->This thread is for simple puzzles... You can post some too if you want, but don't post until the last one was answered... I'll start off with something easy...
A kid is hit by a bus.
DRIVER: My child!
KID: I am your child but you're not my father!
How are the two related?
†
The driver is the mother
->Was it too easy? :xD: Took me a minute before I figured it out... Maybe because I'm a guy? :P
Well, here's another one:
Jose is the son of Mario.
Therefore, Mario is the ____________ of Jose's father.
Could be easier than the first one... Meh... :xD:
†
Lol I could make so many answers for this one.
Mario is actually a woman - She's the wife of Jose's father.
Mario is really a man - He's the husband of Jose's father~ :D
Or he's just Jose's father.
->LOL, no... :xD: No theoretical stuff here, just pure facts... :haha:
Fact: I'm glad you weren't able to answer it on your first try... :D
†
->:D
The blank is something that Mario possesses
†
name
->Cool... :D
Would someone want to post the next one? :haha:
†
This sucks
Quote from: Diokatsu on September 21, 2010, 06:34:34 pm
This sucks
Hm, intersting puzzle. I'm going to guess...
Dio's mouth?
Quote from: winkio on September 21, 2010, 07:37:03 pm
Quote from: Diokatsu on September 21, 2010, 06:34:34 pm
This sucks
Hm, intersting puzzle. I'm going to guess...
Dio's mouth?
Correct.
I'm sorry, but these aren't even that interesting. I mean, why did you do this?
->You're making me feel bad... (~_~)
I've got nothing better to do... :P
Wanna post one?
†
EDIT:
->Since it's recently been established that this thread SUCKS, I decided to kick it up a notch. This is a puzzle that I borrowed from the Professor Layton series. I whipped up something quick in MS Paint:
(http://i844.photobucket.com/albums/ab9/giogsgs12/RPG%20Maker%20XP/LinePuzzle.png)
These drawings are nothing but simple patterns, but if you try to draw them without lifting your pen or drawing over a line that was already drawn, some of them become impossible. You can, however, draw ACROSS but not ON a line. Each picture can either be impossible or possible. How many possible drawings are there?

†
There are 0 possible drawings.
->RIGHTY-O! (Lobstrosity asked me to confirm the answer right away)
Post the next one dude! :D
†
This is some random puzzle that I have modified. It shouldn't be too difficult.
You are sadly imprisoned by some random people whom you have never met before. They give you 9 fruit and a set of scales. All of them look the same. One of them isn't poisonous. The non-poisonous fruit weighs slightly more than the rest, but the others (Which are poisonous) all weigh the same. The random and unfriendly people that you just met, say they will kill you if you don't take a bite out of one of the fruit. They also say they'll kill you if you use the scales more than twice. To be sure that you do not get yourself killed by taking a bite out of one of the fruit, how will you figure out which one isn't poisonous? (It is highly unlikely that anyone would be in this situation, but you never know, the answer may come in handy...

)
EDITED!
I weight all of them, and devide it by 9. Then I single one out abd weight it. If it weights more, it's the poisonous one and I eat one of the others. If it weights less, I eat it because the poisonous one is among the other 8.
You suck, this is easy. >:U
->Scales... The one with two plates.. :P
I already have the answer... :haha:
†
Sorry, my mistake. I accidentally typed it up wrong. D:
Quote from: ShadowPierce on September 22, 2010, 08:55:45 am
->RIGHTY-O! (Lobstrosity asked me to confirm the answer right away)
I just want to add this:
If there are more than 2 nodes with an odd number of edges joint by that node, then the drawing is not possible. You just have to count the nodes, it's that simple. Basic graphics theory. :P
Quote from: Blizzard on September 22, 2010, 10:29:56 am
Quote from: ShadowPierce on September 22, 2010, 08:55:45 am
->RIGHTY-O! (Lobstrosity asked me to confirm the answer right away)
I just want to add this:
If there are more than 2 nodes with an odd number of edges joint by that node, then the drawing is not possible. You just have to count the nodes, it's that simple. Basic graphics theory. :P
->It's the trick to getting these puzzles right. :haha: High five! :D
†
Quote from: Lobstrosity on September 22, 2010, 09:14:39 am
This is some random puzzle that I have modified. It shouldn't be too difficult.
