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Sep 15, 2015 10:06 PM

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6539
Then, because B always calls the same side as she flips, she will be wrong.

There are only 2 outcomes. Either they flip the same or they flip differrently. As long as they guess differently between these 2 outcomes, 1 will always be wrong.
Sep 15, 2015 10:32 PM

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Zymf said:
There are only 2 outcomes. Either they flip the same or they flip differrently. As long as they guess differently between these 2 outcomes, 1 will always be wrong.

Okay, now I understand.
You will be what you will to be.
Sep 17, 2015 4:29 AM

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It's been 48 hours so here's a new one. This one's easy so no complaints -_-

What's the next umber in the following sequence:

5961, 1596, 6159...?

Lulu ❤ | My MALoween Candy
Sep 17, 2015 7:09 AM

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3671
9615
You will be what you will to be.
Sep 17, 2015 7:15 AM

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6539
^ Oh yea, of course - Because the leftmost digit will be moved to the front :D
Sep 17, 2015 7:17 AM
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gone bai bai
Sep 17, 2015 1:31 PM

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9885
Interesting math bit with the riddle.

5961 - 1596 = 4365
6159 - 1596 = 4563
9615 - 6159 = 3456
Sep 17, 2015 3:16 PM

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Switch-kun said:
9615

You're right ^^

Lulu ❤ | My MALoween Candy
Sep 18, 2015 1:03 AM

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3671
Okay, so here's the next riddle -
Four jolly men sat down to play, they played all night until the break of day.
They played for cash and not for fun, with a separate score for every one.
When it came time to square accounts, they all made quite fair amounts.
Not one lost and all gained.

Now that you've heard, can you explain?
You will be what you will to be.
Sep 18, 2015 4:49 AM

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12441
Hmm, I've been thinking about this for a few minutes now. What were they playing? What does the separate score part mean? Is the no one lost part an overall result?

If it's cards they're, then they weren't playing with each other.

If it's another kind of playing, they're in an orchestra? because then they'd all get paid and each would have a musical score (sheet music) in front of them (separate score).

I can't think of other explanations, unless they all played cards/other types of gambling games and won more than they lost. So by the end of the night, there were no losses, only gains.

┐(‘~`;)┌

Lulu ❤ | My MALoween Candy
Sep 18, 2015 5:06 AM

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grave_robber said:
If it's another kind of playing, they're in an orchestra? because then they'd all get paid and each would have a musical score (sheet music) in front of them (separate score).

Well, I just picked up a riddle in a hurry, so sorry for such a riddle. This is the correct answer.
You will be what you will to be.
Sep 18, 2015 5:51 AM

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It was a good riddle, I like ones that require a bit of thinking ^^

New Puzzle
"A sultan tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. Upon receiving the advice, they jump on the camels and race to the city as fast as they can.

What did the wise man say to them?"

Lulu ❤ | My MALoween Candy
Sep 18, 2015 6:05 AM

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He asked them to switch their camels.
Sep 19, 2015 3:00 AM

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El-Psy_Congroo said:
He asked them to switch their camels.

That is correct.

Lulu ❤ | My MALoween Candy
Sep 19, 2015 4:06 AM

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919
[b]So there's this king. Someone breaks into his wine cellar where he stores 1000 bottles of wine. This person proceeds to poison one of the 1000 bottles, but gets away too quickly for the king's guard to see which one he poisoned or to catch him.

The king needs the remaining 999 safe bottles for his party in 4 weeks. The king has 10 servants who he considers disposable. The poison takes about 3 weeks to take effect, and any amount of it will kill whoever drinks it. How can he figure out which bottle was poisoned in time for the party?
Sep 19, 2015 4:22 AM

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Dec 2012
6539
But if the servants start drinking of the 999 safe bottles to test if it's poisoned, then there will not be 999 full bottles for the party?

