b*k 发帖数: 27 | 1 A magician has one hundred cards numbered 1 to 100. He puts them into three
boxes, a red one, a white one and a blue one, so that each box contains
at least one card.
A member of the audience selects two of the three boxes, chose one card
from each and announces teh sum of the numbers on the chosen cards. Given
this sum, the magician identifies the box from which no card has been chosen.
How many ways are there to put all the cards into the boxes so that this
trick always works?(Tow ways are con | C******a 发帖数: 115 | 2 答案是12个解,实质上只有两种:B={1}, C={100}, A=其它;
A={1,4,7,10,...}, B={2,5,8,11...}, C={3,6,9,12...}。
由条件可以推出若任意一个集合中有两个数相差为n,则其他两个集合各取
一数得到的两个数相差不可能是n(因为会出现魔术师无法判断的情况)。
不可能有两个集合,比如说A和B中各有相邻的一对数,因为如果那样的话,
由上面的结果,C中的数既不会和B中的数相邻,也不会和A中的数相邻,矛盾。
于是最多有一个集合,比如说A,中有一对相邻的数。于是B和C中都没有相邻
的数对,B和C各选一数也不相邻。令D=B并C,则D中的任两个数都是不相邻的。
假设n=min(|b-c|:b\in B, c\in C),且n=b_0-c_0, b_0\in B, c_0\in C。
那么n>1,在b_0和c_0之间有n-1个A中的数。如果在b_0或c_0不是1或100,
则在b_0或c_0的外侧有一个A中的数,这样就能找到两个A中的数相差为n,矛盾。
这其实也得到了一个解:B={1}, C={100}, A=其它。颠倒顺序即得6个解。
【在 b*k 的大作中提到】 : A magician has one hundred cards numbered 1 to 100. He puts them into three : boxes, a red one, a white one and a blue one, so that each box contains : at least one card. : A member of the audience selects two of the three boxes, chose one card : from each and announces teh sum of the numbers on the chosen cards. Given : this sum, the magician identifies the box from which no card has been chosen. : How many ways are there to put all the cards into the boxes so that this : trick always works?(Tow ways are con
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