The worlds first riddle!

Relative

Neat!

I got as far as 4: 13 and stopped looking.

Anyway, my lady-friend was driving and she is smart enough for both of us. She knew which party we were going to and wrote the address down.

THAT is how really smart people do it! :wink:

The answer continued:

Relative's next statement is more telling still; he already knew Lacomus could not figure it out. So whatever sum, let's call it R, Relative remembers, it can't be made by adding two prime numbers together- if it could, then there would have been the possibility Lacomus knew the answer.

When Relative announces his prescience, Lacomus then knows the answer. This means that out of all the possible addresses that multiply to make P, only one has the property of R described above.
Relative makes the same logical deductions we have, and solves the problem. Once you have figured out the logical steps needed to solve the problem, described above, it is still quite a bit of work to actually find the unique pair of numbers that satisfies these properties.

Incidentally, the party's address is 4-13.

Well done! I hope you had a great party.



This riddle is for ?'Genius' science minds only.

Millennia into the future, millions of miles away, the Chief Science Officer of an interstellar explorer ship makes an exciting discovery.
Deep in space a silent probe is drifting along emitting a steady EM pulse signalling its presence. After determining it carries no armaments, he has it brought aboard. Engineers determine that it has been drifting an indeterminate period of time, its systems powered by fusion, refuelling with the rarefied hydrogen in deep space gathered in by magnetic sails.
The contents of the probe are enigmatic:

A laser

A small metal cylinder composed of a platinum-iridium alloy

A container of water

A small black box with a steady blinking light. A chemical analysis of the box indicates it contains titanium, copper, various complex hydrocarbons, silicon, and a cesium isotope.

A large collection of thin metal sheets inscribed with symbols and drawings, apparently referring to the other items.

Presumably the last item is a symbolic language which explains the usage of the other items.

If you were aboard the explorer, what help could you offer the CSO on how these items might be used together

Try

The space craft.

I've p-m'd an answer that I feel is in the range of 'Yeah, well - maybe - could be' (perhaps)

Hi!

What about 3,8 - to me it seems a solution as well.

About spacecraft : i will look into that matter

Uh, 3,8 is not a solution

The solution to the 'light' puzzle:
The trick is to use other means to transfer information besides observing if the light is on. This is achieved by heat: Turn the 1st switch on for 10 minutes, then turn it off. Turn on 2nd switch and immediately go to check the lamp.
if the lamp is on, it is the 2nd switch.
if the lamp is off, and warm, it is 1st switch
if the lamp is off and cold, it is switch 3.

It was a good idea to PM the answers since they really spoil the riddle.
Good work Iacomus & Tryagain

?'Deep space' answer.

If you recognized the significance of the objects, you would realize that this probe was sent from Earth. The probe is a tutorial for extraterrestrials, like the interstellar explorer, teaching the understandings of our physics. The clues are the collection of items, each of which is used for determining one of our standard physical units.

The laser allows calculation of distance. Our base (metric) unit for distance is the meter. A meter is defined as the distance light travels in 1/29979245 of a second. Why not 1/30000000 of a second? Previously, a meter had been defined as 1,650,763.73 wavelengths of the red-orange line of the spectrum of krypton-86. Prior to that, it was defined by marks on a physical bar of platinum-iridium. The goal of the redefinition was to come up with a physical definition that could be precisely measured, remain constant, and was a close a possible to the previous definition. The historical basis of the meter was ~1/10,000,000 of the distance from the North Pole to the equator.

The metal cylinder represents the kilogram. Presently, the standard calibration for the kilogram is a cylinder of platinum-iridium kept at the International Bureau of Weights and Measures in Paris. A near-identical copy exists in the United States at the National Bureau of Standards. The kilogram is the only metric unit still defined by measure against a real object.

The water defines temperature. The Kelvin degree is 1/273.16 of the temperature at which water transitions between solid liquid and gas, also referred to as the triple point.

