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

#### 万精油

• Hero Member
• Posts: 1831
##### 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

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

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。

#### testpost

• Newbie
• Posts: 23
##### 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

• Newbie
• Posts: 23
##### 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