Post by: katsumoto on January 22, 2004, 07:41:00 AM
No legs have I to dance,
No lungs have I to breathe,
No life have I to live or die
And yet I do all three.
What am I?
This is an easy riddle to solve. I will reveal the answer in a few days. Let's see who cracks this first! As soon as someone solves this riddle, I will continue to add more riddle to this thread. As the series goes on, the riddle will get harder. Good Luck.
Post by: tomguns on January 22, 2004, 07:58:00 AM
Post by: katsumoto on January 22, 2004, 08:14:00 AM
EPISODE II
What has roots that nobody sees, and is taller than trees.
Up, up it goes, and yet it never grows.
What is it?
Crack this one! Good luck!
Post by: scotchtape on January 22, 2004, 08:33:00 AM
Post by: DermicSavage on January 22, 2004, 11:17:00 AM
| QUOTE (scotchtape @ Jan 22 2004, 10:33 AM) |
| a building with a basement, nobody can see a basement. |
but it could also be referring to the foundation
Post by: katsumoto on January 22, 2004, 12:17:00 PM
EPISODE III
I have many feathers to help me fly. I have a body and head, but I'm not alive. It is your strength which determines how far I go. You can hold me in your hand, but I'm never thrown. What am I
Good Luck!
"Will hipstergk27 do it again?"
Post by: sk8ermike6789 on January 22, 2004, 01:40:00 PM
| QUOTE |
| Up, up it goes, and yet it never grows. |
pfft, thats y i didnt get the riddle, mountains DO grow.. everest is getting 2inches taller each year cause the continential plates or whatever are moving together...
Post by: Large Dopant white on January 22, 2004, 02:11:00 PM
| QUOTE (sk8ermike6789 @ Jan 22 2004, 11:40 PM) | ||
pfft, thats y i didnt get the riddle, mountains DO grow.. everest is getting 2inches taller each year cause the continential plates or whatever are moving together... |
You're not supposed to look at a riddle like it's a math question, Mike. You're supposed to use simple logic and common sense (paradoxally, thinking hard is what makes a riddle harder
).Not trying to jump down your throat ('cause if that was the case, I'd just diss you and be over with it
), but maybe that'll put you in the right mindset for riddles and puzzles.Not that I'm any good at 'em...
Post by: sk8ermike6789 on January 22, 2004, 04:28:00 PM
| QUOTE (Large Dopant white @ Jan 22 2004, 04:11 PM) | ||||
You're not supposed to look at a riddle like it's a math question, Mike. You're supposed to use simple logic and common sense (paradoxally, thinking hard is what makes a riddle harder ).Not trying to jump down your throat ('cause if that was the case, I'd just diss you and be over with it ), but maybe that'll put you in the right mindset for riddles and puzzles.Not that I'm any good at 'em... |
lol, i was jay pee'ing
Post by: sk8ermike6789 on January 22, 2004, 04:32:00 PM
Post by: XeroKitsune on January 22, 2004, 04:59:00 PM
Post by: sk8ermike6789 on January 22, 2004, 05:41:00 PM
| QUOTE (XeroKitsune @ Jan 22 2004, 06:59 PM) |
| ... a laser. Since you are not required to leave the string in tact and the bottle is glass <transperent/semitransparent> burn the string. |
well... i was looking for magnifying glass and you could burn the string with that... a laser would cut the bottle also, and thats against the rules
Post by: DuDeR MaN on January 22, 2004, 05:43:00 PM
| QUOTE (katsumoto @ Jan 22 2004, 05:17 PM) |
| WOW! hipstergk27 is on a roll. Okay, this is a dumb one: EPISODE III I have many feathers to help me fly. I have a body and head, but I'm not alive. It is your strength which determines how far I go. You can hold me in your hand, but I'm never thrown. What am I Good Luck! "Will hipstergk27 do it again?" |
a SHUTTLECOCK!?
Post by: geniusalz on January 22, 2004, 05:45:00 PM
Post by: sk8ermike6789 on January 22, 2004, 05:49:00 PM
| QUOTE (geniusalz @ Jan 22 2004, 07:45 PM) |
| Go to sunlight (I know you'll probably say: sunlight: wtf iz taht? ), get a magnifying glass, and burn the string. |
LMFAO your 4 minutes too late
BTW: wtf iz sunlight?
Post by: XeroKitsune on January 22, 2004, 06:05:00 PM
| QUOTE (sk8ermike6789 @ Jan 22 2004, 09:41 PM) |
| a laser would cut the bottle also |
That would depend entirely on the intensity of the laser, it's wavelength and the color of the glass. In this case it would be prity much identical since your focusing sunlight with a magnifing glass. With a laser, you just using a artifical source of light being focused normaly with lenes much like a magnifing glass.
Rember a low power laser is an integral part of the XBox and any cd/dvd disk reading device.
Post by: katsumoto on January 23, 2004, 08:04:00 AM
| QUOTE |
| i got one..... Picture an empty wine bottle with a cork secured at the top in the usual way. Inside the bottle a metal ring hangs suspended by a string. How is it possible to make the metal ring drop to the bottom of the bottle without touching the ring, the thread, the cork, or the bottle while leaving the cork in place and intact? |
It took me awhile to solve this. Dang...
Anyway, try solving this:
EPISODE IV
Five hundred begins it, five hundred ends it,
Five in the middle is seen;
First of all figures, the first of all letters,
Take up their stations between.
Join all together, and then you will bring
Before you the name of an eminent king.
Good Luck!
Post by: Mr Deez on January 23, 2004, 09:26:00 AM
Post by: katsumoto on January 23, 2004, 09:34:00 AM
Actually that's not the correct answer. However, you are on the right track.
Post by: Zero on January 23, 2004, 11:40:00 AM
Post by: katsumoto on January 23, 2004, 11:51:00 AM
| QUOTE |
| DuDeR MaN's answer also works for #3 |
I didn't realize it until you mentioned it. I guess #4 is hard to solve. No one has responded yet. I'll give it more time.