Disregarding this fact I'd do the following:
Day 1: Split the 1000 bottles in 10 groups of 100 and have the servants take a sip from every single bottle in each group.
Day 2: Split each group in half and make 10 new groups and have the servants take a sip from every single bottle in the new groups.
Day 3: Split the groups in half again and make 10 new groups so that only 25 of the bottles from Day 1 remains (you get the idea) and have the servants take a sip from every single bottle in the new groups.
Day 4-6: Keep splitting the groups in half, forming new unique groups.
Day 7: Split the groups in half one last time and have the servants take a sip from every single bottle in the new groups. Since there initially was 100 bottles in each group, there will only be 100*0.5^7 ≈ 1 bottle from from Day 1 remaining in each group.

Now I'd wait 2 weeks till my servants would begin dying 1 by 1 and on the same day as the party, I'd be able to work out which bottle contained poison.
ZymfSep 19, 2015 4:43 AM
Sep 19, 2015 4:38 AM

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8797
Empty the wine from all the bottles since it only says he needs the bottles?
I've been here way too long...
Sep 19, 2015 10:17 AM

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Zymf said:
But if the servants start drinking of the 999 safe bottles to test if it's poisoned, then there will not be 999 full bottles for the party?

Disregarding this fact I'd do the following:
Day 1: Split the 1000 bottles in 10 groups of 100 and have the servants take a sip from every single bottle in each group.
Day 2: Split each group in half and make 10 new groups and have the servants take a sip from every single bottle in the new groups.
Day 3: Split the groups in half again and make 10 new groups so that only 25 of the bottles from Day 1 remains (you get the idea) and have the servants take a sip from every single bottle in the new groups.
Day 4-6: Keep splitting the groups in half, forming new unique groups.
Day 7: Split the groups in half one last time and have the servants take a sip from every single bottle in the new groups. Since there initially was 100 bottles in each group, there will only be 100*0.5^7 ≈ 1 bottle from from Day 1 remaining in each group.

Now I'd wait 2 weeks till my servants would begin dying 1 by 1 and on the same day as the party, I'd be able to work out which bottle contained poison.

Does all the servants taste the group?
In the second step you split it into half, are you disregarding 500 bottles ?

I think the way you think is correct, can you explain your solution properly?

TheConquerer said:
Empty the wine from all the bottles since it only says he needs the bottles?
Lol No
Sep 19, 2015 10:42 AM

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Dec 2012
6539
No, each servant only drink from 100 bottles each day.

Also, when you split the groups in half and make new groups, you have to split them "evenly".

To make it more simple, lets say there are 10 different colors and 100 bottles of each color. Day 1, you will have each servant choose a color and sip from every bottle of that color. When you split the groups, there have to be half as many red, half as many green and so on. The rest will be passed on to the next servant in the line.

When the first servant die, you'll know what color the poisonous drink has and you have effectively limited the number of dangerous bottles to 100. However, because the poison takes 2 weeks to take effect, you have to split all 1000 bottles anew on Day 2

I can't really explain my solution in a simpler way xD
Sep 19, 2015 11:02 AM

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10014
The bottle with a broken seal is the poisoned one. Or do you really plan to open one thousand wine bottles and expect the wine to stay good for a party that will happen in four weeks?
Sep 19, 2015 11:57 AM

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lupadim said:
The bottle with a broken seal is the poisoned one. Or do you really plan to open one thousand wine bottles and expect the wine to stay good for a party that will happen in four weeks?

Ah...the main focus of the puzzle is not about the container.
It's about how you identify the poisonous container with a strategy involving 10 people.
Sep 19, 2015 12:03 PM

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Jun 2015
919
@Zymf the answer is similar so I guess it's fine.

Answer:
Sep 20, 2015 12:41 PM

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9885
I'm the desire of all men, rich and poor. I've caused countless deaths, and no doubt will cause many more. I'm the one thing man can't live without, yet the one thing they all doubt. What am I?
Sep 20, 2015 1:32 PM

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6539
@El-psy_congroo: My solution was similar in some ways but there was some key differences. Your solution would only take 2 weeks and mine would take 3 weeks.

Also, I'm not sure if my solution is actually valid because of the fact that you can't divide odd numbers with 2...

@Astro: So deep :o!
Sep 20, 2015 2:52 PM

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Astros said:
I'm the desire of all men, rich and poor. I've caused countless deaths, and no doubt will cause many more. I'm the one thing man can't live without, yet the one thing they all doubt. What am I?