The black box is a timer, blinking off seconds. The second is defined as "the duration of 9,192,631,770 cycles of the radiation associated with a specified transition of the cesium-133 atom" (from the National Bureau of Standards).


The collection of metal plates would describe how these items are used to calculate our constants and our system of mathematics. Together, these would give an alien race the knowledge necessary to understand our system of sciences, and potentially communicate scientific knowledge with us.


Win a big prize. (some hope).
It is the year 2525, and man is still alive. Advancements in science have enabled humans to engage in dumber and dumber activities and survive. One example is the popular cannonball dive. Participants don a suit of steel electro-shielding armour, jump off a diving board, curl into a ball, and plunge to the bottom of a pool.

Anyway, as part of the end of year celebrations seven participants will be cannonballing in various situations.

A prize of $1,000,000 will be awarded to anyone who can successfully guess the order in which the cannonballers will reach the bottom of the pool. You would like that $1,000,000.

First, the ground rules: Unless otherwise noted, all participants may be assumed to be 100kg perfect spheres with a diameter of half a meter (mankind is kind of out of shape). All dives start from one kilometre above the surface of the pool. All pools are 10 meters deep. If a diver is wearing a parachute, the chute detaches automatically on contact with the water.

Cannonballer #1 is doing a "Classic Earth" jump. No special conditions and the air and water temperature is 50° F.

Cannonballer #2 will be doing the same, but he is wearing a parachute that deploys immediately.

Cannonballer #3 is doing a "Classic Moon" jump, starting 1km above the moon into a pool of 50° F water (moon pools have a thin force-field over them that doesn't affect the cannonballer, keeping water pressure equal to Earth sea level).

Cannonballer #4 is doing a "Classic Earth" in a more southern climate, with an air and water temperature of 75° F.

Cannonballer #5 is doing a moon jump with a parachute into a pool of 75° F water.

Cannonballer #6 is doing an Earth jump with a parachute with a 25° F air and water temperature.

Cannonballer #7 is doing a "Classic Earth" jump, but he has put on a little padding over the holidays, and has a diameter of one meter, and a mass of 500kg. Air and water temperature is 50° F.

Can you correctly guess the order in which divers will arrive at the bottom of their respective pools

If you can, that $1,000,000 is yours, because the rest of humanity is out of shape and bad at science. :wink:


How many humans would you need to ask at random before there is a-better-than- average chance that at least two of them have the same birthday

How many Martians would you need to ask at random before there is a-better-than-average chance that at least two of them have the same birthday

For the purposes of this puzzle pretend that all birthdays are equally likely, and that Martians exist.

Tryagaing : you really are a source of wonderful stories and puzzles!

My (rough) guess for the "cannonballers" would be
5 < 3 < 7 < 4 < 1 < 2 < 6

without detailed calculations, absolutely not good enough for landing a MER And I think I'm in better physical shape than my guess is.

Relative, number 2 is correct. 6&7 tie for last place. Keep up with the exercise.

Answer to the ?'greasy spoon café'

The easiest way to ensure that you won't be stuck with dish duty is to go first. Place the first plate in the exact center of the table. Wherever the other waiter places his plate on his turn, place your next plate at the exact opposite setting.
Since every setting has a mirror opposite, at any stage in the game if, he can still place a plate, so can you.

Ok, now for the answers: to the 7-question problem. How many did you get?

1. The rope is 60 feet long. The wording makes it tricky, but the algebraic equation would be 30 + x/2 = x. Doubling both sides of the equation gives us 60 + x = 2x. Subtract x from both sides, and there you go.

2. Well this was certainly tricky. The pattern had absolutely nothing to do with any kind of mathematical progression. Write out the numbers in English: thousand, billion, octillion, hundred, zero, four, eight, three. The elements correspond to the letters of the alphabet. The value of an element is the first integer to contain that letter its name. So the next element is five.

