[問題]數學競試題

[問題]數學競試題

galaxylee 於 星期三 九月 14, 2005 9:20 am


左鍵: 點擊縮放; 右鍵: 觀看原圖
(91年全國高中數學能力競試台中區賽題)

galaxylee
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qeypour 於 星期三 九月 14, 2005 1:38 pm


可否請galaxylee兄提供全國高中數學能力競試之連結
謝謝

qeypour
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galaxylee 於 星期三 九月 14, 2005 1:49 pm


qeypour 寫到:可否請galaxylee兄提供全國高中數學能力競試之連結
謝謝



高雄大學應數系游森棚教授網站有2003~2005年的試題
在中學數學參考資料分頁下
http://umath.nuk.edu.tw/~senpengeu/

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Re: [問題]數學競試題

chanjunhong 於 星期三 九月 14, 2005 3:42 pm


galaxylee 寫到:左鍵: 點擊縮放; 右鍵: 觀看原圖
(91年全國高中數學能力競試台中區賽題)


分析後的性質:
p and q are primes

f(pXq)=f(p)Xf(q)
f(p^(n))=f(p)Xp^(n-1)

所以結論是:
f(p)Xf(q)/(pXq)

chanjunhong
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chanjunhong 於 星期四 九月 15, 2005 9:08 am


抱歉,因為時間的關係沒法一次證完,利用偷懶的時間打。
p is prime and >2
assume that set(p)={1,2,3,4,......p-2,p-1,p}
                   fun(p)={x|x is element of set(p), such that (x,p)=(x+1,p)=1}
                   f(p):means that counting elements of  fun(p)
than:f(p)=(p-2) ,and f(p^(n))=f(p)Xp^(n-1)

(1):f(p)=p-2
    Because that p-1 and p are not elements of fun(p),
    but, for all i if 1<= i <= p-2
    (i,p)=(i+1,p)=1, so  1<= i <= p-2 is element of fun(p).
    hence, f(p) =p-2  for all p is a prime without  p=2
ex  f(3)=1, f(5)=3

(2):f(p^(2))=f(p)Xp
     set(p^(2))={1,2,..p-2,p-1,p,p+1,p+2,..2p-2,2p-1,2p,..pxp-2,pxp-1,pxp}
     Because that (i,p)=(p+i,p)=(nxp+i,p)=1 for all i=1,2,....p-1
                                                             and for all n=0,1,2,.....
     so all of nxp+i is an element of fun(p)  if n=0,1,.p-1
                                                             and i=1,2,3,....p-2
     hence f(p^(2))=f(p)xp

(3): use this we can get the formula :f(p^(n))=f(p)xp^(n-1)

chanjunhong
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chanjunhong 於 星期四 九月 15, 2005 10:45 am


證明:p and q are two different primes without 2, such that
       f(pq)=f(p)xf(q)

also assument that set(pq)={1,2,3,........,pq}
in the set of set(pq), there are [pq/p] integers which are multiples of p.
also there are [pq/q] integers which are multiples of q
,but pq is counting twice.

must counting two sets { p-1, 2p-1,.....,pq-1} and {q-1, 2q-1, 3q-1,.....,pq-1}

#{ p-1, 2p-1,.....,pq-1} =[pq/p]
#{q-1, 2q-1, 3q-1,.....,pq-1}=[pq/q]

but also some integers counting twice
{ p-1, 2p-1,.....,pq-1} has multiples of q,
there are [[pq/p]/q]

{q-1, 2q-1, 3q-1,.....,pq-1} has multiples of p
there are [[pq/q]/p]

hence f(pq)=pq-2[pq/p]-2[pq/q]+4
                =pq-2p-2q+4=(p-2)(q-2)=f(p)Xf(q)

ex set(15)={1,2,.....,15}
               ={3,6,9,12,15}U{5,10,15}
               U{2,5,8,11,14}U{4,9,14}U{1,7,13}

f(15)=15-2[15/3]-2[15/5]+4
       =15-2x5-2x3+4
       =(3-2)x(5-2)=f(3)xf(5)


finially, we also can get another formula:
f(p^(a)q^(b))=p^(a)q^(b)X(1-2/p)X(1-2/q)

chanjunhong
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qeypour 於 星期五 九月 16, 2005 8:01 am


chanjunhong 寫到:證明:p and q are two different primes without 2, such that
       f(pq)=f(p)xf(q)

also assument that set(pq)={1,2,3,........,pq}
in the set of set(pq), there are [pq/p] integers which are multiples of p.
also there are [pq/q] integers which are multiples of q
,but pq is counting twice.