You are sadly imprisoned by some random people whom you have never met before. They give you 9 fruit and a set of scales. All of them look the same. One of them isn't poisonous. The non-poisonous fruit weighs slightly more than the rest, but the others (Which are poisonous) all weigh the same. The random and unfriendly people that you just met, say they will kill you if you don't take a bite out of one of the fruit. They also say they'll kill you if you use the scales more than twice. To be sure that you do not get yourself killed by taking a bite out of one of the fruit, how will you figure out which one isn't poisonous? (It is highly unlikely that anyone would be in this situation, but you never know, the answer may come in handy...

)
EDITED!
Weigh 4 on each scale. If they weigh the same the odd one out is poison, don't eat it. Otherwise the side that tips has a poison fruit, eat from the other side.
I got's no puzzles, atm.
->I think you didn't read it right... All apples are poisoned except one... BTW, I've got the right answer (Lob confirmed it) but am still waiting who else gets it... Interesting puzzle... :D
†
It's an old puzzle, I'm actually surprised that more people don't know it:
1. Weigh three fruits on each scale. If they weigh the same, weigh the other 3 fruits. Otherwise, weigh the heavier 3.
2. Weigh the two of the three fruits that you chose from the last weighing. If they weigh the same, eat the third fruit. Otherwise, eat the heavier one.
Here's a riddle that's similar, but more complicated:
You have 12 cases of beer. Each case contains 12 bottles of beer. Each bottle of beer weighs 12 ounces. One of the cases has bottles filled with water instead, which weighs slightly less than the beer, so the bottles weigh 11 ounces. With just one weighing on a numeric scale (it will display the weight in pounds and ounces of what you are weighing), how do you determine which case contains the water?
Hint:
you can move bottles from one case to another.
Quote from: winkio on September 22, 2010, 02:35:55 pm
Here's a riddle that's similar, but more complicated:
You have 12 cases of beer. Each case contains 12 bottles of beer. Each bottle of beer weighs 12 ounces. One of the cases has bottles filled with water instead, which weighs slightly less than the beer, so the bottles weigh 11 ounces. With just one weighing on a numeric scale (it will display the weight in pounds and ounces of what you are weighing), how do you determine which case contains the water?
Hint: you can move bottles from one case to another.
Number the boxes one to twelve. From each box, set aside a # of bottles corresponding to the box's #. Keep the bottles appart, so you can place them back in the right boxes. Now put all the boxes on the scale.
Take the weight in ounces and divide by 12. The remainder subtracted from 12 gives you the box # which had the bottles of water. You can now replace the bottles in the containers.
Hope I'm right with this one, didn't get the chance to correct my last one. :p
yup
->Since no one posted yet, I'll post another one... :haha:
One cold & dark night, Pedobear "surprise adopts" you. He tells you that you can make one final statement that will determine how you die.
If what you say is true, you will die of "surprise butt sex".
If you say something false, you die naked in a public area.
Both statements are disadvantageous to you. What should your final statement be to prevent you from suffering from one of the humiliating deaths?
Humor added for more fun... :^_^':
†
"God exists"
->Since that is related to religion, it is considered an opinion. You need a statement that can't be true nor false. :D
†
technically, opinions can't be proved true or false. Of course, you are looking for a logical paradox, something like "This statement is false." Or the quote in my sig, "I am lying."
EDIT: Oops, posted it before I was done. wait one second...
EDIT2: Or, you are looking for the answer "I will die naked in a public area", which would contradict whatever punishment is given to you.