Post by: DuDeR MaN on January 23, 2004, 06:44:00 PM
| QUOTE (katsumoto @ Jan 23 2004, 04:51 PM) | ||
Zero...you are right.
I didn't realize it until you mentioned it. I guess #4 is hard to solve. No one has responded yet. I'll give it more time. |
I'm so smart
Post by: shavedrat on January 25, 2004, 11:31:00 AM
Post by: styxx_78 on January 26, 2004, 06:09:00 AM
Do you have anymore?
Post by: katsumoto on January 26, 2004, 06:18:00 AM
Okay, here's a hard one. This is a thinker. Let's see who can solve this.
EPISODE V
If you have six men and they each had six baskets. Each basket has six cats inside and each cat has six kittens. Assuming all are whole and healthy, how many legs are there?
Good Luck!
Post by: XxmutinyxX on January 26, 2004, 07:22:00 AM
Post by: moistness on January 26, 2004, 07:31:00 AM
Post by: katsumoto on January 26, 2004, 09:27:00 AM
Here's why:
There are six men who each have six baskets. This equals 36 baskets. Each basket has six cats which equals 216 cats. Each cat has six kittens which equals 1,296 kittens. Therefore the kittens represent 5,184 legs. The cats represent 864 legs. The men represent 12 legs. The sum total is 6,060 legs.
Alright, even harder. Good Luck!
EPISODE VI
During WWII, there was a bridge connecting Germany and Switzerland, and on the German side, there was a sentry tower with a guard in it. He would come out every three minutes to check on the bridge, and he had orders to turn back anyone who tried to get into Germany, and shoot anyone trying to escape without a pass. There was a woman who desperately needed to get into Switzerland, and she knew she didn't have time to get a pass. It would take her at least six minutes to cross the bridge, but she managed to do it. How?
Post by: moistness on January 26, 2004, 01:38:00 PM
Wait til I tell my father, who was fascinated by your riddles and was absolutely convinced that the answer he begged me to post in an excited schoolboy sort of way was correct! He said "If that answer is not correct then i shall want to know why!" He is a retired research chemist and has BsC, MsC and PhD After his name, Maths was VERY important in his work!
Post by: XxmutinyxX on January 26, 2004, 02:20:00 PM
Post by: moistness on January 26, 2004, 02:39:00 PM
Post by: JLR on January 27, 2004, 12:38:00 AM
| QUOTE (sk8ermike6789 @ Jan 23 2004, 01:32 AM) |
| i got one..... Picture an empty wine bottle with a cork secured at the top in the usual way. Inside the bottle a metal ring hangs suspended by a string. How is it possible to make the metal ring drop to the bottom of the bottle without touching the ring, the thread, the cork, or the bottle while leaving the cork in place and intact? |
get someone else to turn the bottle upside down
Post by: JLR on January 27, 2004, 12:47:00 AM
Post by: XxmutinyxX on January 27, 2004, 01:03:00 AM
lol, j/ki heard this before but i forget the answer, maybe someone else remembers
Post by: katsumoto on January 27, 2004, 06:31:00 AM
Hey XxmutinyxX, I think you're right on...
| QUOTE |
| rubbing a girls clit with a towel lol, j/k i heard this before but i forget the answer, maybe someone else remembers |
The answer is "TOWEL", right? The more the towel dries, it gets wetter.
Okay solve this one. I don't think this is hard to solve, but this riddle will make you think. Good Luck!
EPISODE VII
1. There are 5 houses in five different colors.
2. In each house lives a person with a different nationality.
3. These five owners drink a certain drink, smoke a certain band of cigar and keep a certain pet.
4. no owners have the same pet, smoke the same brand of cigar nor drink the same drink.
THE QUESTION IS - WHO OWNS THE FISH?
The Britt lives in the red house.
The Swede keeps dogs as pets.
The Dane drinks tea.
The green house is on the left of the white house.
The green house owner drinks coffee.
The person who smokes Paul Mall rears birds.
The owner of the yellow house smokes Dunhill.
The man living in the house right in centre drinks milk.
The Norwegian lives in the first house.
The man who smokes blends lives next to the one who keeps cats.
The man who keeps horses lives next to the man who smokes Dunhill.
The owner who smokes Bluemaster drinks beer.
The German smokes Prince.
The Norwegian lives next to the blue house.
The man who smokes Blend has a neighbor who drinks water.
Post by: moistness on January 27, 2004, 06:44:00 AM
Post by: XxmutinyxX on January 27, 2004, 08:29:00 AM
| QUOTE (katsumoto @ Jan 27 2004, 04:31 PM) | ||
Hey XxmutinyxX, I think you're right on...
|
i made a joke, and actually got it right? wow
Post by: katsumoto on January 27, 2004, 09:08:00 AM
| QUOTE |
| that was posted by someone else a week or so ago, was german in his one!!! |
Damn...was it? Well, yeah... the answer is the German and here's why (in respective order):
1st house 2nd house 3rd house 4th house 5th house
Norwegian Dane Brit German Swede
Yellow Blue Red Green White
Dunhill Blend Pallmall Prince Bluemaster
Water Tea Milk Coffee Beer
Cats Horses Birds Fish Dogs
Here's another riddle. Hopefully this wasn't posted. Good Luck!!!
EPISODE VIII
Andy dislikes the catcher. Ed's sister is engaged to the second baseman. The center fielder is taller than the right fielder. Harry and the third baseman live in the same building. Paul and Allen each won $20 from the pitcher at pinochle. Ed and the outfielders play poker during their free time. The pitcher's wife is the third baseman's sister. The pitcher, catcher, and infielders except Allen, Harry, and Andy, are shorter than Sam. Paul, Andy, and the shortstop lost $50 each at the racetrack. Paul, Harry, Bill, and the catcher took a trouncing from the second baseman at pool. Sam is involved in a divorce suit. The catcher and the third baseman each have two children. Ed, Paul, Jerry, the right fielder, and the center fielder are bachelors. The others are married. The shortstop, the third baseman, and Bill each cleaned up $100 betting on the fight. One of the outfielders is either Mike or Andy. Jerry is taller than Bill. Mike is shorter than Bill. Each of them is heavier than the third baseman.
