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标题: Generalizing 野菜花's problem [打印本页]

作者: fzy 时间: 2005-3-14 18:28
标题: Generalizing 野菜花's problem

This is a problem given by 野菜花 last week:

在六张纸片的正面分别写上 1， 2， 3， 4， 5， 6，在它们的反面也随意地写上这六个数，然后计算每张纸片两面数字之差的绝对值。证明这六个绝对值中至少有两个是相同的。

Now let's generalize it: Is this true for any N? It is obviously false for N = 1. How about N>1? If not, for what N is it true?

作者: 野菜花 时间: 2005-3-14 19:20
标题: 回复：Generalizing 野菜花's problem

Suppose there are no two numbers are the same, then the N absolute values are 0, 1, 2...

(N-1)

sum(a-a')=0

sum|a-a'|=0+1+...+(N-1)=N(N-1)/2

As (a-a') and |a-a'| are either both ever or both odd, if N(N-1)/2 is odd, then the assumption is wrong.

if N is even, N=2m, N(N-1)/2=2m*(2m-1)/2=m*(2m-1), m must odd,

so N=2m=2(2k+1)=4k+2

If N is odd, N=2m+1, N(N-1)/2=(2m+1)*2m/2=(2m+1)*m, m must be odd,

so N= 2m+1=2(2k+1)+1=4k+3

Therefore N=4k+2, or 4k+3

作者: fzy 时间: 2005-3-14 19:29
标题: Right answer, but

can you prove it is false for 4k and 4k+1?

作者: 野菜花 时间: 2005-3-14 20:28
标题: 回复：Right answer, but

看来你今天想把我累死啊？那么那道三位数的题你还没证明

42不行呢！

作者: yma16 时间: 2005-3-14 21:09
标题: just find some counterexample, such as

1,2,3,4

2,4,3,1

The difference is 1,2,0,3.

作者: 野菜花 时间: 2005-3-14 21:12
标题: need prove for all k [:%]

need prove for all k

作者: fzy 时间: 2005-3-16 00:10
标题: Examples for 4k, 4k+1

For N=12:

(1,12), (2,11), (3,10), (4,4), (5,9), (6,8), (7,6), (8,5), (9,3), (10,2), (11,1), (12,7)

For N=13:

(1,13), (2,12), (3,11), (4,4), (5,10), (6,9), (7,8), (8,6), (9,5), (10,3), (11,2), (12,1), (13,7)

Same pattern should work for all k.

作者: 野菜花 时间: 2005-3-16 00:23
标题: 回复：Examples for 4k, 4k+1

I found counter exampls for N=4,5,8,9,12,13, but I could not find a clear pattern like yours.

作者: fzy 时间: 2005-3-16 00:40
标题: 回复：回复：Examples for 4k, 4k+1

I know you will get it. I just don't want it to fall too low. (Too many new problems )

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