->Though both answers can be accepted, I was specifically looking for the second one... :P
GJ! :^_^':
†
EDIT:
->Damn, gotta think of harder ones... :xD:
†
Quote from: ShadowPierce on September 23, 2010, 07:30:10 pm
->Since that is related to religion, it is considered an opinion. You need a statement that can't be true nor false. :D
†
This statement can't be proven true or false. Just face the fact thatI won this one.
I feel he is not capable of accepting it.
a husband and wife are celebrating their 20th birthday together their children, grand children, and great grand children attend the party. how is this possible
It's not at all. And if it is, what the fuck... :xD:
Quote from: Ryexander on September 25, 2010, 02:36:39 am
a husband and wife are celebrating their 20th birthday together their children, grand children, and great grand children attend the party. how is this possible
They're in their genes.
Adoption.
Good one. Haven't thought of that.
They were both born on Feb 29th.
20 years of Feb 29th would mean they're 80 years of age.
Quote from: Aqua on September 25, 2010, 04:31:45 pm
They were both born on Feb 29th.
20 years of Feb 29th would mean they're 80 years of age.
That's not how age works. >:U
Are they not human?
Like, are they fruit flies or something?
Quote from: Ryexander on September 25, 2010, 02:36:39 am
a husband and wife are celebrating their 20th birthday together their children, grand children, and great grand children attend the party. how is this possible
Their children are the actual children of the couple, while the grand children and great grand children happen to belong to different grandparents and great-grandparents, so they could be any age.
Quote from: Aqua on September 25, 2010, 04:31:45 pm
They were both born on Feb 29th.
20 years of Feb 29th would mean they're 80 years of age.
She's talking about the leap year.
Quote from: Ryexander on September 25, 2010, 02:36:39 am
a husband and wife are celebrating their 20th birthday together their children, grand children, and great grand children attend the party. how is this possible
Quote from: Subsonic_Noise on September 25, 2010, 04:45:34 pm
Quote from: Aqua on September 25, 2010, 04:31:45 pm
They were both born on Feb 29th.
20 years of Feb 29th would mean they're 80 years of age.
That's not how age works. >:U
->Age was never mentioned dude... :P
I also think the Feb29 thing is the most likely answer... :haha:
†
Shit, I got it. ShadowPierce's comment that age wasn't mentioned gave me the idea.
They are celebrating their 20th birthday together. That doesn't mean that they're 20, it means that it's the 20th birthday they are celebrating together. Let's say last year they didn't celebrate together and the year before that it was their 19th time together. So they could be 80 without problems.
->Alright... Since so many possible solutions were pointed out, I'll post the next one...
One night, a man & his wife are going for a cross-country drive when the car they're driving went out of control in a corn field. It crashed to a tree & damaged the car enough that it won't be able to run again. Fortunately, both were safe. The man decided to go somewhere nearby to ask for help. To make sure that his wife will be safe, he rolled up the windows & locked the doors & the trunk/compartment.
A few hours later, he returns. To his surprise, he found his wife bloody & lifeless inside the car. The doors were still locked & no windows were destroyed. Inside was someone he has never seen before.
How did the woman die? Who is the killer?
Creepy puzzle... I wasn't able to solve it on my first try... Maybe someone here can? :haha:
†
She died giving birth to a baby.
She gave dying baby to a birth.
Quote from: ShadowPierce on September 26, 2010, 11:41:14 am
->Alright... Since so many possible solutions were pointed out, I'll post the next one...
One night, a man & his wife are going for a cross-country drive when the car they're driving went out of control in a corn field. It crashed to a tree & damaged the car enough that it won't be able to run again. Fortunately, both were safe. The man decided to go somewhere nearby to ask for help. To make sure that his wife will be safe, he rolled up the windows & locked the doors & the trunk/compartment.