must counting two sets { p-1, 2p-1,.....,pq-1} and {q-1, 2q-1, 3q-1,.....,pq-1}

#{ p-1, 2p-1,.....,pq-1} =[pq/p]
#{q-1, 2q-1, 3q-1,.....,pq-1}=[pq/q]

but also some integers counting twice
{ p-1, 2p-1,.....,pq-1} has multiples of q,
there are [[pq/p]/q]

{q-1, 2q-1, 3q-1,.....,pq-1} has multiples of p
there are [[pq/q]/p]

hence f(pq)=pq-2[pq/p]-2[pq/q]+4                =pq-2p-2q+4=(p-2)(q-2)=f(p)Xf(q)

ex set(15)={1,2,.....,15}
               ={3,6,9,12,15}U{5,10,15}
               U{2,5,8,11,14}U{4,9,14}U{1,7,13}

f(15)=15-2[15/3]-2[15/5]+4
       =15-2x5-2x3+4
       =(3-2)x(5-2)=f(3)xf(5)


finially, we also can get another formula:
f(p^(a)q^(b))=p^(a)q^(b)X(1-2/p)X(1-2/q)


個人覺得紅字部份的4是2+2得來的
第一個2代表pq及pq-1這兩個數
第二個2代表px-1及py這兩個數(滿足px-1=qz,py=qt-1)
不知筆者的4是如何產生
可否詳敘?

qeypour
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文章: 431
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訪客 於 星期五 九月 16, 2005 9:43 am


qeypour 寫到:
chanjunhong 寫到:證明:p and q are two different primes without 2, such that
       f(pq)=f(p)xf(q)

also assument that set(pq)={1,2,3,........,pq}
in the set of set(pq), there are [pq/p] integers which are multiples of p.
also there are [pq/q] integers which are multiples of q
,but pq is counting twice.

must counting two sets { p-1, 2p-1,.....,pq-1} and {q-1, 2q-1, 3q-1,.....,pq-1}

#{ p-1, 2p-1,.....,pq-1} =[pq/p]
#{q-1, 2q-1, 3q-1,.....,pq-1}=[pq/q]

but also some integers counting twice
{ p-1, 2p-1,.....,pq-1} has multiples of q,
there are [[pq/p]/q]

{q-1, 2q-1, 3q-1,.....,pq-1} has multiples of p
there are [[pq/q]/p]

hence f(pq)=pq-2[pq/p]-2[pq/q]+4                =pq-2p-2q+4=(p-2)(q-2)=f(p)Xf(q)

ex set(15)={1,2,.....,15}
               ={3,6,9,12,15}U{5,10,15}
               U{2,5,8,11,14}U{4,9,14}U{1,7,13}

f(15)=15-2[15/3]-2[15/5]+4
       =15-2x5-2x3+4
       =(3-2)x(5-2)=f(3)xf(5)


finially, we also can get another formula:
f(p^(a)q^(b))=p^(a)q^(b)X(1-2/p)X(1-2/q)


個人覺得紅字部份的4是2+2得來的
第一個2代表pq及pq-1這兩個數
第二個2代表px-1及py這兩個數(滿足px-1=qz,py=qt-1)
不知筆者的4是如何產生
可否詳敘?