3. This apparently was the hardest one of the lot. As I mentioned in the discussion forum hint, "Seven of Nine" is a red herring. Star Trek had nothing to do with the solution, I threw Seven of Nine in as a false lead.In Kit Williams's classic puzzle book, "Masquerade" (see links page for details) one of the clues to the treasure's location was "One of Six to Eight". The solution to this clue was the first wife of six wives to Henry VIII, Catherine of Aragon.

4. The answer is 2. To see this, square both sides of the equation. Notice that x + the original series now equals four.

5. There are 7 1's 3 2's 2 3's 1 4's 1 5's 1 6's 2 7's 1 8's 1 9's and 1 0's in this sentence. There are a few key pieces of insight that help solve the problem. First, there can only be 1 zero, because there aren't zero of any number. Next there can't be more than 1 nine. Then one can recognize that most of the larger numbers will have value of 1. The only except will be whatever number fills 1's slot. Trial and error can solve the rest quickly.

6. The next number is 132. This series is known as the Catalan series.

7. The answer is about 3.9 x 10-14. This is, of course, a trivial problem in unit conversion. Trivial if you happen to know what c, petasmoots and femtotribulations are. The symbol "c" is the speed of light, about 3 x 108 meters per second. Most people probably figured out that "peta" and "femto" prefixes are power of ten modifiers, like "mega" and "micro". Peta means "a quadrillion times", femto means "one quadrillionth".
A tribulation is a biblical term referring to the Book of Revelations. Depending on your reading it was either seven years, or half that. I was prepared to accept either answer.

Finally, a smoot is unit of measurement used only for the length of the Harvard Bridge in Boston, Massachusetts.

Now, you know why I only got two!


For some reason, Skinny, Monty and yourself need to cross the Amazon river with three monkeys. These are delinquent monkeys, so there must be at least as many humans in one place as monkeys, or else the monkeys will gang up on the outnumbered human(s) and force them to dance to organ music.

Your group has a single two-primate canoe at your disposal, which can be piloted by any of the humans or Ed, the one smart monkey.

Can everyone cross the river in safety In addition, how can they do it

Hi : In only really got 1,4&5. I thought I had 7, but turns out it is really 1/3 of the way there - i didn't know about tribulation and smoot
I struggled with 6 but couldn't get it, i got pretty close but not to Catalan numbers : I looked that one up later using google.
Now for the One of six to eight - I was totally on the wrong track, trying to solve it logically.
I just KNEW there was something lateral in number two, BUT ..

And "cannonballers" certainly deserve a detailed calculation one of these days .. ah well.
I was lucky to get 2.

You can do it in 4 return trips plus the last 1 way trip.

As long as a monkey in a boat doesn't count as in the same place as one on the shore.

Regarding the matching birthdays, you would need to ask 23 people.
Don't know how long a martian year is so don't know about them.

Water Lillies. Good one Try.

Adrian, at first count you score 1 ½ out of three. But I have been wrong before

Water Lillies it is. (Remember, two I's, ?'lies' at the end?)

Now if only I could understand;
"My first are raven plumes dipped deep in inky well,"?

Birthdays. 23.
With one human there is a 0 percent chance that you'll have two humans with the same birthday.
With two humans the probability that they won't share a birthday is 364/365. The probability that they will share a birthday is therefore 1 - (364/365).
With three humans the probability that they won't share a birthday is the same as for two humans, times 363/365. So the probability that three humans will share a birthday is 1 - (364/365) * (363/365). Notice that with each additional person added, the probability that he or she shares a birthday with one of the previous persons goes up, because there are fewer "free" days remaining.
Following this progression, the probabilities are:
Number of Humans Probability of two shared birthdays
1 0
2 0.00273972602739725
3 0.00820416588478134
4 0.0163559124665502
5 0.0271355736997935
6 0.0404624836491114
7 0.0562357030959754
8 0.074335292351669
9 0.0946238338891667
10 0.116948177711078
11 0.141141378321733
12 0.167024788838064
13 0.194410275232429
14 0.223102512004973
15 0.252901319763686
16 0.28360400525285
17 0.315007665296561
18 0.34691141787179
19 0.379118526031537
20 0.41143838358058
21 0.443688335165206
22 0.47569530766255
23 0.507297234323986

So 23 humans will have a better than average chance of sharing a birthday.

