[jjsaw5]

A guy at work this morning gave us a brain teaser/logic problem.

There are 25 race cars. Only 5 cars can race at one time. How many times would it take to find the fastest three cars? He said it should be less then ten times. A couple people at work got 11, i think i figured it out but I'm curious to see what you guys come up with.

my answer:

25 cars 5 at a time = 5 sessions

1st place finisher from each session races thats 5 cars. Take the top 3 cars from that race. It only takes 6 races.

This post has been edited by jjsaw5: 15 April 2008 - 05:56 AM

[SebKom]

What about clocking these 5 sessions? You only need 5 times!

[baavgai]

Five races of five gets you the five fastest, then one last race for pole position, so six. It there some kind of trick question in here? Of course, if you were smart enough to bother with a stop watch, you woudln't need a last race...

[jjsaw5]

baavgai, on 15 Apr, 2008 - 10:26 AM, said:

Five races of five gets you the five fastest, then one last race for pole position, so six. It there some kind of trick question in here? Of course, if you were smart enough to bother with a stop watch, you woudln't need a last race...

Thats the answer i got as well. I'm not really sure if it is a trick question or not like I said, a co-worker asked us this morning and i haven't had the time to go and talk to him about it yet.

[1lacca]

baavgai, on 15 Apr, 2008 - 04:26 PM, said:

Five races of five gets you the five fastest, then one last race for pole position, so six. It there some kind of trick question in here? Of course, if you were smart enough to bother with a stop watch, you woudln't need a last race...

Nope, if the three fastest are in the first group, and you only qualify the first from each group, then you won't have the 3 fastest in the second round, just the fastest and two lucky ones.

[baavgai]

1lacca, on 15 Apr, 2008 - 10:59 AM, said:

Nope, if the three fastest are in the first group, and you only qualify the first from each group, then you won't have the 3 fastest in the second round, just the fastest and two lucky ones.

Ack, you are correct. Five and a stop watch... which makes it far more entertaing if you don't have a stop watch...

If, in each race you take the top three, you'll be able to elimiate two per race.
Round 1 = 5 races of 5 (25)
Round 2 = 3 races of 5 (15)
Round 3 = 2 races of 5 (9) ( one from the first participates in the next)
Round 4 = 1 race of 5 (5) ( pole position! )

Looks like 11, no less, sans watch. Make sense?

[1lacca]

I would go with 8 races.

Spoiler

1-5: race in groups of five
6: 5 winner -> we have the fastest here
7: 4 runner up from 6th, and the 2nd from 1-5 where the winner comes from -> second fastest is found
8: 4 runner up from 7th, and the 2nd from 1-5 where the winner of 7 comes from (or the third if first and second fastest came from the same original group ) -> 3rd fastest

[baavgai]

Excellent! You win.