15,14,9,5 都扣兩次,有四個重複

訪客

 

chanjunhong 於 星期五 九月 16, 2005 10:34 am


Anonymous 寫到:
qeypour 寫到:
chanjunhong 寫到:證明:p and q are two different primes without 2, such that
       f(pq)=f(p)xf(q)

also assument that set(pq)={1,2,3,........,pq}
in the set of set(pq), there are [pq/p] integers which are multiples of p.
also there are [pq/q] integers which are multiples of q
,but pq is counting twice.

must counting two sets { p-1, 2p-1,.....,pq-1} and {q-1, 2q-1, 3q-1,.....,pq-1}

#{ p-1, 2p-1,.....,pq-1} =[pq/p]
#{q-1, 2q-1, 3q-1,.....,pq-1}=[pq/q]

but also some integers counting twice
{ p-1, 2p-1,.....,pq-1} has multiples of q,
there are [[pq/p]/q]=1

{q-1, 2q-1, 3q-1,.....,pq-1} has multiples of p
there are [[pq/q]/p]=1

hence f(pq)=pq-2[pq/p]-2[pq/q]+4                =pq-2p-2q+4=(p-2)(q-2)=f(p)Xf(q)

ex set(15)={1,2,.....,15}
               ={3,6,9,12,15}U{5,10,15}
               U{2,5,8,11,14}U{4,9,14}U{1,7,13}

f(15)=15-2[15/3]-2[15/5]+4
       =15-2x5-2x3+4
       =(3-2)x(5-2)=f(3)xf(5)


finially, we also can get another formula:
f(p^(a)q^(b))=p^(a)q^(b)X(1-2/p)X(1-2/q)


個人覺得紅字部份的4是2+2得來的
第一個2代表pq及pq-1這兩個數
第二個2代表px-1及py這兩個數(滿足px-1=qz,py=qt-1)
不知筆者的4是如何產生
可否詳敘?

以例子來說,15,14,9,5 都扣兩次,有四個重複 ,就如同你所說的

chanjunhong
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qeypour 於 星期五 九月 16, 2005 1:21 pm


f(pq)=f(p)*f(q) require that p,q are primes
Why can you apply this formula to f(p^a*q^b)  while  p^a,q^b
both   are  not  primes?

qeypour
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訪客 於 星期五 九月 16, 2005 1:47 pm


qeypour 寫到:f(pq)=f(p)*f(q) require that p,q are primes
Why can you apply this formula to f(p^a*q^b)  while  p^a,q^b
both   are  not  primes?


It is good question,you can use the same way to extend the formula.
Or, first, you can finish the formula f(p^(a)q)=f(p)Xp^(a-1)Xf(q).
Then, do f(p^(a)q^(b))=f(p^(a))f(q^(b))=f(p)xp^(a-1)Xf(q)q^(-1)


the formula like f(n) defined
f(n)=#{x|1<=x<=n,(x,n)=1}

the function have some theorems.
1. f(p)=p-1;
2. f(pq)=f(p)f(q)
3  if n=p*q*r with  p,q,r are different primes
    f(n)=n*(1-1/p)*(1-1/q)*(1-1/r)

訪客

 

qeypour 於 星期五 九月 16, 2005 3:19 pm


Anonymous 寫到:
qeypour 寫到:f(pq)=f(p)*f(q) require that p,q are primes
Why can you apply this formula to f(p^a*q^b)  while  p^a,q^b
both   are  not  primes?


It is good question,you can use the same way to extend the formula.
Or, first, you can finish the formula f(p^(a)q)=f(p)Xp^(a-1)Xf(q).
Then, do f(p^(a)q^(b))=f(p^(a))f(q^(b))=f(p)xp^(a-1)Xf(q)q^(-1)


the formula like f(n) defined
f(n)=#{x|1<=x<=n,(x,n)=1}

the function have some theorems.
1. f(p)=p-1;
2. f(pq)=f(p)f(q)
3  if n=p*q*r with  p,q,r are different primes
    f(n)=n*(1-1/p)*(1-1/q)*(1-1/r)


The  most  important part  of this question lies in the extention  of
the formula  f(pq)=f(p)*f(q)  to aonther one   f(p^a*q^b)=f(p^a)*f(q^b)

Unfortunately,you  have omit  this  part.
May  you  present it out  again?  Thanks a lot.
Here we define f(n)=#{x|1<=x<=n,(x,n)=(x+1,n)=1}

The difficulty comes from the action of extention,if we extend p ,it becmes
not prime,we can not apply the formula f(pq)=f(p)*f(q) any more.
I am interested in how you manage the dilemma.

qeypour
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訪客 於 星期五 九月 16, 2005 4:47 pm


My way is right.  Try some example , you can understand the whole ideal.  I did not understand your question.  
If you think that my way is wrong.  Create a counterexample without even number, or write you questions down clear.

