搬运工系列2：由一道最少取多少可以满足XX条件的老题引出秒杀该类型题目的解题思路

Nievesma

作为数学渣，深夜投毒。这是一道老题，钱永强书里看到的，数学高手都惜字如金，渣渣只能到处搜索，CD里有很好的讨论，大家可以版内搜索，我在上次的网站上又找到了大神的解法（最坏打算法），百试百灵，见招拆招，所向无敌了····，特搬运至此与大家分享。弄懂一道题不是关键，关键是思路！思路！思路！NN们我就不多说了，渣渣们还是要做帖子里给出链接的那几道类似的题，巩固技能，加深对这种思路的理解和运用。按照惯例，链接回复可见。希望大家可以有所收获~希望我的数学能有所起色（哭），希望妈妈可以再也不用担心我的数学（哭）·····题目：Each of the integers from 0 to 9, inclusive, is written on a separate slip of blank paper and the ten slips are dropped into hat. If the slips are then drawn one at a time without replacement, how many must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10?

A. 3
B. 4
C. 5
D. 6
E. 7，


Bunuel的解法：You should consider the worst case scenario: if you pick numbers 0, 1, 2, 3, 4, and 5 then no two numbers out of these 6 add up to 10.

Now, the next, 7th number whatever it'll be (6, 7, 8, or 9) will guarantee that two number WILL add up to 10. So, 7 slips must be drawn to ensure that the numbers on two of the slips drawn will have a sum of 10.


原帖子链接如下：
http://gmatclub.com/forum/each-of-the-integers-from-0-to-9-inclusive-is-written-on-a-68961.html

bigfeng

看看

gongxiaorong

slip选E

lorischan

看下

lookyiqi

看一下！               

wwang134

看一下！               

hermionezc

看一下！               

澳大利亚

同意！               

RUG1931350

thanks a lot

Augustus8

Mark一下！