Author Topic: Riddles!  (Read 133584 times)

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Re: Riddles!
« Reply #320 on: August 30, 2019, 04:24:56 PM »
It depends on which ones are fastest. I can see getting it in between 6 and 8.
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Re: Riddles!
« Reply #321 on: August 30, 2019, 04:26:03 PM »
That's what I thought at first but that really just gets you the one fastest one (I think). Pretty sure I can do it in 9, not sure if there is a faster way.
Why only the fastest? After you get the 5 winners you make the 6th race and get who's in first second and third. No?

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Re: Riddles!
« Reply #322 on: August 30, 2019, 04:26:52 PM »
Why only the fastest? After you get the 5 winners you make the 6th race and get who's in first second and third. No?


What if the second of the first race is faster than the first of the second race?
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Re: Riddles!
« Reply #323 on: August 30, 2019, 04:28:34 PM »
What if the second of the first race is faster than the first of the second race?
+1
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Re: Riddles!
« Reply #324 on: August 30, 2019, 04:29:32 PM »
Or for that matter what if all 5 of the first race are the fastest of all?
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Re: Riddles!
« Reply #325 on: August 30, 2019, 04:29:43 PM »
What if the second of the first race is faster than the first of the second race?
Oh I see
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Re: Riddles!
« Reply #326 on: August 30, 2019, 04:49:19 PM »
Here's a solution:
You go through the Door between the letters H & M.

The logic - since the sequence is the clue, and the letters appear on the wall sequentially, it would make sense that the person who painted them on the walls entered through the non-booby trapped door and began painting. When they completed the task with "M", they walked out of the room from the door they started with.

That was kind of my first thought when I saw this but I feel like it's too simplistic

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Re: Riddles!
« Reply #327 on: August 30, 2019, 05:01:25 PM »
The three fastest horses:

A horse farmer takes you to his track and asks you to find the 3 fastest horses out of the 25 he owns. There are three parameters/givens to this test:

- Regardless of how many times you run them, the horses are relatively consistent - meaning if horse "A" is faster than horse "B" in one race, he is ALWAYS faster than Horse "B".
- You can't use a stop watch
- You can only race five horses at a time on the track.

What is the minimum number of races you need to run to find horses 1,2, and 3, and how do you set them up?

@yitzgar @etech0 @aygart @Definitions you guys are on the right track.

The answer is 7 - and here is why:
You group them in groups of 5, and Run 5 different races - let's call the groups A,B,C,D,E; and within their groups as 1-5; So A1 is the fastest horse in group A and A5 is the slowest.

In the 6th race, you run all of the first place horses, let's say that the Order of the 6th race is A1,B1,C1,D1,E1

After race 6, given that they're all relatively consistent speed was, you know the following:

- A1 is the fastest horse
- All of the horses in Groups D and E are out
- A2 and A3 might be the 2nd and 3rd fastest, but A4 and A5 are out
- B1 and B2 might be the 2nd and 3rd fastest, but B3,4,5 are out (since B1 can be 2nd at best, B3 can be 4th at best)
- C1 could be third at best - C2-5 are out.

So 2nd and 3rd place have 5 remaining horses:
A2, A3, B1,B2,C1

And they race to determine the places.

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Re: Riddles!
« Reply #328 on: August 30, 2019, 05:07:18 PM »
That was kind of my first thought when I saw this but I feel like it's too simplistic
Also why start from H
@yitzgar @etech0 @aygart @Definitions you guys are on the right track.

The answer is 7 - and here is why:
You group them in groups of 5, and Run 5 different races - let's call the groups A,B,C,D,E; and within their groups as 1-5; So A1 is the fastest horse in group A and A5 is the slowest.

In the 6th race, you run all of the first place horses, let's say that the Order of the 6th race is A1,B1,C1,D1,E1

After race 6, given that they're all relatively consistent speed was, you know the following:

- A1 is the fastest horse
- All of the horses in Groups D and E are out
- A2 and A3 might be the 2nd and 3rd fastest, but A4 and A5 are out
- B1 and B2 might be the 2nd and 3rd fastest, but B3,4,5 are out (since B1 can be 2nd at best, B3 can be 4th at best)
- C1 could be third at best - C2-5 are out.

