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
galaxylee 寫到:
(91年全國高中數學能力競試台中區賽題)
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)
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?
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
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?
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)
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)
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)
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.