Using these facts, determine the names of the men playing the various positions on the baseball team.
Post by: katsumoto on January 28, 2004, 05:43:00 AM
Post by: katsumoto on January 28, 2004, 12:49:00 PM
Post by: DuDeR MaN on January 29, 2004, 12:50:00 PM
Post by: DuDeR MaN on January 29, 2004, 12:58:00 PM
Post by: XxmutinyxX on January 29, 2004, 03:36:00 PM
Post by: DuDeR MaN on January 29, 2004, 04:07:00 PM
I'll post some more riddles later in the other thread.
Post by: katsumoto on January 30, 2004, 07:32:00 AM
Or should I wait a bit so others can try to solve this riddle?
Post by: DuDeR MaN on January 30, 2004, 07:30:00 PM
Post by: Morglum on February 01, 2004, 02:15:00 AM
Post by: DuDeR MaN on February 01, 2004, 08:47:00 AM
| QUOTE (veg @ Feb 1 2004, 09:36 AM) |
| 7.The pitcher, catcher, and infielders except Allen, Harry, and Andy, are shorter than Sam. This is a biggie...First this makes Sam an outfielder so strike him from all infield, pitcher and catcher positions. Then strike out Allen, Harry, and Andy from all the outfield positions. |
I assumed that Allen Harry and Andy HAD to be infielders. (and all the infielders except..) I mean it doesn't make sense if say harry is the pitcher and then say, "the pitcher except for harry".
| QUOTE |
| 10.Sam is involved in a divorce suit. The catcher and the third baseman each have two children. Ed, Paul, Jerry, the right fielder, and the center fielder are bachelors. The others are married. Makes Sam Married. Catcher and third base also married. Also makes the only position Sam can hold as Left field because right and center are Bachelors. Eliminate Sam from all positions other then left field. Eliminate Ed, Paul, Jerry from center and right field. Eliminate Ed, Paul, Jerry from any known Married position - Third, catcher, pitcher. |
This is another thing I saw. I don't like to assume things in riddles (i assumed the above because of how it ws written) but I did not think that the were all married with children. So after the first time I tried it witht hem married and still messed up.
I like your method of drawing all the positions, I just did an old school graph. Thanks for the explaination.
Post by: veg on February 01, 2004, 08:55:00 AM
The way the sentences are structured will mess you up. And you have to make assumptions. Sometimes you assume wrong and it dorks up your answer.
I always thought the pitcher and the catcher were concidered infielders but in this case they separate them from the other infielders.
I tried putting everthing into a table but I kept getting lost so I came up with the chart idea.
Post by: cleary on February 02, 2004, 01:37:00 PM
bill - center fielder
mike - right fielder
sam - left fielder
ed - shortstop
andy - third baseman
jerry - second baseman
paul - first baseman
allen - catcher
harry - pitcher
Post by: DuDeR MaN on February 02, 2004, 06:12:00 PM
| QUOTE (p027867 @ Feb 2 2004, 04:43 PM) |
| What is more powerful than god, more evil than the devil, the rich want it, the poor have it and if you eat it, you will die? |
please read past posts man. this has already been said.
Post by: cleary on February 03, 2004, 05:28:00 AM
Post by: katsumoto on February 03, 2004, 06:55:00 AM
Yes, this riddle was hard. So, are you ready for another one? Good Luck!!!
EPISODE IX
A wealthy man named Richard Ellis had been counting his money. When he finished, he accidentally left a $100.00 bill on his desk. But when he returned for it a short while later, it was gone. Only two other persons could have seen the bill. One was the maid; the other was the butler. The maid told him that she had hidden it for safekeeping under a green book that was on the desk. But when they looked the bill was not there. The butler said he had found the bill where the maid had left it. He had placed it inside the book, where he thought there was less chance that somebody would find it. He had written down the page numbers so that he would not forget them. The bill was between pages 35 and 36, he said. But when they looked, there was no money in the book. After Mr. Ellis had talked to the maid and the butler, he called the police. He was sure he knew who had taken the money.
Who was it, and how did he know?
Post by: Krill123 on February 03, 2004, 09:24:00 AM
| QUOTE (p027867 @ Feb 2 2004, 09:43 PM) |
| What is more powerful than god, more evil than the devil, the rich want it, the poor have it and if you eat it, you will die? |
Nothing
Post by: Krill123 on February 03, 2004, 09:26:00 AM
back of page 35 so the butler couldn't have put the
bill between the two pages.
Post by: katsumoto on February 03, 2004, 01:32:00 PM
EPISODE X
There are 16 books of the Bible in the following paragraph...can you find them?
I once made a remark about the hidden books of the Bible. It was a lulu; kept people looking so hard for facts and for others it was a revelation. Some were in a jam especially since the names of the books were not capitalized, but the truth finally struck home to
numbers of readers. To others, it was a real job. We want it to be a most fascinating few moments for you. Yes there will be some really easy ones to spot. Others may require judges to help them. I will quickly admit it usually takes a minister to find one of them, and there will be loud lamentations when it is found. A little lady says she brews a cup of tea so she can concentrate better. See how well you can compete. Relax now for there really are sixteen names of the books in the Bible in this story
Post by: katsumoto on February 03, 2004, 01:38:00 PM
Have fun!!!
Post by: cleary on February 03, 2004, 02:06:00 PM
note: this took too long to do
Post by: katsumoto on February 04, 2004, 06:44:00 AM
For those who did not see the 16 books of the Bible, here they are:
I once made a remark about the hidden books of the Bible. It was a lulu; kept people looking so hard for facts and for others it was a revelation. Some were in a jam especially since the names of the books were not capitalized, but the truth finally struck home to numbers of readers. To others, it was a real job. We want it to be a most fascinating few moments for you. Yes there will be some really easy ones to spot. Others may require judges to help them. I will quickly admit it usually takes a minister to find one of them, and there will be loud lamentations when it is found. A little lady says she brews a cup of tea so she can concentrate better. See how well you can compete. Relax now for there really are sixteen names of the books in the Bible in this story.