So 2nd and 3rd place have 5 remaining horses:
A2, A3, B1,B2,C1

And they race to determine the places.
I was just in middle of writing it down :) this is what I narrowed it down to so far ( 8 ).



I should've remembered that I only have to find the top three. Oh well. It's a good riddle
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Re: Riddles!
« Reply #329 on: August 30, 2019, 05:43:22 PM »
@yitzgar @etech0 @aygart @Definitions you guys are on the right track.

The answer is 7 - and here is why:
You group them in groups of 5, and Run 5 different races - let's call the groups A,B,C,D,E; and within their groups as 1-5; So A1 is the fastest horse in group A and A5 is the slowest.

In the 6th race, you run all of the first place horses, let's say that the Order of the 6th race is A1,B1,C1,D1,E1

After race 6, given that they're all relatively consistent speed was, you know the following:

- A1 is the fastest horse
- All of the horses in Groups D and E are out
- A2 and A3 might be the 2nd and 3rd fastest, but A4 and A5 are out
- B1 and B2 might be the 2nd and 3rd fastest, but B3,4,5 are out (since B1 can be 2nd at best, B3 can be 4th at best)
- C1 could be third at best - C2-5 are out.

So 2nd and 3rd place have 5 remaining horses:
A2, A3, B1,B2,C1

And they race to determine the places.
Nice!
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Re: Riddles!
« Reply #330 on: September 03, 2019, 08:35:09 AM »
Also why start from H

Based on my opinion of the solution, the actual letter you start with is irrelevant information. The relevant information is the sequence, and where it begins and ends.

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Re: Riddles!
« Reply #331 on: September 04, 2019, 03:10:39 PM »
An oldie but Goodie:

In front of you are 9 marbles and a balance (a scale that has two sides where you can weigh one side against the other). The marbles are all of identical size, and 8 of the 9 have an identical weight.

Using the balance, you need to find the odd-weight marble, and identify if it is heavier or lighter than the other 8 - but you can only use the balance 3 times.

How do you do it?

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Re: Riddles!
« Reply #332 on: September 04, 2019, 03:20:13 PM »
Would I start with four on each side? 
Only on DDF does 24/6 mean 24/5/half/half

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Re: Riddles!
« Reply #333 on: September 04, 2019, 03:47:51 PM »
Would I start with four on each side? 
no, 3 on each side.

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Re: Riddles!
« Reply #334 on: September 04, 2019, 03:48:48 PM »
no, 3 on each side.
Why wouldn't 4 work?
Only on DDF does 24/6 mean 24/5/half/half

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Re: Riddles!
« Reply #335 on: September 04, 2019, 03:51:28 PM »
An oldie but Goodie:

In front of you are 9 marbles and a balance (a scale that has two sides where you can weigh one side against the other). The marbles are all of identical size, and 8 of the 9 have an identical weight.

Using the balance, you need to find the odd-weight marble, and identify if it is heavier or lighter than the other 8 - but you can only use the balance 3 times.

How do you do it?
Oh I remember this one. It took my classmates like 2 days of bein hasedarim to figure it out in 11th grade.

I don't remember the answer off hand though
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Re: Riddles!
« Reply #336 on: September 04, 2019, 03:52:00 PM »
Why wouldn't 4 work?
What's the second step if you do 4-4?

you can only use the balance 3 times.

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Re: Riddles!
« Reply #337 on: September 04, 2019, 03:54:06 PM »
What's the second step if you do 4-4?
You hope it balances.  :)
What is the second step with 3-3?
Only on DDF does 24/6 mean 24/5/half/half

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Re: Riddles!
« Reply #338 on: September 04, 2019, 04:05:57 PM »
What is the second step with 3-3?
if it balances you know the odd one is in the 3rd group, so put on 2 from group 3.

If it doesn't balance you replace one of the current groups with the 3rd group. Lets say you take the lighter 3 off, if it balances now you know one of the first 3 is lighter but if the heavier one is still heavier you know one of those three is heavier.
Put on 2 of the three with the odd-weighted one to determine which of the three it is.

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Re: Riddles!
« Reply #339 on: September 04, 2019, 04:06:06 PM »
You hope it balances.  :)
What is the second step with 3-3?

If one is heavier then weigh two of the heavier group. If they are equal weigh two of the unweighed group. If one is heavier then you have it if not then the unweighed one is it. Not sure why you would need a third time.
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