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Pearls
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icon Re: Puzzle "tag" (+1)  
Keeping with the geometry theme, take a perfectly made washer. That is, a circle of radius R with a hole in it of radius r that share the same center.

Your boss is lazy and wants you to come up with a way you can take a single measurement with a straight object, like a ruler, and figure our the surface area of a side of washer. That is, pi*R^2 - pi*r^2.

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icon Re: Puzzle "tag" (+2)  
quote:
Pearls wrote:
Keeping with the geometry theme, take a perfectly made washer. That is, a circle of radius R with a hole in it of radius r that share the same center.

Your boss is lazy and wants you to come up with a way you can take a single measurement with a straight object, like a ruler, and figure our the surface area of a side of washer. That is, pi*R^2 - pi*r^2.


This is a neat little geometry puzzle.

Two hints for those trying to work it out on their own:

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Full solution:
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[Last edited by Insoluble at 04-10-2015 01:43 AM]
04-09-2015 at 10:09 PM
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karlpopper
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icon Re: Puzzle "tag" (0)  
Really cute ... my buddy Pythagoras helped me.

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04-10-2015 at 09:38 PM
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disoriented
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icon Re: Puzzle "tag" (+1)  
Sorry if I'm going out of turn.

A friend recently showed me this. This will be very easy for some of you.

1)  ______
2)  ______
3)  ______
4)  ______
5)  ______
6)  ______
7)  ______
8)  ______
9)  ______
10) ______


On blank 1, write any number you wish.
On blank 2, write any number you wish.

Now on blank 3, put the sum of the previous two numbers.
On blank 4, put the sum of just the previous two numbers.

Continue filling in the blanks in this "Fibonacci" pattern until all ten are filled in.

Let x = (value 10) divided by (value 9)
I predict x =
Click here to view the secret text


Let y = sum of all 10 blanks.
I predict y =
Click here to view the secret text


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[Last edited by disoriented at 04-18-2015 02:36 AM]
04-18-2015 at 02:34 AM
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jamesdenem
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icon Re: Puzzle "tag" (0)  
The answer is actually very simple.

Click here to view the secret text


Click here to view the secret text


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[Last edited by jamesdenem at 04-18-2015 05:36 PM]
04-18-2015 at 05:36 PM
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disoriented
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icon Re: Puzzle "tag" (0)  
You got it :D

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04-18-2015 at 05:48 PM
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TripleM
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icon Re: Puzzle "tag" (0)  
quote:
disoriented wrote: On blank 1, write any number you wish.
On blank 2, write any number you wish.

I pick -8 and 5 :glasses
04-18-2015 at 09:53 PM
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TripleM
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icon Re: Puzzle "tag" (0)  
quote:
[Post Deleted] Nuntar wrote: Then the sequence goes -8, 5, -3, -2, -5, -7, -12, -19, -31, -50. The total is -132.

5 + -3 = 2, not -2. The sequence (at the 10th term) ends up with a ratio of 2.

[Last edited by TripleM at 04-18-2015 11:08 PM]
04-18-2015 at 11:06 PM
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disoriented
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icon Re: Puzzle "tag" (0)  
I probably should have restricted this to positive integers ;)

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04-18-2015 at 11:11 PM
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jamesdenem
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icon Re: Puzzle "tag" (+1)  
Time for a tricky and tough logic puzzle.

Tales have been heard of a group of three men who are very wise sages. They are so wise they have come up with their own language, which only a very few have managed to learn enough of to act as translators.
The three men have been given strange nicknames for the uncanny abilities they possess.
All three of them will answer any question posed them truthfully, but in their own language.
Because they are very knowledgeable, they have devised a test to keep everyone from bothering them.
Only one person is allowed inside, told by the interpreter the following:

One of the men is nicknamed 'Past', who will always answer the previous question asked in the chamber. (If he is asked the first question, he will answer randomly.)
One is nicknamed 'Future', who will always answer the next question that will be asked in the chamber. (If he is asked the last question, he will answer randomly.)
The last one is nicknamed 'Present', who will always answer the current question being asked.
They will stand side by side in a chamber, and you must figure out who is who.
Before entering the chamber, you must write down 3 yes/no questions to ask them.
You must ask the questions in the order you write them down, and you can only ask one of the men at a time, but you can choose who you want to ask each question to.
The words for yes and no are "ya" and "da" in some order.

Can you come up with 3 questions to ask to figure out who is who?

EXAMPLE:
Assuming "da" means yes, and your questions are
Is 2+2=4
Is 3+4=9
Is your favorite color red

If you asked each man the second question, Past would answer 'da', Present would answer 'ya', and Future would answer based on if his favorite color were red.