A few hours later, he returns. To his surprise, he found his wife bloody & lifeless inside the car. The doors were still locked & no windows were destroyed. Inside was someone he has never seen before.
How did the woman die? Who is the killer?
Creepy puzzle... I wasn't able to solve it on my first try... Maybe someone here can? :haha:
†
The man realized his wife had autonomous control and she must've opened the door for the killer, who locked it again so she would not escape. Or the guy had a duplicate key. The woman died from blood loss. The killer is whoever is in the car.
lol
Quote from: Aqua on September 25, 2010, 04:31:45 pm
They were both born on Feb 29th.
20 years of Feb 29th would mean they're 80 years of age.
the answer I was looking for
Quote from: Blizzard on September 26, 2010, 09:23:09 am
Shit, I got it. ShadowPierce's comment that age wasn't mentioned gave me the idea.
They are celebrating their 20th birthday together. That doesn't mean that they're 20, it means that it's the 20th birthday they are celebrating together. Let's say last year they didn't celebrate together and the year before that it was their 19th time together. So they could be 80 without problems.
a consequence of the wording of the riddle, I've never realized that this could be a possible answer
I got a riddle...
Your in your house in the basement and there are three switches. In the Attic there are three lights, each connected to its own switch. If you could go in the Attic only once and you can't see the lights from the basement, how would you know what lights go to which switches.
NOTE: this one is not that hard.
Turn one on (A).
Wait a bit.
Turn A off & turn a different one on (B).
Go upstairs to check.
The lightbulb that is on is connected to switch B.
Feel the other two.
The warm lightbulb is connected to switch A.
Quote from: Aqua on September 26, 2010, 09:31:13 pm
Turn one on (A).
Wait a bit.
Turn A off & turn a different one on (B).
Go upstairs to check.
The lightbulb that is on is connected to switch B.
Feel the other two.
The warm lightbulb is connected to switch A.
HAHAHAHAHA. THE TRICK IS YOU BURN THE HOUSE DOWN. THAT WILL TEACH THOSE FUCKING ELECTRICIANS TO MAKE SHITTY PLACEMENT DECISIONS.
Quote from: Diokatsu on September 26, 2010, 09:34:26 pm
Quote from: Aqua on September 26, 2010, 09:31:13 pm
Turn one on (A).
Wait a bit.
Turn A off & turn a different one on (B).
Go upstairs to check.
The lightbulb that is on is connected to switch B.
Feel the other two.
The warm lightbulb is connected to switch A.
HAHAHAHAHA. THE TRICK IS YOU BURN THE HOUSE DOWN. THAT WILL TEACH THOSE FUCKING ELECTRICIANS TO MAKE SHITTY PLACEMENT DECISIONS.
->LOL, lvlup! :xD:
Seriously though, Aqua's pwning puzzle after puzzle... :xP:
†
Quote from: ShadowPierce on September 26, 2010, 09:36:33 pm
Quote from: Diokatsu on September 26, 2010, 09:34:26 pm
Quote from: Aqua on September 26, 2010, 09:31:13 pm
Turn one on (A).
Wait a bit.
Turn A off & turn a different one on (B).
Go upstairs to check.
The lightbulb that is on is connected to switch B.
Feel the other two.
The warm lightbulb is connected to switch A.
HAHAHAHAHA. THE TRICK IS YOU BURN THE HOUSE DOWN. THAT WILL TEACH THOSE FUCKING ELECTRICIANS TO MAKE SHITTY PLACEMENT DECISIONS.
->LOL, lvlup! :xD:
Seriously though, Aqua's pwning puzzle after puzzle... :xP:
†
Protip: Make harder puzzles that people haven't seen before. I've read all those puzzles before.
Quote from: Diokatsu on September 26, 2010, 09:42:23 pm
Quote from: ShadowPierce on September 26, 2010, 09:36:33 pm
Quote from: Diokatsu on September 26, 2010, 09:34:26 pm
Quote from: Aqua on September 26, 2010, 09:31:13 pm
Turn one on (A).