-----------------------------
Okay, so far I've posted 10 riddles. The riddles started with a few sentences to becoming a whole paragraph. Some where easy and some where...hard? Now, the riddles will become harder to solve. I want to stomp someone. Hopefully I could do that. I had people thinking in Episode VIII. Are you guys ready? Here we go...
EPISODE XI
A legendary king possessed a huge amount of gold. He hid this treasure carefully: in a building consisting of a number of rooms. In each room there were a number of boxes; this number was equal to the number of rooms in the building. Each box contained a number of golden coins that equaled the number of boxes per room. When the king died, one box was given to the royal barber. The remainder of the coins had to be divided fairly between his six sons.
Is a fair division possible in all situations? Good Luck!!!
Post by: cleary on February 04, 2004, 01:02:00 PM
Post by: katsumoto on February 04, 2004, 01:50:00 PM
Post by: cleary on February 04, 2004, 02:35:00 PM
)- the royal barber was givin one box with "s" coins
- the 6 brothers, S³ - "s" coins remain
with all this done you get this problem: S(S - 1)(S + 1).
- the last part is divisible by 6, if a number is divisible by 6, it is also divided by 3 and even. ( which this problem is )
- so we got "s" whatever it is. the problem is: S(S - 1)(S + 1), and this always has 3 successive numbers. in this problem one of them is always divisible by 3, and another one is always even.
- so this is S=1 ( which doesn't give the boys jack shit
)thats how i did it. i was gonna scan the sheet of paper i did it on, but my scanner is broken.
Post by: Yoko They Cryed on February 04, 2004, 02:42:00 PM
Post by: hipstergk27 on February 05, 2004, 04:35:00 AM
| QUOTE |
| ok, so we got a number of rooms. we'll call this "s" - each room has "s" boxes - each box has "s" coins so this also equals: SxSxS=S³ ( "x" represents times ) - the royal barber was givin one box with "s" coins - the 6 brothers, S³ - "s" coins remain with all this done you get this problem: S(S - 1)(S + 1). - the last part is divisible by 6, if a number is divisible by 6, it is also divided by 3 and even. ( which this problem is ) - so we got "s" whatever it is. the problem is: S(S - 1)(S + 1), and this always has 3 successive numbers. in this problem one of them is always divisible by 3, and another one is always even. - so this is S=1 ( which doesn't give the boys jack shit ) |
hmm...this answer seems remarkably similar to this one Solution to: King Midas
and if you did work it out all on your own, you still got the answer partially wrong. yes, fair division is always possible, but the only case where this...
| QUOTE |
| each son will get nothing |
is true, is if S=1. any other number will result in the sons receiving some amount of gold equally.
tsk tsk tsk
Post by: katsumoto on February 05, 2004, 06:39:00 AM
| CODE |
| ,. o .,<>., o |\/\/\/\/| '========' (_ SSSSSSs )a'`SSSSSs /_ SSSSSS .=## SSSSS .#### SSSSs ###::::SSSSS .;:::""""SSS .:;:' . . \\ .::/ ' .'| .::( . | :::) \ /\( / /) ( | .' \ . ./ / _-' |\ . | _..--.. . /"---\ | ` | . | -=====================,' _ \=(*#(7.#####() | `/_.. , ( _.-''``';'-''-) ,. \ ' '+/// | .'/ \ ``-.) \ ,' _.- (( `-' `._\ `` \_/_.' ) /`-._ ) | ,'\ ,' _.'.`:-. \.-' / <_L )" | _/ `._,' ,')`; `-'`' | L / / / `. ,' ,|_/ / \ ( <_-' \ \ / `./ ' / /,' \ /|` `. | )\ /`._ ,'`._.-\ |) \' / `.' )-'.-,' )__) |\ `| : /`. `.._(--.`':`':/ \ ) \ \ |::::\ ,'/::;-)) / ( )`. | ||::::: . .::': :`-( |/ . | ||::::| . :| |==[]=: . - \ |||:::| : || : | | /\ ` | ___ ___ '|;:::| | |' \=[]=| / \ \ | /_ ||``|||::::: |; | | | \_.'\_ `-. : \_``[]--[]|::::'\_;' )-'..`._ .-'\``:: ` . \ \___.>`''-.||:.__,' SSt |_______`> <_____:::. . . \ _/ `+a:fi......jrei''' |
GETTER DONE!!! Well, here's another one! This one may be very easy. Good Luck!!!
EPISODE XII
There is a scientific method to determining what these words have in common. Can you figure it out?
GAGE, HOER, NO, LACE, OF, SIPS, ARK
Post by: G-Prime975 on February 05, 2004, 12:45:00 PM
[EDIT] Hehe, I misspelled a few elements.
[/EDIT]
Post by: katsumoto on February 05, 2004, 02:03:00 PM
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Did you enjoy the image? Anyway, here's another one. Good Luck!
EPISODE XIII
Triangle numbers are formed by adding successive integers, i.e.
1 = 1+0,
3 = 1+2,
6 = 1+2+3,
10 = 1+2+3+4,
etc.
So the first few triangle numbers would be:
1,3,6,10,15,21,28, ...
Suppose somebody tells you a random integer. How can you quickly find out whether or not it is a triangle number?
Post by: DuDeR MaN on February 05, 2004, 05:28:00 PM
Post by: SKoT on February 06, 2004, 06:53:00 AM
Ok,
n! Which is the same as n C k
k!(n-k)!
Where n is the row number and k is the position in that row, counting from top to bottom and from left to right. Also, we start counting from zero, so the 6 above would be in the row 4 and column 2.
The rows of Pascal's triangle also represent coefficients of a binomial raised to a power. Consider:
(a+b )4
If we expand this, we get:
a4+4a3b+6a2b2+4ab3+b4
The coefficients are the 4th row of Pascal's triangle:
1 4 6 4 1
Post by: SKoT on February 06, 2004, 06:56:00 AM
Post by: katsumoto on February 10, 2004, 12:38:00 PM
Multiply the number by 8 then add 1. If the result is a perfect square, then the original number is a triangle number.
