🙏🙏🙏

每周数学(二):The CHOCOLATE BARS。10/01/2015

每周数学(二)10/01/2015
Maths Questions for your children
👍誏儿孙们思考的数学!

Q1. A Mathematics teacher had a certain number of CHOCOLATE BARS, he gave them to the top three students as prizes.
The top students received two third of the chocolate bars plus one third of a bar , the second student received two third of the remaining plus one third of a bar, and the third student received  two third of the remaining plus one third of a bar and had received only one bar of chocolate.
How many chocolate bars did the teacher have initially?

Q2 Similar to question 1,
If the teacher have to give away chocolate bars as prizes to 4 or 5 top students, how many chocolate bars must he have?
Q3 Similarly, what happen if the pattern of 2/3 changes to 3/4 or 4/5?

数学每周(二)10/01/2015
给孩子的数学题
👍誏儿孙们思考的数学!

 一位数学老师有一定数量的巧克力棒,他把它们作为奖品发给了前三名的学生。
最优秀的学生得到所有巧克力棒的三分之二加上一塊巧克力棒的三分之一,第二名学生得到剩余的三分之二和一塊巧克力棒的三分之一,第三名学生得到剩下的三分之二和一块巧克力棒的三分之一, 只收到了一块巧克力。
老师最初有多少块巧克力?

Q2 与问题 1 类似,
如果老师必须将巧克力棒作为奖品分发给 4 或 5 名优等生,那么他必须拥有多少块巧克力棒?
Q3 同样,如果 2/3 的模式变为 3/4 或 4/5 会发生什么?



💝
请享受解题的美妙过程!
Please enjoy the thinking process, the approaches and concepts.  😄
祝福安康

Solutions:
*Chocolate bars*

1.       3.         9.        27.  ....
1.       4.         13.      40 ....

13x(2/3)+1/3 =9
13-9=4

4x(2/3)+1/3 =3
4-3=1

1x(2/3)+1/3 =1

每周数学(一): PSLE2000 03/01/2015

每周数学(一)
Maths Questions for ....
👍给....的数学题目!
Q1. What is the last digit of the sum from 1 to 97?
有几个解法?
(Year 2000 PSLE Q15)👌

Q2. Magic SQ 3x3, and 4x4 怎样解?
💝
请享受解题的美妙过程!
Please enjoy the process, the approaches and concepts.  😄



Q1 
1。成人解法
S=n(n+1)/2.......

2。儿童解法

1
2
3+97=100
4+96=100
5+95=100
.
.
48+52=100
49+51=100
50

Answer: 3

.
Q2.

1+2+......+9=9x10/2=45
45/3=15

9+1+5 =15
9+2+4 =15

“一道小六会考数学题所引发的思考和理念”

公元二千年
小六会考数学有五十道题目
其中第十五题是
15. What is the last digit of the sum from 1 to 97?
求 1至97 总和的个位数

请问您怎么解答这样的数学题目?思考的过程又是如何呢?

1. 考试的策略;生活中处理问题的方法.
这是第十五道题;有学生为了这解的焦头烂额;没有剩余的精力去回答其余的三十五道题目
每一个考卷;总会有几道题目特别困难有趣;这是为了分辨顶尖优等生而设置的;不会做也没什么大不了.
应考如果采取较好的策略;遇到十分困难的题目, 稍微考虑一下;应该暂时搁置一边;等做好了其他四十多道题之后;再回头面对具有挑战性的难题;换位思考一下;问题也许就迎刃而解也说不定. 生活上的难题也应如是;暂时放下並不意味放弃;总要面对接受问题,然后处理放下. 並不一定要完美的解决.

2. 等差级数方程式的联想

一般同学都会应用等差级数的代数方程式去求的总和;但是往往死背方程式, 忘了初衷, 忘了概念;不知道如何研算方程式;忘了以图形巩固概念.

S=1+2+3+.....+n
S=n+(n-1)+......+1
2S=nx(n+1)

S=97x98/2=97x49=....3

用图解那就更为明确了(见附图)
如果记著基本概念, 随时能够演算出方程式;那就百无一失了,岂不妙哉!

3. 换位思考 活学活用

灵活的运用概念, 就能够以简单明快的方法来解答问题了!

1
2
3+97=100
4+96=100
....
49+51=100
50

答案也就一目了然了!

要求得级数的总和也不难
一共有(49-2)个100

总和
= 47x100+50+3
= 4753
筒单清晰,多么好呀!

总结:
趣味数学 老少皆宜 乐趣盎然; 还可以增进亲子关系;不时锻练一下脑筋, 也许会延后推迟老年痴呆症的问题
...

Mathematics Tuition with Enrichment Program 数学培训课程

Mathematics Tuition with Enrichment  program. 
数学培训班
@16, Phoenix Rise Singapore 668218

1. Secondary 3 & 4, (O level)
with Additional Maths. 
Thinking Skills, Problem Solving Strategies,趣味数学,Logics. 
Tuesday, 4-6 pm
Full fees: $600 for 4 lessons
starting 31/03/2015

2. Primary 5 & 6 (PSLE)
With Mathematical Magic, Games and Puzzles. 奥数。
Learning Mathematics in an Enjoyable way. 
Full fees: $500 for every 4 lessons
Commencing: 02/04/2015

Please note:
1.Personally coached by former Polytechnic senior lecturer with excellent in Teaching award. One of the best teachers in motivating students to learn. 
2.Registration: tengchongheng@gmail.com or
 before 26/03/2015
3. Limited places (5)available
Only one group each. 
4. Small group training(3-5)
5. discount to be given to Phoenix, Bukit Gombak, Chao Chu Kang residents and Children ofCatholic High School and Mountbatten School Alumni and children of Amtek/EM Tools old friends
6. Special Concessions or even FOC for students from lower income families.

(这是我退休后的活动之一)
请介绍有志于学,但家境贫困的孩子们!谢谢!