What about our misunderstood friends the Martians, who have a year of nearly 670 days? (As someone in the discussion forums pointed out, they may have leap days as well, but it turns out not to matter.)
Surprisingly, only eight more Martians will be needed.

As for the Monkeys, please remember only two in the boat at any one time.


The Cannonballers.

With a little bit physics and math you can lay claim to that $1,000,000. The correct order of arrival is as follows:

#4 is the first to arrive. Without air resistance, he would arrive in about 14 seconds. Is the air resistance enough to allow the moon divers to pass him? Probably not, air resistance rises with velocity and this diver won't be anywhere near his terminal velocity of 300m/s when he hits the water.

#1 comes in second. His condition is the same as #4's but at a cooler temperature. Cooler air is denser, so more air resistance.

#5 comes in third. The moon's gravity is much weaker than the Earth's, so doesn't catch up with his terrestrial competitors even though he doesn't have to overcome air resistance. And of course, his parachute does nothing without an atmosphere. He lands about 35 seconds after jumping (gravity increases slightly as he descends, but not by much).

#3 comes in fourth. Again, the water temperature difference mean he trails #5 by a slight margin, almost certainly less than half a second later.

#2 comes in fifth. We don't specify the kind of parachute, but assuming it's one that slows the diver to a "safe" speed, he can't be moving more than a few meters per second. Even at 10 m/s, he'll come in over a minute later.

#6 and #7 tie for last, by failing to reach the bottom of the pool. #6 hits a massive block of ice with a bang (water is a solid at 25° F). #7 has a lot more mass than his competitors, but also a lot less density. In fact, he floats.

With these results, you can collect the $1 million jackpot. If inflation between now and then averaged 2.5% per year, this sum would have a pre-tax value of over two dollars in today's money.


Problems for today;
This is so easy, even I was able to do it.

Deep in the grey cave, you have discovered a bowl filled with the Water of Truth. Legend has it that drinking exactly four cups of this illuminating liquid will confer great wisdom and reason upon the imbiber.
However, drink one ounce too little and you will lose all wisdom and reason you had to begin with.

Drink one ounce too much and you will gain a weeklong hangover.
Beside the bowl are, inconveniently, two vessels measuring three and five cups respectively.

Do you have enough reason to measure out exactly four cups without wasting the precious Water


Nefarious Ned, the archetypal villain, is planning to blow up the old silver mine tomorrow morning with Perilous Pauline trapped inside. Historically, his method has been to tie up Pauline, light the dynamite with an hour-long fuse, and get out of town. Lately, he has noticed that a few minutes before he is about to demolish the bank/dam/old sawmill with Pauline inside; Heroic Hank arrives and rescue her.

Ned has a new plan. This time he is going to use a forty-five minute fuse. With a little luck, the dynamite will dispose of both Hank and Pauline.
Unfortunately, Ned sent his non-alliterative comic relief henchman, Slow Jim to buy his fuses. Jim came back with a half-dozen of the usual hour-long fuses.

The obvious solution would be to take three-quarters of one of the fuses, but Ned's observed that while the hour-long fuses will always burn up in exactly one hour, their rate of burn is not constant. In fact, it is not uncommon for him to light a fuse and have it burn like crazy for the first fifty minutes, and then slow to a crawl during the last ten minutes. This, oddly enough, is when Pauline is usually rescued.

Another possibility would be to light one fuse, wait fifteen minutes, and then extinguish it. If only Ned had a watch, that would work.
Can you help Ned find a forty-five minute fuse

Try PM.