For example: 10*8+1 = 81. 81 = 9^2. So 10 is a triangle number.
11*8+1 = 89. 89 is not a perfect square, so 11 is not a triangle number.
Okay, I don't think that was really a riddle. But here's one:
EPISODE XIV
The king dies and two men, the true heir and an impostor, both claim to be his long-lost son. Both fit the description of the rightful heir: about the right age, height, coloring and general appearance.
Finally, one of the elders proposes a test to identify the true heir. One man agrees to the test while the other flatly re-fuses. The one who agreed is immediately sent on his way, and the one who re-fused is correctly identified as the rightful heir.
Can you figure out why? Good Luck!!!
Post by: DuDeR MaN on February 10, 2004, 01:44:00 PM
Post by: cleary on February 10, 2004, 04:44:00 PM
| QUOTE (NooberTehGod @ Feb 11 2004, 01:43 AM) |
| he asks for a blood test or something like that. The king's son was a hemopheliac, or whatever, and therefor would never submit to being cut for fear of dying of blood loss |
yup
Post by: Ripasser on February 10, 2004, 09:26:00 PM
Post by: katsumoto on February 11, 2004, 07:14:00 AM
Post by: DuDeR MaN on February 11, 2004, 10:22:00 AM
| QUOTE (katsumoto @ Feb 11 2004, 12:14 PM) |
| what position in line gives him the greatest chance of being the first duplicate birthday? |
in front of the line, he won't get the free ticket but he has a better chance to be the first duplicate birthday.
and for that last one.... I thought about a blood test but I figured the one guy wouldn't do it since it would show he isn't the king's son. Whatever.. keep them coming.
Post by: Ripasser on February 11, 2004, 03:19:00 PM
Post by: cleary on February 11, 2004, 03:21:00 PM
Post by: DuDeR MaN on February 12, 2004, 01:57:00 AM
Post by: katsumoto on February 12, 2004, 07:02:00 AM
Here's why:
Suppose he was the Kth person in line. Then he wins if and only if the K-1 people ahead all have distinct birtdays AND his birthday matches one of theirs. Let
A = Fred's birthday matches one of the K-1 people ahead
B = those K-1 people all have different birthdays
Then
Prob(Fred wins) = Prob(B ) * Prob(A | B )
(Prob(A | B ) is the conditional probability of A given that B occurred.)
Now let P(K) be the probability that the K-th person in line wins, Q(K) the probability that the first K people all have distinct birthdays (which occurs exactly when none of them wins). Then
P(1) + P(2) + ... + P(K-1) + P(K) = 1 - Q(K)
P(1) + P(2) + ... + P(K-1) = 1 - Q(K-1)
P(K) = Q(K-1) - Q(K) <--- this is what we want to maximize.
Now if the first K-1 all have distinct birthdays, then assuming uniform distribution of birthdays among D days of the year, the K-th person has K-1 chances out of D to match, and D-K+1 chances not to match (which would produce K distinct birthdays). So
Q(K) = Q(K-1)*(D-K+1)/D = Q(K-1) - Q(K-1)*(K-1)/D
Q(K-1) - Q(K) = Q(K-1)*(K-1)/D = Q(K)*(K-1)/(D-K+1)
Now we want to maximize P(K), which means we need the greatest K such that P(K) - P(K-1) > 0. (Actually, as just given, this only guarantees a local maximum, but in fact if we investigate a bit farther we'll find that P(K) has only one maximum.) For convenience in calculation let's set K = I + 1. Then
Q(I-1) - Q(I) = Q(I)*(I-1)/(D-I+1)
Q(I) - Q(I+1) = Q(I)*I/D
P(K) - P(K-1) = P(I+1) - P(I)
= (Q(I) - Q(I+1)) - (Q(K-2) - Q(K-1))
= Q(I)*(I/D - (I-1)/(D-I+1))
To find out where this is last positive (and next goes negative), solve
x/D - (x-1)/(D-x+1) = 0
Multiply by D*(D+1-x) both sides:
(D+1-x)*x - D*(x-1) = 0
Dx + x - x^2 - Dx + D = 0
x^2 - x - D = 0
x = (1 +/- sqrt(1 - 4*(-D)))/2 ... take the positive square root
= 0.5 + sqrt(D + 0.25)
Setting D=365 (finally deciding how many days in a year!),
desired I = x = 0.5 + sqrt(365.25) = 19.612 (approx).
The last integer I for which the new probability is greater then the old is therefore I=19, and so K = I+1 = 20.
...viola!
EPISODE XVI
Land of the giants, it is often called,
Where warriors and wizards play with wolves;
Its suns and stars get hot and cold,
Whether one is at home or on the road.
Fourteen tribes live in the west, we are told,
One less than those in the eastern world;
The weakest thirteen will give up and fold.
While the strongest continue their quest for gold.
The kings are powerful and the magic is bold,
Others can fly and are a site to behold;
But when the season is past and the story is told,
New stars are born and old tribes evolve.
What is the Land of the Giants?
Have fun witht his riddle. Good Luck!!!