Click here to view the secret text


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04-19-2015 at 08:32 PM
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Lucky Luc
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icon Re: Puzzle "tag" (0)  
quote:
Can you come up with 3 questions to ask to figure out who is who?

No. I can't. Because I'm too stupid. And even thogh there is probably a solution to this puzzle, I consider this a valid answer ;)

Seriously, though, I can't see a way, and I don't see where my logic is flawed, so I might just as well post what I've come up with:
Click here to view the secret text

I would be kind of surprised if this was actually the correct solution, but right now I don't see the flaw in my reasoning. As always, I'll be happy to be proven wrong :) .
04-19-2015 at 09:18 PM
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TripleM
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icon Re: Puzzle "tag" (0)  
One flaw in your reasoning is:
Click here to view the secret text


What's the difference between the main puzzle and the 'harder' puzzle? Is it that a person who answers randomly may say something other than 'yes' or 'no'?

[Last edited by TripleM at 04-20-2015 02:44 AM]
04-20-2015 at 02:43 AM
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RyTracer
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icon Re: Puzzle "tag" (+2)  
I have a solution. First of all, I am going to change the man named, "Present", to "Current", so I can have unique identifying letters. I'm also going to use Y for Yes and N for No. The questions have been altered so "ya" will always mean "yes".

Click here to view the secret text


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[Last edited by RyTracer at 04-20-2015 06:59 PM]
04-20-2015 at 06:50 AM
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RyTracer
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icon Re: Puzzle "tag" (0)  
quote:
TripleM wrote:
What's the difference between the main puzzle and the 'harder' puzzle? Is it that a person who answers randomly may say something other than 'yes' or 'no'?


I think the difficulty in the second part is that you don't know whether "ya" and "da" are yes or no. I thought he was wanting that to be the case in the first place, but his example implied that in the easier version, you know what yes and no mean.

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04-20-2015 at 07:25 AM
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jamesdenem
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icon Re: Puzzle "tag" (0)  
The difference between the puzzle I gave and the harder version is that you know beforehand that they will answer a yes/no question with "ya" or "da", but you don't know which one means yes.

The harder version is that you are not given the knowledge of the words ahead of time.
In my version, you know before you write the questions the words they will use, as long as it is phrased as a yes/no. The harder version is when you don't have that information.

RyTracer:
Click here to view the secret text


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[Last edited by jamesdenem at 04-20-2015 11:28 AM]
04-20-2015 at 11:22 AM
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Nuntar
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icon Re: Puzzle "tag" (+1)  
Well, the ya/da complication is easy to deal with. Just add "...if and only if ya means yes" to each question. Then a "ya" answer will work exactly as a "yes" answer in RyTracer's solution.

By the way, I think there's a minor slip in RyTracer's solution (which is otherwise correct):
quote:
If they answer YN, possible outcomes are PFC and CFP. Ask the third question to the man on the left. Y=CFP, N=PFC.

Since Q2 is "Are you Current?", Past will always answer No to Q3. In the order CFP, Current will also answer No to Q3, so you don't gain any information.

So instead, in this case, ask Q3 to the man on the right. Yes = PFC, No = CFP.

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[Last edited by Nuntar at 04-20-2015 02:15 PM]
04-20-2015 at 01:38 PM
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RyTracer
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icon Re: Puzzle "tag" (0)  
What Nuntar said. Imagine both changes in my solution.

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04-20-2015 at 04:16 PM
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RyTracer
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icon Re: Puzzle "tag" (0)  
Here's an answer for the harder puzzle:

Click here to view the secret text


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[Last edited by RyTracer at 04-20-2015 10:42 PM]
04-20-2015 at 09:48 PM
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jamesdenem
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Might as well post another puzzle, since there hasn't been a new one in a while...

