solution: s=1921-1+[(1921-1) 4-(1921-1) 100+(1921-1) 400]+(31+28+31 +30+31+30+1)...

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solution: S=1921-1+[(1921-1)4-(1921-1) 100+(1921-1)400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since the remaining is 5.So ,1st July is Friday.

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Page 1: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

solution:

S=1921-1+[(1921-1)4-(1921-1) 100+(1921-1)400]+(31+28+31

+30+31+30+1)

=1920+480-19+4+182

=2567

2567/7=366……5

So,the answer is NO.Since the remaining is 5.So ,1st July is Friday.

Page 2: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

• How the postman should go?

One postman had to go to post a letter. The area that he had to go just liked the area below:

Everyday,he starts from the post office (o) and he had to go through all the streets in this area.To take the shortest distance,we had to consider how can the postman goes? Actually,this is a “one line draw question”

( 一 筆劃 ).Its ending point and starting point is the same in the post office (o).

Page 3: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

According to the theory of “one line draw question”( 一 筆劃 ),if the route is not repeated,the special point in this diagram is at most 2 .

But the diagram below is A,C,E,G these points are special points. So, to walk through all the streets ,it is not possible not to repeat the road.

But ,what is the shortest distance?

Page 4: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

Answer:

The way to go is oABCDEGAGCDEFO

Page 5: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

• The reason that the iron gate can be small with a slightly push .

Although the gate in the shop are heavy,it can be closed easily.

If you observed the gate carefully, it is not difficult to find a correct answer----the gate is made up of rhombus or parallelogram. But why the rhombus or parallelogram can make the gate close and open easily?

Page 6: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

It is because the properties of these shapes are called as the unstable polygon. This property can make the gate close and open easily and it is useful in daily life.

Page 7: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

4) Literature

In the Chinese literature , we can find many poems that is related to Mathematics . Such as

湯顯祖《牡丹亨》十年窗下,遇梅花凍九才開。夫貴妻榮八字安排。敢你七香車穩情載,六官查有你朝拜,五花誥封你非分外。論四德似你那三從願偕。二指大泥金報喜,打

個傘蓋飛來。This poem is arranged with numbers from 1 to 10 in a

descending order.

Page 8: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

蘇軾春色三分,二分塵土,一分流水,細看來,不是楊花,點點行人淚。

The composer shows his sorrow in a Mathematical way ----3+2=1.

<<南史 謝靈運傳 >>

天下才共一石,曹子建獨得八斗,我得一斗,自古及今共用一斗。

The tricky point of this poem is that it included a sum 10=8+1+1

Page 9: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

紀盷一片兩片三四片,五片六片七八片,九片十片片片

飛,飛人蘆花皆不見。

When you read this poem you can feel the rhythm is very easy since some numbers are added in the poem.

This two poems below also use the numbers for composing.

李賀《夢天》黃塵清水三山下,更變千年如走馬。遙望齊州九點

煙,一泓海水杯中瀉。

Page 10: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

杜甫《絕句》兩個黃鸝鳴翠柳,一行白鷺上青天。窗含西嶺千秋雪,門泊東吳萬里

Page 11: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

《水仙子》程大位元宵十五鬧縱橫來往觀燈街上行我見燈上下紅光映繞三遭,數不真從頭兒三數無零五數時四甌不盡七數時六盞不停端的是幾盞明燈The fifth to the seventh sentences mean

all nights will be taken if you take three a time,

four will be left if take 5 a time ,

and six will be left if take seven a time.

Page 12: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

After look at some many Chinese poems, let look at this English poem, it is a poem and also a question.See weather you can find the answer.

JUST FOR TODAY

If I were to tell you that......

When the day before yesterday was

referred to as “the day after tomor-

row “,the day that was then called “yes-

terday”was as far away from the day we

now called “tomorrow”as yesterday is

from the day on which we shall be able to

speak of last Monday as “a week ago

yesterday”.

Page 13: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

On what day of the week would I be making that statement? Of course that I was telling the truth.

The answer is:

THURSDAY

This poem can only be true on Thursday because at that day,the distance between last Saturday and tomorrow (Friday ) is 6 days,and the distance between yesterday (Wednesday) and next Thursday are also 6. So today is Thursday!

Page 14: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

Have you watch the movie << 城南事 >> ?

There is a song in the movie:蟲蟲蟲蟲,飛!蟲子,蟲子,一大堆!A Taiwan teacher after watching this movie,he have formed a formula :

蟲蟲蟲蟲 x 飛 = 蟲子 x 蟲子 + 一大堆

Page 15: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

In this formula, one Chinese word represents one number .

For example , 蟲 represents 6

飛 represents 7

then 蟲蟲蟲蟲飛 is 66667.

