称11个球（3.5星）

constant · 发表于 2007-2-2 04:57:11

有11个球，从中任意取出1个球，剩下的10个都能分成两组，每组5个，使得两组的重量相同。证明11个球的重量相等。

注：这个题用到的数学比其他称球题都多。

野菜花 · 发表于 2007-2-2 17:00:04

设这

11个球的重量分别为X1, X2, ..., X11.

将球X1拿开, 剩下10个分成5个一组，两组重量相等得一方程，把各项都移到方程左边，右边为零。

再类似地对球 X2, X3, ..., X11 列出方程，共得11个未知数11个线性方程。www.ddhw.com

此线性方程组的系数矩阵的对角线都是0，每行有5个1，5个-1。www.ddhw.com

我们只要证明此系数矩阵的秩是10. 将所有列加到最后一列，最后一列全是0. 考虑前10行前10列的矩阵，证明此行列式不等于0。www.ddhw.com

不妨假设，每行都有5个1，4个-1 (若不然，将那行乘-1，这样并不改变行列式是否为0)。www.ddhw.com

将所有列加到第一列，第一列就全是1，再每行减起第一行。第一列除第一个1就全是0，那10阶行列式值就与这9阶行列式相同。www.ddhw.com

而这9阶行列式的对角线元素都是1或-1，非对角线处都是0或偶数(2 or -2)。由行列式定义可知，行列式的值等于所有不同列不同行的元素乘积的代数和，在这代数和里只有一个乘积是奇数，所以这9 阶行列式必不为0。www.ddhw.com

这样所有的球的重量都等于X11.

constant · 发表于 2007-2-2 18:08:08

比我的证明好多了

野菜花 · 发表于 2007-2-2 19:40:48

不会吧

你的方法大概思路是怎样的？

constant · 发表于 2007-2-3 01:25:16

原来的题中重量都是整数，这时可以用奇偶性来证。我自己推广到实数的，方法是把这11个实数表成整系数向量，然后用整数时的方法，证明每一项的系数都相等。

野菜花 · 发表于 2007-2-5 01:27:36

Thanks!

野菜花 · 发表于 2007-2-19 21:42:47

The matrix is not necessarily like that.

VerySimple · 发表于 2007-2-19 22:35:06

有11个球，从中任意取出1个球，剩下的10个都能分成两组，每组5个，使得两组的重量相同。证明11个球的重量相等。

1) Label from x1, x2, ..., x11
2) 从中任意取出1个球, x11 must be one of them.
3) 剩下的10个都能分成两组, so x1 x2 x3 x4 x5 could be one of the two group
4) 从中任意取出1个球, x10 must be one of them.
5) 剩下的10个都能分成两组, so x1 x2 x3 x4 x5 could be one of the two group
.....
4) 从中任意取出1个球, x5 must be one of them.
5) 剩下的10个都能分成两组, so x1 x2 x3 x4 x6 could be one of the two group
...

The order of the matrix does not have to be equal to the order of the ball selected.

Am I right?

VerySimple · 发表于 2007-2-19 22:42:47

Assuming there are only 5 balls

1) x5 is possible to be taken out, when it happens then we have
x1+x2-x3-x4=0
x1+x3-x2-x4=0
x1+x4-x2-x3=0
ignore all of them except that x1 + x2 - x3 - x4 = 0
2) x4 is possible to be taken out, when it happens then we have
x1+x2-x3-x5=0
...
ognore all of them except that x1 + x2 - x3 - x5 = 0
...

At last, we will have a matrix that is easily to be resolved.

野菜花 · 发表于 2007-2-19 23:20:51

3) 剩下的10个都能分成两组, so x1 x2 x3 x4 x5 could be one of the two group

It is not necessarily true, as we only know there is one (at least) way to divide the10 balls.

When we have ordered the 11 balls, we can not change the order arbitrarily. (at least it is hard to explain in the proof).

野菜花 · 发表于 2007-2-19 23:23:43

1) x5 is possible to be taken out, when it happens then we have
x1+x2-x3-x4=0
x1+x3-x2-x4=0
x1+x4-x2-x3=0
ignore all of them except that x1 + x2 - x3 - x4 = 0

We only know there is one equation true.

本贴由[野菜花]最后编辑于：2007-2-19 15:59:26

VerySimple · 发表于 2007-2-19 23:30:11

We only know there is one equation is true.

If only one equation is true, then there will be no answer.

For example, if x1 + x2 != x3 + x4, then how can that be that all of the 11 balls has equal weight?

VerySimple · 发表于 2007-2-19 23:31:15

We only know there is one equation is true.

If only one equation is true, then there will be no answer.

For example, if x1 + x2 != x3 + x4, then how can that be that all of the 5 balls has equal weight?

野菜花 · 发表于 2007-2-19 23:33:18

"We only know there is one equation true." does not mean "there is only one equation true"

本贴由[野菜花]最后编辑于：2007-2-19 15:50:40

VerySimple · 发表于 2007-2-19 23:37:45

"We only know there is one equation is true." does not mean "there is only one equation is true"

Then if there are at least one equation is true, how can we prove that our selection is not true?

野菜花 · 发表于 2007-2-19 23:46:09

If the problem is as you thought: no matter how divide the 10 balls, the scale is balanced as long as 5 on each side, maybe constant would have ranked it at 1星 problem.

VerySimple · 发表于 2007-2-19 23:51:01

I believe that constant has raised an problematic (-3.5星) problem.

撼蚁之旅 · 发表于 2007-10-8 06:59:14

由题得: a+(b+c+d+e)=f+(g+h+i+j) f+(b+c+d+e)=a+(g+h+i+j)

设(b+c+d+e)为x,(g+h+i+j)为y,得 a+x=f+y

f+x=a+y

解得 x=y,所以a=f.同理,a=k.

其余类推.

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称11个球（3.5星）

回复：称11个球（3.5星）

比我的证明好多了[:B][:-Q][>:D<]

不会吧[:>] 你的方法大概思路是怎样的？

回复：不会吧[:>] 你的方法大概思路是怎样的？

Thanks! [@};-][@};-]

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称11个球（-3.5星）[:-M][:)][;)]

太简单了.a+(b+c+d+e)=f+(g+h+i+j)