数学奥赛题 - 珍珠湾ART - Powered by Discuz!

If f(x1,x2,...,xn)=k=f(k,k,...,k)

Then from f(x1,x2,...,xn)=f(k,k,...,k), there at least is i, such that xi=k by condition 2 (otherwise f(x1,x2,...,xn)>

So f(x1,x2,...,xn)=xi

Did I misunderstand your question?

i should be independent of (x1,..., xn).

Did you post your problems in WXC? Thank you.

I did. Do you want to do it regularly? Then I will hold off my answeres here.

I just have some rough idea. When n=2, it can be proven very easy. Then I want to go to n=4, n=8, ...., This method probably work, but anyway, I guess I can not get it tonight, because I am very tired today. Besides, I do not like this idea very much, because I think induction is like cheating sometimes.

Anyway, since this is a pretty good problem, you may hold your solution for some time. I think some people here and in WXC might work it out. (Some people in WXC often write solutions with many jumps, sigh...)

I will write down my proof (if I verify it) ASAP even though it is ugly.

What I mean is that I posted her problems to WXC, althought that should be your job. And if YOU want to "post her problems to WXC" regularly, I will hold off answering her problems here.

Sorry. Thanks a lot.

I was really tired then. Just had a short sleep.

If you saw I did not post her problems, please do. Thanks in advance.

For today's three problems, I see that one is solved. You may give the answer for that problem? You may hold the answer for the other two.

Claim1: f(x1,x2,...,xn)=xi (i is not fixed)

As If f(x1,x2,...,xn)=k=f(k,k,...,k),

then from f(x1,x2,...,xn)=f(k,k,...,k), there at least is i, such that xi=k by condition 2 (otherwise f(x1,x2,...,xn)>

So f(x1,x2,...,xn)=xi

（转自：顶顶华闻 www.TopChineseNews.com ）

We have to prove the i is independent from (x1,x2,...,xn).

For convenience, wlog, we may assume there are distinct integers y1,y2,...,yn, such that

f(y1,y2,...,yn)=y1 , then we must prove f(x1,...,xn)=x1 for any (x1,...,xn).

let Z1={z1 in Z: f(z1,k,k,...k)=z1, for some k<>z1}

let Z2={z2 in Z: f(k,z2,z2,...,z2)=z2, for some k<>z2}

Clearly, the intersection of Z1 and Z2 is empty, otherwise , if z is a common integer

f(z,k,k,...,k)=z=f(l,z,z,...,z), for some k<>z, l<>z, contradicts with condition2.

Also we have: Z1UZ2=Z. (By Claim 1)

（转自：顶顶华闻 www.TopChineseNews.com ）

Let A={(x,y) in Z^2: f(x,y,y,...,y)=x}

Let B={(x,y) in Z^2: f(x,y,y,...,y)=y}

Clearly the intersection of A and B is {(x,x):xin Z}

AUB=Z^2

（转自：顶顶华闻 www.TopChineseNews.com ）

let z1 in Z1, z2 in Z2

If (z2,z1) is in A, f(z2,z1,z1,...,z1)=z2, then z2 is in Z1, contradiction.

If (z2,z1) is in B, f(z2,z1,z1,...,z1)=z1, then z1 is in Z2, contradiction

So one of Z1, z2 is empty.

（转自：顶顶华闻 www.TopChineseNews.com ）

If Z1 is empty:

f(z,k,k,...,k)=k for any z, k

then f(m,y1,y1,...,y1)=y1=f(y1,y2,...,yn) for any m<>y1, as y1,...,yn are distinct, contradiction with condition2.

So Z2 must be empty.

f(z,k,...,k)=z for any z,k.

If f(x1,x2,...,xn)=xi<>x1,

then f(xi, m,m,...,m)=xi=f(x1,x2,...,xn) for any m beyond {x1,x2,...,xn}, contradiction with condition2.

Therefore f(x1,x2,...,xn)=x1//

（转自：顶顶华闻 www.TopChineseNews.com ）
（转自：顶顶华闻 www.TopChineseNews.com ）
（转自：顶顶华闻 www.TopChineseNews.com ）
（转自：顶顶华闻 www.TopChineseNews.com ）
（转自：顶顶华闻 www.TopChineseNews（转自：顶顶华闻 www.TopChineseNews.com ）
（转自：顶

We may prove it by induction on n

n=1 trivial

Define g(x2,x3...,xn)=f(x2,x2,...,xn)

then g(k,k,...,k)=f(k,k,...,k)=k

if g(x2,x3,...,xn)=g(y2,y3,...,yn), then f(x2,x2,x3,...,xn)=f(y2,y2,y3,...,yn), thus at lease one pair of xi=yi, i>1

So g satisfy the two conditions, by assumption of induction, g(x2,...,xn)=xi for some fixed i>1. That is, f(x2,x2,...,xn)=xi, where i>1 is independent from x2,x3,...,xn.

顶华闻 www.TopChineseNews.com ）

（转自：顶顶华闻 www.TopChineseNews.com ）

I'm sure you can make the Iranian national team. This is the last problem in the test.

it is not true if the domain becomes infinite dimensional (Z^infinity -> Z).

scared! I am just joking, of course， your problems are very welcomed, they at least let us 开开眼界。

I should say I don't know. The original problem was Zk^n -> Zk. In that case it cannot be extended to Zk^infinity -> Zk, since a function satisfying the condition is then uniquely determined by an untra filter. I did not think carefully after I changed the problem to Z^n -> Z.

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