Post by: DuDeR MaN on February 12, 2004, 11:28:00 AM
| QUOTE (katsumoto @ Feb 12 2004, 12:02 PM) |
| Well...clblanch1622 solved the problem. Fred should try to be the 20th person in line. Congrats. Here's why: Suppose he was the Kth person in line. Then he wins if and only if the K-1 people ahead all have distinct birtdays AND his birthday matches one of theirs. Let A = Fred's birthday matches one of the K-1 people ahead B = those K-1 people all have different birthdays Then Prob(Fred wins) = Prob(B ) * Prob(A | B ) (Prob(A | B ) is the conditional probability of A given that B occurred.) Now let P(K) be the probability that the K-th person in line wins, Q(K) the probability that the first K people all have distinct birthdays (which occurs exactly when none of them wins). Then P(1) + P(2) + ... + P(K-1) + P(K) = 1 - Q(K) P(1) + P(2) + ... + P(K-1) = 1 - Q(K-1) P(K) = Q(K-1) - Q(K) <--- this is what we want to maximize. Now if the first K-1 all have distinct birthdays, then assuming uniform distribution of birthdays among D days of the year, the K-th person has K-1 chances out of D to match, and D-K+1 chances not to match (which would produce K distinct birthdays). So Q(K) = Q(K-1)*(D-K+1)/D = Q(K-1) - Q(K-1)*(K-1)/D Q(K-1) - Q(K) = Q(K-1)*(K-1)/D = Q(K)*(K-1)/(D-K+1) Now we want to maximize P(K), which means we need the greatest K such that P(K) - P(K-1) > 0. (Actually, as just given, this only guarantees a local maximum, but in fact if we investigate a bit farther we'll find that P(K) has only one maximum.) For convenience in calculation let's set K = I + 1. Then Q(I-1) - Q(I) = Q(I)*(I-1)/(D-I+1) Q(I) - Q(I+1) = Q(I)*I/D P(K) - P(K-1) = P(I+1) - P(I) = (Q(I) - Q(I+1)) - (Q(K-2) - Q(K-1)) = Q(I)*(I/D - (I-1)/(D-I+1)) To find out where this is last positive (and next goes negative), solve x/D - (x-1)/(D-x+1) = 0 Multiply by D*(D+1-x) both sides: (D+1-x)*x - D*(x-1) = 0 Dx + x - x^2 - Dx + D = 0 x^2 - x - D = 0 x = (1 +/- sqrt(1 - 4*(-D)))/2 ... take the positive square root = 0.5 + sqrt(D + 0.25) Setting D=365 (finally deciding how many days in a year!), desired I = x = 0.5 + sqrt(365.25) = 19.612 (approx). The last integer I for which the new probability is greater then the old is therefore I=19, and so K = I+1 = 20. ...viola! |
my head hurts now. your "riddles" have gotten way too hard for me to comprehend... i give up
Post by: cleary on February 12, 2004, 01:47:00 PM
the nba ( national basketball association )
Post by: katsumoto on February 13, 2004, 06:40:00 AM
Okay, here's another one. This may be easy. Let's have fun! Duder Man, don't give up yet. The fun just begun... EPISODE XVII
Some know me as two of twenty-six
Distanced by seven,
Plus the first between odds,
Its left neighbor twice for eleven.
I can travel 3 states in a fraction of an hour,
Some 90 of your air bags, 70 of your thinking power.
What am I? Good Luck!
Post by: cleary on February 13, 2004, 05:54:00 PM
Post by: rastareaper on February 14, 2004, 09:52:00 PM
"I can travel 3 states in a fraction of an hour," water can change to gas, liquid or solid in a fraction of an hour.
Post by: rastareaper on February 14, 2004, 10:06:00 PM
Post by: cleary on February 14, 2004, 10:09:00 PM
Post by: rastareaper on February 15, 2004, 06:42:00 AM
Post by: MeNaCe911 on February 15, 2004, 10:22:00 AM
Post by: katsumoto on February 19, 2004, 11:56:00 AM
Post by: lordvader129 on February 19, 2004, 12:15:00 PM
| QUOTE (katsumoto @ Feb 19 2004, 03:56 PM) |
| How old is the PRIEST?? |
36
just so you know, i didnt cheat or know the answer prior, i figured it mathematically
once you figure out what Fifty and Ten Dozens Twenty is your good
Post by: katsumoto on February 24, 2004, 02:07:00 PM
HINT: Read the riddle very carefully!
Post by: MeNaCe911 on February 24, 2004, 04:38:00 PM
Post by: katsumoto on February 25, 2004, 06:39:00 AM
Post by: chilin_dude on February 25, 2004, 09:51:00 AM
or possibly something to do with their is only one stranger that is 3 years old? But couldn't figure from here
Post by: geniusalz on February 25, 2004, 10:54:00 AM
Would that be you, or me?
Post by: lordvader129 on February 25, 2004, 10:59:00 AM
Post by: darth_jarret on February 27, 2004, 06:01:00 PM
Post by: darth_jarret on February 27, 2004, 06:04:00 PM
You have been zapped into this room by means unknown. There are two doors on one side of the room, there are two computers. It even seems they are each guarding a door.
There are two signs, and a leper in the room with you.
The first sign reads:
"Behind one door is God realization. Behind the other door is a room painted entirely green. The only things in that room are a man and a ladder, both of which are painted entirely green. The man has been hired by the coalition of Gangsta Rappers to hurl a non-stop barrage of insults at anyone who enters the room. " (end sign 1)
The second sign sits between the two computers (also the two doors) and reads as follows:
"One of these computers is programmed to lie to you regardless of how friendly you are to the user interface, the other computer is programmed to tell the truth, regardless of your emotional state."
It is about this time that the gangrenous leper begins to speak "I get you deal on phone time $.08 minutes weekdays all odd hours. $.10 on Thursday except the hour between 2:00 and 3:00, or any registered birthday . . . "
He begins to - quite slowly - lurch towards you, he's already reaching out towards you, you get the sense he's a "touchy-feely" sorta person. You quickly calculate you have time to ask one computer one question before he reaches you. The riddle before you is this: What question do you ask, and to which computer do you pose your question? Remember, you don't know whether the computer's answer will be a lie, and you've got to figure out which door to go out.
Hints:
1) You've heard great things about God realization, 2) It doesn't matter which computer you ask, i.e. there's an answer for both.
Post by: Moleman on February 27, 2004, 06:16:00 PM
Time for my math riddle.
What is the Xbox RSA key used to sign XBE files?
Go!
Post by: lordvader129 on February 27, 2004, 07:35:00 PM
| QUOTE (Moleman @ Feb 27 2004, 10:16 PM) |
| What is the Xbox RSA key used to sign XBE files? |
42
Post by: Moleman on February 28, 2004, 08:16:00 AM
| QUOTE (lordvader129 @ Feb 28 2004, 12:35 AM) | ||
42 |
That's just the answer to life, the universe, and everything. But what's the question?