Wait a bit.
Turn A off & turn a different one on (B).
Go upstairs to check.
The lightbulb that is on is connected to switch B.
Feel the other two.
The warm lightbulb is connected to switch A.
HAHAHAHAHA. THE TRICK IS YOU BURN THE HOUSE DOWN. THAT WILL TEACH THOSE FUCKING ELECTRICIANS TO MAKE SHITTY PLACEMENT DECISIONS.
->LOL, lvlup! :xD:
Seriously though, Aqua's pwning puzzle after puzzle... :xP:
†
Protip: Make harder puzzles that people haven't seen before. I've read all those puzzles before.
->I use ones that I've heard before... :^_^': I think I'll try making one of my own... :D
†
two puzzles for the price of zero:
using four '9's and addition, subtraction, multiplication, and division, find and expression that evaluates to 100.
(everything is in base 10)
there are 100 prisoners standing in a line, facing front. Each prisoner has either a white or a black hat on. Since they are in the line, each prisoner can see the hat of every man in front of him (the person at the back can see 99 hats). Starting from the back of the line, and going to the front, each is asked what color hat they have. If they are right, they live, wrong, they die. How many prisoners can be saved, and how should they do it?
Quote from: winkio on September 26, 2010, 09:47:39 pm
two puzzles for the price of zero:
using four '9's and addition, subtraction, multiplication, and division, find and expression that evaluates to 100.
(everything is in base 10)
->Do I have to use all operations or should it be like this?
BTW, I'm still having a hard time figuring out the second one... :P
†
Quote from: ShadowPierce on September 26, 2010, 09:16:22 am
Quote from: Ryexander on September 25, 2010, 02:36:39 am
a husband and wife are celebrating their 20th birthday together their children, grand children, and great grand children attend the party. how is this possible
Quote from: Subsonic_Noise on September 25, 2010, 04:45:34 pm
Quote from: Aqua on September 25, 2010, 04:31:45 pm
They were both born on Feb 29th.
20 years of Feb 29th would mean they're 80 years of age.
That's not how age works. >:U
->Age was never mentioned dude... :P
I also think the Feb29 thing is the most likely answer... :haha:
†
What now?
its not realy a puzzle but it is puzzling me ...
If I raced a friend of mine from the second floor to the eight floor (I used the stairs he used the elevator) and I won. Is it not fair that i pressed the button for the elevator every 2 floors,
Puzzle
Why? I really want to see what people will say
Quote from: ShadowPierce on September 26, 2010, 10:12:43 pm
Quote from: winkio on September 26, 2010, 09:47:39 pm
two puzzles for the price of zero:
using four '9's and addition, subtraction, multiplication, and division, find and expression that evaluates to 100.
(everything is in base 10)
->Do I have to use all operations or should it be like this?
BTW, I'm still having a hard time figuring out the second one... :P
†
you got the first one.
Quote from: winkio on September 26, 2010, 09:47:39 pm
there are 100 prisoners standing in a line, facing front. Each prisoner has either a white or a black hat on. Since they are in the line, each prisoner can see the hat of every man in front of him (the person at the back can see 99 hats). Starting from the back of the line, and going to the front, each is asked what color hat they have. If they are right, they live, wrong, they die. How many prisoners can be saved, and how should they do it?
They can save half the prisoners for sure. Each "odd" prisoner says the color of the hat of the person in front of him so that one can live while each "even" prisoner says the color he has just heard from the prisoner saying that before him.
That would be my first thought.
Nope, you can save more.
How about this:
You can save all for sure except for one. The first person counts how many people with the same hats in front of him there are in a row and adds one. If there are 5 people with black hats immediately in front of the first person to tell, then that person will say black 6 times. Now the next 5 people know that they have black hats and the one after them knows that he has a white hat. When its the white hat guy's turn, he has to count the white hats in front of him including his own (since he wants to live, lol). Let's say there are 2 more people with white hats, so he says 3 times white. So the next 2 people know that they have white hats and the one after has a black one. This can go on until the end. Only the first guy might not survive since he might have a different hat then the people after him.