A woman? :p

Lulu ❤ | My MALoween Candy
Sep 20, 2015 3:05 PM

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9885
grave_robber said:
A woman? :p
No.
Sep 20, 2015 3:06 PM

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15934
Astros said:
I'm the desire of all men, rich and poor. I've caused countless deaths, and no doubt will cause many more. I'm the one thing man can't live without, yet the one thing they all doubt. What am I?
money?

Where there is no imagination there is no horror. || Arthur Conan Doyle || Happy Halloween!
Sep 20, 2015 3:09 PM

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Astros said:
I'm the desire of all men, rich and poor. I've caused countless deaths, and no doubt will cause many more. I'm the one thing man can't live without, yet the one thing they all doubt. What am I?

True Love?
Cheesy reply
I've been here way too long...
Sep 20, 2015 3:14 PM

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9885
Suzune-chan said:
money?
No.

TheConquerer said:

True Love?
Cheesy reply
No. Like Swiss
Sep 20, 2015 3:33 PM

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May 2014
8797
Hmm heres my second shot

Freedom?

Thats all from me :P
I've been here way too long...
Sep 20, 2015 3:37 PM

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Dec 2013
9885
TheConquerer said:
Hmm heres my second shot

Freedom?

Thats all from me :P
Ding Ding. Winner winner chicken dinner. I tried making it tricky in using the archaic definition of doubt in the last sentence. Either it was that or the ambiguity that made it difficult.
Sep 20, 2015 4:12 PM

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May 2014
8797
Woooooo
Yeah the doubt definitely threw me off since it sounds strange to say people doubt freedom.

Anyways I'll post a simple one and go to bed.

"Poor Edmund Dantes has been captured by evil forces and thrown in prison, there he is put in the torture chamber and faces the evil torturer, Maldovak. Being a man who enjoys teasing his victims Maldovak walks up to Edmund and bellows!
" Edmund! I will let you tell me any statememt, if it is a true one I will burn you till your skin peels and if it false I will drown you till you die! What will you choose?"
Edmund choose his words very carefully and Maldovak was forced to let him free being the simple brute he was."

What did Edmund say?
I've been here way too long...
Sep 20, 2015 4:23 PM

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Dec 2013
9885
TheConquerer said:
Woooooo
Yeah the doubt definitely threw me off since it sounds strange to say people doubt freedom.

Anyways I'll post a simple one and go to bed.

"Poor Edmund Dantes has been captured by evil forces and thrown in prison, there he is put in the torture chamber and faces the evil torturer, Maldovak. Being a man who enjoys teasing his victims Maldovak walks up to Edmund and bellows!
" Edmund! I will let you tell me any statememt, if it is a true one I will burn you till your skin peels and if it false I will drown you till you die! What will you choose?"
Edmund choose his words very carefully and Maldovak was forced to let him free being the simple brute he was."

What did Edmund say?
You're a liar.
Sep 20, 2015 6:32 PM

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Feb 2015
3671
I'll drown
You will be what you will to be.
Sep 20, 2015 6:38 PM

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15934
Hm, there is an easy answer to this one, but I just can't figure it out

Where there is no imagination there is no horror. || Arthur Conan Doyle || Happy Halloween!
Sep 20, 2015 10:26 PM

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6539
Switch-kun said:
I'll drown
I second this :D
Sep 21, 2015 12:04 AM

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8797
Switch-kun said:
I'll drown

Correct though its less elegant than I would've wished...
I've been here way too long...
Sep 21, 2015 12:45 AM

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Feb 2015
3671
:P
Okay, so here's another two level riddle (if you solved one, then solve the next, more difficult one) -

You have 10 bags full of coins, in each bag are 1,000 coins.
But one bag is full of forgeries, and you can't remember which one.
But you do know that a genuine coin weighs 1 gram, but forgeries weigh 1.1 grams.
To hide the fact that you can't remember which bag contains forgeries, you plan to go just once to the central weighing machine to get one accurate weight.
How can you identify the bag with the forgeries with just one weighing?
2nd part - What if you didn't know how many bags contain forgeries?
You will be what you will to be.
Sep 21, 2015 3:27 AM