Post by: lordvader129 on February 28, 2004, 08:49:00 AM
(wouldnt it be sick if that was true? lol)
Post by: katsumoto on March 01, 2004, 08:35:00 AM
Post by: katsumoto on March 01, 2004, 08:39:00 AM
EPISODE XIX
To the bold, I am a weapon;
To the gallant, I am a pin;
To the gracious, I am a service;
To the bridesmaid, I am a win.
To the traitor, I am a bribe;
To the star, I am a screen;
To the swift, I am an arrow;
To the bride, I am a dream.
What am I?
Post by: Parrappa on March 01, 2004, 10:08:00 PM
| QUOTE (Emc1683 @ Mar 1 2004, 08:38 PM) |
| Silver. |
yeap, your right, took me a while to figure it out, but it has to be silver
To the bold, I am a weapon=Silver Bullet
To the gallant, I am a pin=Silver Star
To the gracious, I am a service=Silver Service???
To the bridesmaid, I am a win=Silver Medal???
To the traitor, I am a bribe=read the bible
To the star, I am a screen=Silver Screen
To the swift, I am an arrow=???????
To the bride, I am a dream=Silver Wedding
Post by: Raver758 on March 01, 2004, 10:35:00 PM
I'm not racist, I'm black , I have no name but what I was given. As long as I'm turned on no cold no rain no snow will stop me. I'm here to please everyone from America to India?
Who ?????? Answer below
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
Im the XBOX
Post by: lordvader129 on March 02, 2004, 09:52:00 AM
Post by: chilin_dude on March 03, 2004, 12:38:00 PM
Post by: katsumoto on March 04, 2004, 06:22:00 AM
Okay, that may have been an easy one, but what about this:EPISODE XX
My first is double twenty-one but twice of twenty-two,
Decipher it and you will see that statement is quite true.
My next two are just three apart, or, looking at the link,
You'll have the answer easily and will not have to think.
My fourth is so more ways than one, so long as you can spell,
While my fifth splits first and second, so what is it? Can you tell?
It may seem like these words of rhyme are nonsense things to say,
Though the five together are right here (two ways) so what are they?
GOOD LUCK!!!
Post by: beljim on March 04, 2004, 11:32:00 AM
Next!!!!
Post by: katsumoto on March 05, 2004, 06:21:00 AM
WORDS is the correct anwser, here's why:The answer that you seek to find, in words, is now shown here,
To understand the clues read on, I'll try to make it clear.
The twenty-first of letters is U in the alphabet,
Double it, W is the letter you would then get.
Now the twenty-second letter of the alphabet is V,
When written twice together a W it could be.
I hope I've explained the clues to give W as letter one,
Now let's move on, there's four more letters to be done.
The second and third letters are three from each other,
But to scour the alphabet to find them, you needn't bother.
If you were looking carefully you'd have seen in the third line,
The linking word, OR, fits this criteria just fine.
This makes the second O, and R would be the third,
Which gives us W, O and R as letters of this word.
The fourth letter in the alphabet is fourth in the answer too,
Which makes the fourth one D, only one more left to do!
Finally the fifth lies between the first and second,
In the alphabet, S splits W and O, that's what I reckoned.
Combine the five, see that they've been used here a lot?
And in the seventh line, sixth word, the answer even got a spot!
So now you know the answer and those clues are less absurd,
It's amazing what you can do with a word, oops, I mean WORDS!
----------------------------------------------------------------------------
NEXT --- This is an easy one, but sometimes the easiest one is hard to solve! Good Luck!
EPISODE XXI
As far east as east...
As far west as west...
If north is the future
What name fits me best?
Post by: Parrappa on March 05, 2004, 10:52:00 PM
Post by: katsumoto on March 09, 2004, 06:23:00 AM
Alaska has the western-most point in the U.S. AND the eastern-most point in the United States. This is possible because Alaska straddles the international date line.
Alaska's State motto is: North to the future.
Here's another riddle. Can anyone solve this with the quickness? Good Luck
EPISODE XXII
An iron horse with a flaxen tail.
The faster the horse runs,
the shorter his tail becomes.
What is it?
Post by: katsumoto on March 09, 2004, 11:17:00 AM
Post by: katsumoto on March 11, 2004, 07:13:00 AM
Post by: Parrappa on March 11, 2004, 01:11:00 PM
Post by: lordvader129 on March 11, 2004, 03:32:00 PM
"they keep me locked up" = ribs/breastbone
"my neighbors import and export" = lungs
"I got a lotta my own connections" = veins and arteries
"The guy upstairs is the boss of it all" = brain
"Hes also incarcerated" = skull
"so long as we get the messages." = nerves/nervous system
thats all i got right now
edit:
"with thousands collectin'" = aveoli (sp?)
"two dozen guards" = ribs
Post by: katsumoto on March 22, 2004, 08:51:00 AM
| QUOTE |
| EPISODE XXIII Yeah, they keep me locked up, but I guess Im thankful, in short. I hand out the beatings, while my neighbors import and export. We make a good team, especially me as the muscle, But with two dozen guards, Im glad we never tussle. But really, I got a lotta my own connections, Imports, exports, with thousands collectin'. But between us, theres really no competition, I call it harmonizin cause were on the same mission. No one is unnecessary, or, you know, too small, The guy upstairs is the boss of it all. Hes also incarcerated, but for his good I bet it is, Business is good, so long as we get the messages. Whos doing the talking, and who are the other players in this riddle? There are a total of 28, not including the narrator. |
Can anyone solve this riddle? I am still waiting to see who can crack this. So far Parrappa and lordvader129 are on the right track, but remember, there's a total of 28. Good luck.
Post by: LepPpeR on March 22, 2004, 09:02:00 AM
I think I got it all
Post by: katsumoto on March 22, 2004, 09:26:00 AM
EPISODE XXIV
You're sitting at a bar, an explosively-dank hole in Funky Town USA.
