有乘又有除、题目很简单,那就再赠大家一道高难度国外竞赛题吧!

 

“一个两位数与109的乘积为四位数,它能被23整除且商是一位数,这个两位数最大等于几?”

 

好久好久没有遇到这么简单的题了,要不要乐博士再赠大家一道题

 

这个两位数“能被23整除且商是一位数”,我不清楚还有商不是一位数的情况么?那还是两位数么?满足这个条件的两位数有:23、46、69和92。

 

再结合另一个条件“它与109的乘积为四位数”,上面的4个数中92不符合,所以剩下:23、46和69。

 

综上,这个两位数最大等于69!

 

为了表达乐博士的诚意,真的赠大家一道题(顺便学习一下英语):

Problem:NCAA Final Four

问题:NCAA四强

And then there were four: South Carolina in the East bracket, Gonzaga in the West bracket, Oregon in the Midwest bracket and North Carolina in the South bracket. These four teams have earned a coveted spot in the arena known as the Final Four. There were 64 teams (not including play-in games) who entered the NCAA Tournament – 16 teams in each of the four brackets. The tournament is a single elimination, meaning each game played eliminates one team.

 

进入NCAA四强的球队是:东区的南卡罗来纳大学、西区的贡萨加大学、中西区的俄勒冈州大学和南区的北卡罗来纳大学。这四只球队成为了本届NCAA四强——the Final Four。共有64支球队进入了NCAA锦标赛(不包括外围比赛),每区16支球队。锦标赛采取单淘汰赛制,即每场比赛淘汰失败的球队。

 

1) How many games had to be played to determine the Final Four?

问题1:决出四强共需比赛多少场?

 

Since one team from each of the four brackets of 16 teams advances to the Final Four,

2) how many possible Final Four team combinations were possible?

问题2:每区16支球队,只有一个名额进入四强,四强可能的组合有多少种?

 

Assuming equal probability of winning for each team,

3) what is the probability that the championship game is South Carolina against North Carolina, and North Carolina then wins the national championship?

问题3:假设每场比赛双方获胜的几率相等,请问决赛南卡罗来纳大学与北卡罗来纳大学相遇且最终北卡罗来纳大学夺冠的概率是多少?

 

问题1:最直观的解法是分别计算每个区决出1名四强队伍的比赛场次:16进8比赛8场、8进4比赛4场、4进2比赛2场、2进1比赛1场,共:8 + 4 + 2 + 1 = 15(场)。所以64支球队决出四强需要比赛:15 × 4 = 60(场)。

针对这种单淘汰赛制的比赛还有更为简洁的方法:因为每场比赛淘汰1支球队,64支球队剩下4支球队(淘汰了60支球队),所以共比赛:64 - 4 = 60(场)。

 

问题2:每个区决出的队伍有16种可能,所以最终的四强共有:16 × 16 × 16 × 16 = 65536(种)可能的组合。

 

问题3:这就需要考验大家对NCAA的熟悉度了,因为进入四强后,分别是SOUTH-WEST和EAST-MIDWEST对决,所以南卡罗来纳大学与北卡罗来纳大学不会提前相遇,他们分别战胜对手的概率均为1/2、最终北卡获胜的概率也是1/2,所以他们相遇且北卡最终夺冠的概率为:½ × ½ × ½ = 1/8。