Author Topic: 每周一题:平衡天平 (8/2/04--8/8/04)  (Read 6706 times)

万精油

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每周一题:平衡天平 (8/2/04--8/8/04)
« on: 八月 02, 2004, 12:02:27 am »
上周问题讨论

  上周的问题testpost的答案完全正确。当想到用矩阵表达并引进高度以后,很
容易证明这个序列的稳定状态只能是不增序列。比较难一点的是要证明它不但是不
增,而且必须是递减。所以,testpost证明的关键是第十一步。

  这周的问题不是太难,如果想对了方法,也就是几句话。但这个题要用到的原
理很重要,许多别的趣味问题都要用到它,所以,这道题算是为以后的题做准备工
作。

本周问题

  抽屉里有重量是1到100的砝码100个。现在从里面随便取10个出来。
试证明我们总可以从这十个砝码中挑出两组砝码来使得它们在天平上平衡。也就是
说这两组的重量和相等。注意,两组的个数不一定一样,也不一定要用到全部十个
砝码。

testpost

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每周一题:平衡天平 (8/2/04--8/8/04)
« Reply #1 on: 八月 03, 2004, 12:12:34 pm »
Too simple?

Define E is a set of all posible of 10 element
arrays made by 1 and 0. (1024 posibles, you can
count it)

Assume 10 selected numbers are A=(a1,a2,...,a10).
For each member of E, define product p = A*e as:
Sum[1..10] of (a * e). There are 1024 products.

   
value of these products ranges from 0 to 1000
(indeed, 55 - 955). Less than 1024 posibles.

Due to box principle, there are 2 member of E
where e1 != e2 have to same product Ae1 = Ae2.
We then have: A(e1-e2) = 0.

because e1 != e2, So, (e1-e2) will not be all 0s.
Also notice that (e1-e2) is only made up by 0, 1,
and -1.

In product of A(e1-e2), the sum of positives is the
sum of negitives.
testposttestposttestpost

testpost

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每周一题:平衡天平 (8/2/04--8/8/04)
« Reply #2 on: 八月 03, 2004, 02:37:22 pm »
Correct previous post of line 9: (-:

(indeed, 0 - 955). Less than 1024 posibles.

Should make it simpler: (basically the same)

with given 10 numbers, there are 1024 possibles
of summary of a subset range from the minium of
0 to the maximium of 955. Based on box principle,
there are 2 sub sets of this 10 shares the same
sum value. Get rid of the identical member of
these 2 subset and the rest balance the Scales.
testposttestposttestpost