You have an award that says, most likely to be the character in a riddle . . . looking around your attention finds a small chair with a mobile entity covering it with it's lesser half . . . It speaks, you hear "Glass of water please . . . " You see the bartender examine the seated man for a moment. He then reaches under the counter and pulls out a shotgun point-blank in the man's face. The seated man speaks again "Thank You" he says, and walks out of the bar . . . why did the man get what he needed?
Good luck!
Post by: LepPpeR on March 22, 2004, 01:49:00 PM
So my guess is that he had the hiccups and the shot gun got rid of them but the water might have helped as well.
Post by: katsumoto on April 01, 2004, 06:40:00 AM
Post by: SKoT on April 01, 2004, 07:13:00 AM
Sunday: 687/4800
Monday: 685/4800
Tuesday: 685/4800
Wednesday: 687/4800
Thursday: 684/4800
Friday: 688/4800
Lol, this math class is boring btw
Post by: LepPpeR on April 01, 2004, 11:57:00 AM
For a normal year, the distribution of the 13th days over the days of the week is as follows, if January 13th falls on weekday x:
x x+1 x+2 x+3 x+4 x+5 x+6
2 1 1 3 1 2 2
For a leap year, the distribution of the 13th days over the days of the week is as follows, if January 13th falls on weekday x:
x x+1 x+2 x+3 x+4 x+5 x+6
3 1 1 2 2 1 2
For subsequent years, these distributions shift cyclic 1 place to the right for normal years, and 2 places to the right for leap years.
For three normal years followed by one leap year, we can now calculate the distribution of the 13th days:
x x+1 x+2 x+3 x+4 x+5 x+6
2 1 1 3 1 2 2 (normal year, not shifted)
2 2 1 1 3 1 2 (normal year, shifted 1 place to the right)
2 2 2 1 1 3 1 (normal year, shifted 2 places to the right)
+ 2 1 2 3 1 1 2 (leap year, shifted 3 places to the right)
-----------------------------------
8 6 6 8 6 7 7 (3 normal years and 1 leap year, not shifted)
For four normal years, we find:
x x+1 x+2 x+3 x+4 x+5 x+6
2 1 1 3 1 2 2 (normal year, not shifted)
2 2 1 1 3 1 2 (normal year, shifted 1 place to the right)
2 2 2 1 1 3 1 (normal year, shifted 2 places to the right)
+ 1 2 2 2 1 1 3 (normal year, shifted 3 places to the right)
-----------------------------------
7 7 6 7 6 7 8 (4 normal years, not shifted)
In four years, the latter two distributions shift 5 places cyclic to the right. So for a period of 16 years with 4 leap years, we find:
x x+1 x+2 x+3 x+4 x+5 x+6
8 6 6 8 6 7 7 (3 normal years and 1 leap year, not shifted)
6 8 6 7 7 8 6 (3 normal years and 1 leap year, shifted 5 places to the right)
6 7 7 8 6 6 8 (3 normal years and 1 leap year, shifted 10 places to the right)
+ 7 8 6 6 8 6 7 (3 normal years and 1 leap year, shifted 15 places to the right)
-----------------------------------
27 29 25 29 27 27 28 (16 years with 4 leap years, not shifted)
For a period of 100 years with 24 leap years (the years 2001-2100, 2101-2200, and 2201-2299), we can now calculate the distribution of the 13th days:
x x+1 x+2 x+3 x+4 x+5 x+6
27 29 25 29 27 27 28 (16 years with 4 leap years, not shifted)
29 25 29 27 27 28 27 (16 years with 4 leap years, shifted 20 places to the right)
25 29 27 27 28 27 29 (16 years with 4 leap years, shifted 40 places to the right)
29 27 27 28 27 29 25 (16 years with 4 leap years, shifted 60 places to the right)
27 27 28 27 29 25 29 (16 years with 4 leap years, shifted 80 places to the right)
27 28 27 29 25 29 27 (16 years with 4 leap years, shifted 100 places to the right)
+ 8 7 7 6 7 6 7 (4 normal years, shifted 120 places to the right)
-----------------------------------
172 172 170 173 170 171 172 (100 years with 24 leap years, not shifted)
For a period of 100 years with 25 leap years (the years 2301-2400), we can now calculate the distribution of the 13th days:
x x+1 x+2 x+3 x+4 x+5 x+6
27 29 25 29 27 27 28 (16 years with 4 leap years, not shifted)
29 25 29 27 27 28 27 (16 years with 4 leap years, shifted 20 places to the right)
25 29 27 27 28 27 29 (16 years with 4 leap years, shifted 40 places to the right)
29 27 27 28 27 29 25 (16 years with 4 leap years, shifted 60 places to the right)
27 27 28 27 29 25 29 (16 years with 4 leap years, shifted 80 places to the right)
27 28 27 29 25 29 27 (16 years with 4 leap years, shifted 100 places to the right)
+ 7 8 6 6 8 6 7 (3 normal years and 1 leap year, shifted 120 places to the right)
-----------------------------------
171 173 169 173 171 171 172 (100 years with 25 leap years, not shifted)
Now we can calculate the distribution of the 13th days for the period 2001 up to 2400:
x x+1 x+2 x+3 x+4 x+5 x+6
172 172 170 173 170 171 172 (100 years with 24 leap years, not shifted)
170 173 170 171 172 172 172 (100 years with 24 leap years, shifted 124 places to the right)
170 171 172 172 172 170 173 (100 years with 24 leap years, shifted 248 places to the right)
+ 172 171 173 169 173 171 171 (100 years with 25 leap years, shifted 372 places to the right)
------------------------------------
684 687 685 685 687 684 688 (400 years, not shifted)
Since January 13th, 2001 (x in the distribution) is a Saturday, we get the following probability for the 13th days over the days of the week:
Saturday: 684/4800
Sunday: 687/4800
Monday: 685/4800
Tuesday: 685/4800
Wednesday: 687/4800
Thursday: 684/4800
Friday: 688/4800
Conclusion: the probability that the 13th of a certain month in a certain year is a Friday, is the highest.
Have a nice day...... I do not condone the cheating on homework.