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Ark's Perplexing Puzzles

arkvoodle

Diabolical
Joined
Sep 20, 2008
Location
Somewhere
This was a thread that i've been thinking about for a while, and I think it'll be a fun addition to the section.
Unlike some other forum games, this one requires a level of intelligence and a logical sense of thinking!

First off, here are the rules;

1- After 3 incorrect answers, there will be a hint
2- After 3 further incorrect answers, the answer will be presented.
3- Whoever gets the correct answer and presents the solution will be awarded puzzle points!
4- When a certain amount of puzzle points are gathered up, the user will be donated market points by myself as a reward. (Their PP total will be reset after this)
5- (added October 30th) For every wrong answer, the total points available will decrease by 3 D:
6- Have fun puzzling!

So. onto the first puzzle!
~~~~


Ark's Perplexing Puzzles- Number 001- Brain Food- 15 PP

You have three boxes of fruit.
One contains just apples, one contains just oranges, and one contains a mixture of both.

Each box is labeled -- one says "apples," one says "oranges," and one says "apples and oranges."

However, it is known that none of the boxes are labeled correctly.

How can you label the boxes correctly if you are only allowed to take and look at just one piece of fruit from just one of the boxes?

~~~

Good Luck! The deadline for the answer is 1st November!

-----

Current Points Totals-


Durion- 105PP
Steve- 82PP
Michael Heide- 67PP
Athenian200- 62PP
Destiny- 40PP
Locke64- 25PP
Meego7- 25PP
SheikahWarrior- 17PP
TheGreen- 22PP
SuperSilly- 20PP
February Eve- 19PP
Xinnamin- 15PP
Josh- 14PP
Baysiderulez- 12PP
Jedizora- 9PP
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K4King- 5PP
Penguinboy82- 5PP
Hazel- 2PP
 
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Xinnamin

Mrs. Austin
Joined
Dec 6, 2009
Location
clustercereal
Take a fruit out of the box labeled "apples and oranges".

If that fruit is an apple, then the box must be all apples because it cannot be a mixture. The box labeled "oranges" cannot have oranges and thus must be the mixture. The box labeled "apples" must then be oranges.

If that fruit is an orange, then the box must be all oranges because it cannot be a mixture. The box labeled "apples" cannot have apples and thus must be the mixture. The box labeled "oranges" must then be apples.

Did I do that right?
 

Destiny

Single❒Taken❒Assassin✔
Joined
Jul 26, 2010
Location
nowhere
Search the box labelled "Apples and Oranges".
Identify what you obtain; it is either: a Apple, or a Orange.
Use the label for the obtained item on the crate.

So if I pull out an Apple from the mixed crate, Put and Apple Label on it. What to put on the Apple crate is Orange, And the Apple Crate gets an mixed Label.

If I pull out an Orange from the mixed crate, Put a Orange on the mixed crate label. On the Apple crate I put an Orange label, on the Orange crate, I put a mixed label.

Crap I was ninja'd.
 

arkvoodle

Diabolical
Joined
Sep 20, 2008
Location
Somewhere
Aww. Sorry about that Destiny :<

Anyway. NEXT PUZZLE! -For the easier alternative, see below-

Puzzle 002 -Fake Coins- 20/20PP-


You have twelve coins. You know that one is fake. The only thing that distinguishes the fake coin from the real coins is that its weight is imperceptibly different. You have a perfectly balanced scale. The scale only tells you which side weighs more than the other side.

What is the smallest number of times you must use the scale in order to always find the fake coin?

Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.

These are modern coins, so the fake coin is not necessarily lighter.

Presume the worst case scenario, and don't hope that you will pick the right coin on the first attempt.
~~~
ALTERNATIVE!

Puzzle 002- Family Reunion- 5PP

At a family get-together there was a: Son, Daugher, Mother, Father, Aunt, Uncle, Niece, Nephew, and Cousin. However, only 4 people were present. How is this possible?

~~~

Expires 1st November-
 
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penguinboy82

Nature's troll
Joined
Mar 17, 2010
Location
Pacific Northwest
Sorry, couldn't figure out the first one, so I'm doing the second.

The answer is that the son, daughter, neice, and nephew are the same people, and the the mom, dad, aunt and uncle are the same people. The father and mother are siblings. Neither of their spouses came. If the son belongs to the father and the daughter to the mother (or vice versa), then that means the son is the mothers nephew and the mother is the son's aunt. Same thing goes for the daughter. She is the father's neice, and the father is her uncle.

Does that make sense?

EDIT: I may have gotten the answer to the first, and if it's correct I would like THOSE puzzle points instead. Is the answer 144?
 

arkvoodle

Diabolical
Joined
Sep 20, 2008
Location
Somewhere
Sorry, couldn't figure out the first one, so I'm doing the second.

The answer is that the son, daughter, neice, and nephew are the same people, and the the mom, dad, aunt and uncle are the same people. The father and mother are siblings. Neither of their spouses came. If the son belongs to the father and the daughter to the mother (or vice versa), then that means the son is the mothers nephew and the mother is the son's aunt. Same thing goes for the daughter. She is the father's neice, and the father is her uncle.

Does that make sense?

EDIT: I may have gotten the answer to the first, and if it's correct I would like THOSE puzzle points instead. Is the answer 144?

Sorry, but i'll have to take the one you answered first. The answer was correct! :D
 

SuperSilly

Horizon Walker
Joined
Aug 3, 2009
Location
Somewhere
Split the coins into two groups of six. Place the two groups on either side of the scale. Call these groups H and L, for Heavy and Light.