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Dec 2012
6539
You just weigh:
1 coin from the 1st bag
10 coins from the 2nd bag
25 coins from the 3rd bag
50 coins from the 4th
75 coins from the 5th
100 coins from the 6th bag
250 coins from the 7th bag
500 coins from the 8th bag
750 coins from the 9th bag
1000 coins from the 10th bag

If it weighs 2861 it's the 10th bag
If it weighs 2836 it's the 9th bag
If it weighs 2811 it's the 8th bag
... and so on

Second part
Instead of just random number of coins from each bag, I think you'll have to make sure that you take at least twice as many coins for each new bag.

For example you can take:
- 1 coin from the 1st bag
- (1*2) = 2 coins from the 2nd bag
- (2*2) = 4 coins from the 3rd bag
- (4*2) = 8 coins from the 4th bag
... and so on

If both bag 1, 2 and 3 contains forgeries, you'll be able to work it out because (1+2+4)*0.1 doesn't exceed 8*0.1
Sep 21, 2015 3:42 AM

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Feb 2015
3671
Correct
You will be what you will to be.
Sep 21, 2015 5:09 AM

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Dec 2012
6539
Also about the last part :o?

Draw a line that passes through each box edge only once. You can start and end anywhere you want.
Sep 21, 2015 6:27 AM

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Jun 2015
919
Zymf said:
Also about the last part :o?

Draw a line that passes through each box edge only once. You can start and end anywhere you want.

I assume that the line should come out of box in this question.
If it's the case it's not possible.
Reason:
If it's a
_._._
|_|_|_|
|_|_|_|
The line can come out.
But in the diagram the two lines are merged, which makes it impossible.
Need two more edges to make it possible.
I'm not sure about my reasoning.
Sep 21, 2015 7:23 AM

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Dec 2012
6539
El-Psy_Congroo said:
I assume that the line should come out of box in this question.
The line he is drawing is just an example. The actual puzzle itself looks like this:

You can start inside one of the boxes or outside, it's your choice. As long as it's 1 continious line, everything goes.

You are right that it is impossible.. but your reasoning isn't quite what I'm looking for.
Sep 21, 2015 8:27 AM

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704
The question confuses me a bit, are we passing a line through 1 edge of each box maximum? Or passes through all edges but only once?
Sep 21, 2015 8:31 AM

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Dec 2012
6539
All edges but only once :D
Like this:
Sep 21, 2015 10:30 AM

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May 2014
8797
Poke a hole in the paper and draw on the other side?

I'm too lazy to take a picture and show how but it should work.
I've been here way too long...
Sep 21, 2015 10:53 AM

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Dec 2012
6539
No, I told you it was impossible ^^

Every time you enter a box you must also exit a box.
There are only 2 exceptions to this rule and that is when you begin and end your line.
There are 3 boxes with an un-even number of edges and therefore it is theoretically impossible.

Your turn to post a riddle Conquerer :D
Sep 21, 2015 11:32 AM

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May 2014
8797
Well yeah but thats why we poke a hole, its like a wormhole :D

Anyway heres my riddle

"A conquerer is asked to provide a good riddle but can't think of any, what does he do?"

Tell me when you find the answer
I've been here way too long...
Sep 21, 2015 11:36 AM
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10764
TheConquerer said:
Well yeah but thats why we poke a hole, its like a wormhole :D

Anyway heres my riddle

"A conquerer is asked to provide a good riddle but can't think of any, what does he do?"

Tell me when you find the answer


Why, he meditates about the situation until he succesfully thinks of a quirky and smart riddle.
gone bai bai
Sep 21, 2015 11:51 AM

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May 2014
8797
I didnt actually Google any of my riddles, I just remember them from an old book I had.

Speaking of which I remembered another riddle

"Sir Lancelot is on his way to to Saint Ives where he met a man with seven wives,
Each wife had seven sacks,
Each sack had seven cats,
Each cat had seven kits:
Kits, cats, sacks, and wives,
How many were there going to Saint Ives?"
I've been here way too long...
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