Am I getting closer?
each person can only say black or white once. But you are getting close...
->How about tapping the next person's right shoulder when the next one's wearing black, & the left shoulder when the next one's wearing white?
†
Does a prisoner know when the guy behind him gets killed or not?
EDIT: What about this:
The first person says white if the next two people have the same color and black if they have a different color. Now the second person knows their own color from that information because they can see the color of the person in front of them. The third person has heard the first and knows whether the person before them has the same or a different color. Now they only need to listen to what the person before them says and they will know their own color. This tactic would save 66 prisoners.
Quote from: ShadowPierce on September 27, 2010, 10:05:13 am
->How about tapping the next person's right shoulder when the next one's wearing black, & the left shoulder when the next one's wearing white?
†
no other communication is allowed
Quote from: Blizzard on September 27, 2010, 10:22:38 am
Does a prisoner know when the guy behind him gets killed or not?
EDIT: What about this:
The first person says white if the next two people have the same color and black if they have a different color. Now the second person knows their own color from that information because they can see the color of the person in front of them. The third person has heard the first and knows whether the person before them has the same or a different color. Now they only need to listen to what the person before them says and they will know their own color. This tactic would save 66 prisoners.
That is a decent tactic, but there is another strategy where you can save more.
Then what about the dying? Do they know whether the people before died or not?
EDIT: Right now I am considering a completely different perspective. If you put together all answers, you can regard them as 0 (black) and 1 (white). If you can compress 80 bits of information into 20 bits, then you would be able to describe the colors of the last 80 people using the first 20. That way you would save 80 people. Am I going into a good direction here or does it have nothing to do with information theory? xD
No, but you keep getting closer. Here's a hint: you can save 99 people.
Crap. D: I figured out a way to save 93 of them. 7 serve as binary counter for how many prisoners there are with black hats (since 2^7 can count up to 128). The rest is easy. The 8th person know how many black hats there are and since there are 93 people left, he also knows how many white hats there are. Then he just would have to count the hats he can see and do the math to see which hat he is wearing. Everybody else had to do the same thing That way 93 people would survive.
I guess I have to try harder if I want to save 99 of them. xD
EDIT: Ok, this is either the solution you are looking for or a clever alternative solution. The first person can't be saved as he has no previous information. But through his voice tone he can give a hint to the next person. The first person can simply tell the color of the next person so the second person knows their color. Now, the second person says their color in a normal manner if the next person has the same color. If the next person has a different color, they yell out their color. That's it. They tell the correct color at the same time as they transmit the information of the color of the next person's hat. 99 people will be saved.
nope, they are not allowed any other sort of communication like tone changing.
The solution is inside here, if you give up:
The person at the back can't be saved, and thus counts the number of each color hat in front of him. Since he sees 99 hats, there must be an odd number of one color and an even color of another. He says whichever color has an odd number. After that, the next person looks up and sees 98 hats. There will be either even numbers of both or odd numbers of both. If even numbers, he knows that he had one of the hats that the first person saw as an odd number. Otherwise, he knows he has the opposite color of what the first person said. In this manner, the prisoners keep track of what the people before them have said, and whether the numbers of hats are odd or even, and each can guess his own color hat.
*gasp* Winko! you evil person you! you contently left out that there are 50 hats of each color! that was key to solving this riddle!
Um... You can do that with any number of hats of each color o.o
Doesn't have to be 50
Aqua is correct. They could all have the same color hat and this strategy would still work. Or any other combination.
*facepalm* nvm I see that now...
I give up, lol! I somehow knew that "odd" and "even" were the words why you told me that I kept getting closer, but I couldn't find a way to apply them. xD