One side will contain the fake coin and thus both sides will show different weights. Since we don't know yet if the fake is heavier or lighter, we will choose one of the two groups at random, say, group L. (Times scale used: 1) Split group L into two groups of three. Call these L1 and L2. Weigh them, and if the scale shows them to be equal, then you know the fake is not in group L. (Times scale used: 2) You also know that the fake is heavier than normal coins because it was what made group H heavier than L, which would have been normal coins if they were equal in weight.

If L1 and L2 are inconsistent in weight, then you know that the fake coin is lighter than normal coins, and you can eliminate group H.

If the scale was still inconsistent after weighing L1 and L2, take the lighter of these groups and take two from them. Weigh these. If one is lighter than the other, it is the fake. If they are equal, then the remaining coin is fake. (Times scale was used: 3)

If the the fake was heavier, split H into H1 and H2, and weigh them to see which is heavier. (This would be the third time the scale was used if there was no need to proceed with group L.) Take the three that were in the heavier group and take two from it. Weigh these two. If they are equal, then the remaining coin is the fake, because it was what originally imbalanced the scale. If one of these coins is heavier than the other, then that coin is the fake. (Times scale used: 4)

So, you use the scale a minimum of 3 or 4 times.

Was that right? :sweat:
 

arkvoodle

Diabolical
Joined
Sep 20, 2008
Location
Somewhere
Split the coins into two groups of six. Place the two groups on either side of the scale. Call these groups H and L, for Heavy and Light.

One side will contain the fake coin and thus both sides will show different weights. Since we don't know yet if the fake is heavier or lighter, we will choose one of the two groups at random, say, group L. (Times scale used: 1) Split group L into two groups of three. Call these L1 and L2. Weigh them, and if the scale shows them to be equal, then you know the fake is not in group L. (Times scale used: 2) You also know that the fake is heavier than normal coins because it was what made group H heavier than L, which would have been normal coins if they were equal in weight.

If L1 and L2 are inconsistent in weight, then you know that the fake coin is lighter than normal coins, and you can eliminate group H.

If the scale was still inconsistent after weighing L1 and L2, take the lighter of these groups and take two from them. Weigh these. If one is lighter than the other, it is the fake. If they are equal, then the remaining coin is fake. (Times scale was used: 3)

If the the fake was heavier, split H into H1 and H2, and weigh them to see which is heavier. (This would be the third time the scale was used if there was no need to proceed with group L.) Take the three that were in the heavier group and take two from it. Weigh these two. If they are equal, then the remaining coin is the fake, because it was what originally imbalanced the scale. If one of these coins is heavier than the other, then that coin is the fake. (Times scale used: 4)

So, you use the scale a minimum of 3 or 4 times.

Was that right? :sweat:

Uh...you got it...and you took most of my explanation! D:
Congrats!


Puzzle 003- The Prison Warden- 22PP

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?
 
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penguinboy82

Nature's troll
Joined
Mar 17, 2010
Location
Pacific Northwest
I'll edit this post when I get the answer, but are they able to see how many prisoners have gone before them? Hmm, then based on the fact that they CAN see each other, even if not the other's faces, that it would be 69, because they will have all gone the same amount of times, and the leats would be 3. Also, only the first person who went inside would be able to declare it, because they are the one who knows the position of the switches in the beginning. After he/she has gone their third time, they must decide if the switches are in different positions or not. Then they would be free.

Sorry if that didn't make sense... :(

EDIT: total of what? MY point total was 5.
 
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arkvoodle

Diabolical
Joined
Sep 20, 2008
Location
Somewhere
I can't give out a hint or additional info until another two incorrect answers. and please be swift, because place holding posts mayb be disregarded in future.

Sorry. incorrect answer! D:

Previous Total- 25
Current Total- 22
 
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Destiny

Single❒Taken❒Assassin✔
Joined
Jul 26, 2010
Location
nowhere
1. Appoint a "scorekeeper" and the other 22 prisoners as "transmitters".

2. When a transmitter enter the switch room for the first time, he will flick switch B ON if he sees it in the OFF position. Alternatively, he will flick switch A if he sees switch B is in the ON position and waits for his next visit. The transmitter's mission is to find ONE opportunity to flick switch B to the ON position. After he completes his mission, he will flick switch A for all subsequent visits. In short, each transmitter's role is to send a signal to the scorekeeper that he has been to the switch room.

3. The scorekeeper's role is to flick off switch B whenever he sees it in the ON position. Each time he flicks off switch B he adds 1 to his score. When he sees switch B is already OFF, he flicks switch A instead and does not add to the score. As soon as he accumulates 22, he reports to the Warden
 

arkvoodle

Diabolical
Joined
Sep 20, 2008
Location
Somewhere
1. Appoint a "scorekeeper" and the other 22 prisoners as "transmitters".

2. When a transmitter enter the switch room for the first time, he will flick switch B ON if he sees it in the OFF position. Alternatively, he will flick switch A if he sees switch B is in the ON position and waits for his next visit. The transmitter's mission is to find ONE opportunity to flick switch B to the ON position. After he completes his mission, he will flick switch A for all subsequent visits. In short, each transmitter's role is to send a signal to the scorekeeper that he has been to the switch room.

3. The scorekeeper's role is to flick off switch B whenever he sees it in the ON position. Each time he flicks off switch B he adds 1 to his score. When he sees switch B is already OFF, he flicks switch A instead and does not add to the score. As soon as he accumulates 22, he reports to the Warden

Correct~ 23PP~

Puzzle 004- Islands -10PP

There are two beautiful yet remote islands in the south pacific. The Islanders born on one island always tell the truth, and the Islanders from the other island always lie.

You are on one of the islands, and meet three Islanders. You ask the first which island they are from in the most appropriate Polynesian tongue, and he indicates that the other two Islanders are from the same Island. You ask the second Islander the same question, and he also indicates that the other two Islanders are from the same island.

Can you guess what the third Islander will answer to the same question?
 

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