There are 100 prisoners being held in prison. They have been given a chance at freedom by the guards, and if they fail it, will be killed.
The guards have placed in a room 100 boxes numbered from 1 to 100, and have written on slips of paper the names of the prisoners. The guards then place one slip of paper in each box.
Each prisoner has a slip of paper with his name on it.
The guards will allow only one prisoner to enter the room at a time, which is soundproofed so that the other prisoners cannot hear what is going on in the room.
Each prisoner is allowed to only open 50 of the 100 boxes, and the guards will let them go free if all 100 prisoners are able to find the box with their name in it.
After opening boxes, the prisoner is let out another exit and taken to a different room to wait for the other prisoners.
The room is monitored by the guards, and after each prisoner enters, opens the boxes and leaves; a guard enters, closes the boxes, and places them back where they originally were if they had been moved.
To be fair to the prisoners, the guards decide to let the first prisoner entering the room open all 100 boxes, and he can switch any two names if he wishes.
After that, the remaining 99 prisoners will enter the room, open their 50 boxes, and leave. The guards are very cautious, and have sensors fine enough to tell if anyone after the first has moved any names, and they will be placed back to the boxes they were in when that prisoner entered.
Likewise, they will carefully make sure that only up to two names are switched by the first prisoner.
The prisoners have a short time before they will be taken to the room, and devise a plan to ensure that they will all go free.

What plan do the prisoners make?


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04-25-2015 at 03:02 PM
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disoriented
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icon Re: Puzzle "tag" (0)  
This is almost identical to a puzzle posted earlier in this thread.

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[Last edited by disoriented at 04-25-2015 09:23 PM]
04-25-2015 at 09:21 PM
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jamesdenem
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Hopefully the similarity will help someone come up with the solution, since you need a strategy that ensures all prisoners find their name.

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04-26-2015 at 02:29 AM
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icon Re: Puzzle "tag" (+2)  
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04-26-2015 at 12:58 PM
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RyTracer
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icon Re: Puzzle "tag" (0)  
Oh, I didn't realize we were waiting on a puzzle. Jamesdenem never commented if the solution for the harder version of the last puzzle was okay, or could be optimized to 3 questions.

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04-27-2015 at 07:25 PM
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jamesdenem
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icon Re: Puzzle "tag" (0)  
I secreted the harder version since I don't know if it is possible with only 3...
But your 4 question system works well enough.

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04-28-2015 at 10:17 AM
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TripleM
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icon Re: Puzzle "tag" (0)  
There's a famous puzzle about how to divide a cake amongst n people so each person is happy with their slice. (From the most famous for two people - you cut, I pick - onwards).

What about a version where you're not sharing a cake, but trying to guess a number as closely as possible? (Doesn't have to follow any uniformity rules or anything, eg - how much do you think this house will sell for at auction, or what do you think the score will be in this game?)

Usually what happens in this situation is one person guesses X, the next person guesses slightly more than X, and the third person guesses X-1, leaving the first person with no hope. Is there a fair system for guessing here?

[edit]hm, this probably makes no sense, since everyone could just write down their guess on a bit of paper. But that's not what normally happens when I do this in real life :P

[Last edited by TripleM at 05-01-2015 07:32 AM]
05-01-2015 at 07:01 AM
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jamesdenem
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I would think this could work to make it fair:

Let N be the number trying to be guessed, T being the total number of guesses possible.

The first person guessing has a 1/T chance of choosing N with his guess, X.
If he does not choose N, and any 2 other people choose X-1 and X+1, then he has no chance of being closest.

Thus wouldn't it be fair to set a rule that states you cannot choose a number adjacent to previous guesses?

That would limit participants to at most T/2, roughly. It would also give you at most 2 winners.
Each person would have a 1/T chance of choosing N, but if none of them do, then you would have at most 2 people who are closest to the number. ( N-1 and N+1 )


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05-01-2015 at 01:09 PM
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Someone Else
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How about this: Each person picks two numbers that they are 90% confident are around the target number. The winner is the person with the smallest range that had the target in their range. To prevent this case:
 x       y
 x+1   y-1

simply have each person choose their lower bound first, and then they choose their upper bounds in reverse order. Then, the worst case scenario is:
 x      y-1
 x+1      y

05-01-2015 at 01:33 PM
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RyTracer
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icon Re: Puzzle "tag" (+2)  
Wow guys, this topic is dusty. Time for a brief revival.

Imagine a point in three-dimensional space that emits light in all directions. You may place opaque spheres around the point but not intersecting it. The spheres must also not intersect each other. They may be as large or close as you want but not infinite in size or closeness. What is the fewest number of spheres you need to block out all direct light from every angle at an arbitrary distance from the light (obviously beyond the spheres)? Also, roughly describe the configuration.

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01-30-2016 at 05:32 AM
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Lucky Luc
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icon Re: Puzzle "tag" (+3)  
Okay, let's see if I'm better at geometry than logic.

First off, one assumption I made: The spheres' radii may differ. Then I think I have a solution:
Click here to view the secret text

01-30-2016 at 03:47 PM
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RyTracer
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That is correct.

Part 2. What is the answer if all spheres are the same radius?

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01-31-2016 at 07:15 AM
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