訪客

 

qeypour 於 星期五 九月 16, 2005 5:37 pm


Anonymous 寫到:
qeypour 寫到:f(pq)=f(p)*f(q) require that p,q are primes
Why can you apply this formula to f(p^a*q^b)  while  p^a,q^b
both   are  not  primes?


It is good question,you can use the same way to extend the formula.
Or, first, you can finish the formula f(p^(a)q)=f(p)Xp^(a-1)Xf(q).
Then, do f(p^(a)q^(b))=f(p^(a))f(q^(b))=f(p)xp^(a-1)Xf(q)q^(-1)


the formula like f(n) defined
f(n)=#{x|1<=x<=n,(x,n)=1}

the function have some theorems.
1. f(p)=p-1;
2. f(pq)=f(p)f(q)
3  if n=p*q*r with  p,q,r are different primes
    f(n)=n*(1-1/p)*(1-1/q)*(1-1/r)


When you mention out  It is good questionas  above.
It is clear that you  know what I mean.
I am astonished in your answering not understand  my question.

qeypour
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大 師
 
文章: 431
註冊時間: 2005-07-23

qeypour 於 星期五 九月 16, 2005 5:47 pm


chanjunhong 寫到:證明:p and q are two different primes without 2, such that
       f(pq)=f(p)xf(q)

also assument that set(pq)={1,2,3,........,pq}
in the set of set(pq), there are [pq/p] integers which are multiples of p.
also there are [pq/q] integers which are multiples of q
,but pq is counting twice.

must counting two sets { p-1, 2p-1,.....,pq-1} and {q-1, 2q-1, 3q-1,.....,pq-1}

#{ p-1, 2p-1,.....,pq-1} =[pq/p]
#{q-1, 2q-1, 3q-1,.....,pq-1}=[pq/q]

but also some integers counting twice
{ p-1, 2p-1,.....,pq-1} has multiples of q,
there are [[pq/p]/q]

{q-1, 2q-1, 3q-1,.....,pq-1} has multiples of p
there are [[pq/q]/p]

hence f(pq)=pq-2[pq/p]-2[pq/q]+4
                =pq-2p-2q+4=(p-2)(q-2)=f(p)Xf(q)

ex set(15)={1,2,.....,15}
               ={3,6,9,12,15}U{5,10,15}
               U{2,5,8,11,14}U{4,9,14}U{1,7,13}

f(15)=15-2[15/3]-2[15/5]+4
       =15-2x5-2x3+4
       =(3-2)x(5-2)=f(3)xf(5)


finially, we also can get another formula:f(p^(a)q^(b))=p^(a)q^(b)X(1-2/p)X(1-2/q)


May you demonstrate the detail?

qeypour
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大 師
 
文章: 431
註冊時間: 2005-07-23

訪客 於 星期五 九月 16, 2005 5:51 pm


qeypour 寫到:
Anonymous 寫到:
qeypour 寫到:f(pq)=f(p)*f(q) require that p,q are primes
Why can you apply this formula to f(p^a*q^b)  while  p^a,q^b
both   are  not  primes?


It is good question,you can use the same way to extend the formula.
Or, first, you can finish the formula f(p^(a)q)=f(p)Xp^(a-1)Xf(q).
Then, do f(p^(a)q^(b))=f(p^(a))f(q^(b))=f(p)xp^(a-1)Xf(q)q^(-1)


the formula like f(n) defined
f(n)=#{x|1<=x<=n,(x,n)=1}

the function have some theorems.
1. f(p)=p-1;
2. f(pq)=f(p)f(q)
3  if n=p*q*r with  p,q,r are different primes
    f(n)=n*(1-1/p)*(1-1/q)*(1-1/r)


When you mention out  It is good questionas  above.
It is clear that you  know what I mean.
I am astonished in your answering not understand  my question.


no!
Sorry! I really don't know what you mean.
or you can explain what you mean and solve it.
We will  appreciate your answer!!

訪客

 






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