Consider 飛 is a one digit number, 蟲子 is a two digits number, 一大堆 is a three digits number.

It has only one solution, can you find the answer?

Page 16: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

Solution:

蟲蟲蟲蟲 x 飛 ,will the answer be a 5 digits number?

Think carefully ,it is impossible.Since in the R.H.S.,the biggest number of 蟲子 is 98,

then 蟲子 x 蟲子 is 98x98,after added a 3 digits number,though it may be a 5 digits number,but this number is not much different from 10000.Twice 9999 is 19996,is has a great different from 10000, so they cannot be equal.

Page 17: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

Next, look at the 4 digits number 蟲蟲蟲蟲 ,will it be 8888?

If it is true then 飛 must be 1.Since 89 89=7921.8888-7921=967.So the word 一from 一大堆 can only be represented by 9.But 9 has already be used so 89 is not the answer.What about 87?Since 8787=7569 and 8888-7569 is a 4 digits number,it can represent 一大堆 these three words.

Page 18: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

Similarly, if 蟲蟲蟲蟲 is 7777,then 79 79=6241 and the sum of 7777-6241 is a 4 digits number, it must be wrong ,so you don’t have to try the other,like 6666 or 5555 etc.

As the result, the only possible answer is 9999.So the formula is

9999x1=9696+783.

Page 19: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

<<再別康橋 >>is a very famous poem which composed by 徐志摩 ,there is two sentences which almost known by everybody.輕輕的我走了,正如我輕輕的來。Also someone has form a mathematics formula like that

輕輕的 = 我 +走了正 -如我 = 輕輕的 來Use the same rule as the before case, and this also has one so

lution.

Page 20: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

SOLUTION:

First, the numbers which fit 輕輕的 is only

15 15=225 OR

21 21=441

Next,we focus on 我 , the numbers that can satisfy this formula is only 0,1,4or 9.But 0 cannot be accepted.So 我 and 來 cannot be 0. 輕輕的 only can be 15 or 21, both of them are odd numbers,so 來

can only be an odd number. 來 is impossible be 1,because a two digits number divided by a one digit number is still a two digits number.And the answer i

n the L.H.S. can only be a one digit number .

Page 21: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

As the result, we know that 來 can only be 9 and 來 is 3.

Also we can prove that 我 =1 is wrong.

Since 15=1+14

This time 我 and 走 this two different words both are 1, it does not satisfy the rule,so it is wrong.

If we change the number to 441,then 441=1+20.

Now , 的 and 我 both are 1, so it also wrong.

Then the number represents 我 is 4.

Page 22: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

And the final ANSWER is

225= 4+13 and

7-8 4= 225 9

that is

15=2+13 and

7-8 4=15 3

Page 23: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

5) interesting Games1. Can you form a greatest number using

three 9 within 3 seconds?

Answer

999

Page 24: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

2.In a casino a man said to the people there “I close my eyes and you throw two dice, I can guess what the numbers are.If I say the wrong answer, I can give you all my money with me now.”

Then the people do what the man said.

Page 25: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

The man said “you pick one of the dice and multiply the number on the die by 2 and add 5 to it and times 5 .Now add the number of the other die that you throw before to the result you’ve just got.”

The people said “89!”

That man said immediately “The numbers are 6 and 4.”

And the answer is CORRCET.

SOLUTION:

You just have to minus 25 from 89.The result is 64,then the numbers are 6 and 4.

Page 26: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

3.Two friends are trapped in a hill for 3 days but they do not have any food or drinks accept a bottle of 24ml water.They wanted to share the water equally in order to make it fair.One of the people remember that there are one 11 ml and two 5 ml empty bottles in his bag.How can they share the 24 ml water equally with the help of these bottles?

Page 27: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

Solution

Order 1 2 3 4 5 6 7

Bottlewithwater

20 13 13 18 18 23 23 12

11 mlbottle 0 11 6 6 1 1 0 11

5 mlbottle 0 0 5 0 5 0 1 1

0

Page 28: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

4.If you ask a lady what is her age,she must think that you are impolite.But now, doesn’t

matter, because you’ve got an idea.

First, you can say “Let me guess your age, please answer my question, but don’t do the wrong calculation.Twice your age ,then minus your age by 1 and twice it.Finally, minus the second result from the first result.What is the answer?”

When the lady says “41”.

Then you know that she is 21 year old.

Page 29: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

SOLUTION: 41 plus 1 and divided by 2 is 21.

Page 30: solution: S=1921-1+[(1921-1)  4-(1921-1)  100+(1921-1)  400]+(31+28+31 +30+31+30+1) =1920+480-19+4+182 =2567 2567/7=366……5 So,the answer is NO.Since

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