To the land of classical music:

Violins produced on the island of Grxcd have become collectors' items since it sank into the sea two centuries ago. All the island's violins were produced by Bropcs or one of his sons, or by Czwyz or one of his sons. Every violin was labelled ostensibly to reveal its maker but, although Bropcs and his sons always labelled their violins truthfully, Czwyz and his sons always labelled their violins with falsehoods. Both families playfully interfered with collectors' attempts to establish provenances for their violins. For example, collectors figured out that a violin labelled " This violin was not made by any son of Bropcs." was made by Bropcs Sr.; can you see why? The most desirable violins are so labelled that a connoisseur can tell that it must have been made by one of the fathers, either Bropcs Sr. or Czwyz Sr., but cannot tell which. How might such a violin be labelled?

To the land of classical music:

Prof. W. Kahan

I have been asked for some more ?'old fashioned' riddles. (Ones that do not need maths) Try:


If you remove the whole of me, there is still some left over.
What am I

There is an 11-letter word in the English language that is pronounced incorrectly by over 99% of the population. What is this word

Count the number of 'F's in the following text:

FINISHED FILES ARE THE RE- SULT OF YEARS OF SCIENTIF-
IC STUDY COMBINED WITH THE EXPERIENCE OF YEARS.

How many are there

What comes Twice in every day, four times in every week, but only once in a year

What comes next in this sequence?

O, T, T, F, F, S, S,

Using only the letters in the TOP ROW of your keyboard, what is the longest English word you can type And its not (qwerty).

retirwepytee6yltcerrocniemoselohw !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Crossing the amazon river.

Human and monkey cross, monkey gets out human goes back.
Human and Ed cross, human gets out Ed goes back.
Human and Ed cross, human gets out Ed goes back.
Ed and monkey cross, monkey gets out Ed goes back.
Human and Ed cross.

What am I missing?

Fuses, this one has been posted before I think.
You light 2 fuses, 1 at one end and the other at both ends.
When the one burning at both ends has finished 30 minutes have passed.
Light the other end of the second fuse now and when it finishs 45 minutes will have passed.

retirwepytee6yltcerrocniemoselohw !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

First class Adrian, throw another dingo on the barbe for me.

Crossing the amazon river.

Human and monkey cross, monkey gets out human goes back.
Human and Ed cross, human gets out Ed goes back.
Human and Ed cross, human gets out Ed goes back.
Ed and monkey cross, monkey gets out Ed goes back.
Human and Ed cross.

What am I missing?

About 10 trips, quicker than the hard worked answer I have. You are my Hero of the Week.

My rather long winded answer.

You start by having Ed reluctantly ferry his evolutionarily challenged cousins across the river. By themselves, they cause no trouble. When finished, Ed returns and stays behind with a human (all men are equal for the sake of this problem) while the other two human row across.
One human ferries one monkey back to the starting side, then goes across with Ed. This human picks up the other monkey and returns to the first bank while Ed thinks "Why the @#$?!? did they have me bring those monkeys over in the first place?!?"
The two humans now on the first bank cross over and begin afternoon tea, having had enough travel for one day. A now thoroughly disgusted and motion sick Ed is sent across to ferry all the monkeys over once again.

Cup answer.
Fill the three-cup vessel to the top. Pour it all into the five-cup vessel.
Fill the three-cup vessel again. Pour all that will fit into the five-cup vessel; this will be only two cups, leaving one cup in the three-cup vessel.
Empty the five-cup vessel back into the bowl. Pour the one remaining cup into the five-cup vessel. Refill the three-cup vessel and add it to the five-cup. You now have four cups in the five-cup.Drink it all up, and see how easy it was!

Our villainous Ned can use the other fuses in lieu of a watch. Here is how:
First, he takes one fuse and forms a loop so that the two ends touch at point A Call this Fuse AA.
Next, he takes a second fuse and touches one end of it to the two ends of Fuse AA, and straightens it out to point B. Call this Fuse AB.
Then, he takes a third fuse and touches it to the loose end of Fuse AB, at point B. This will be Fuse BC.
Now Ned touches a match to point A. Fuse AA will begin burning at both ends, while Fuse AB will start to burn from A to B.
Because AA is burning at both ends, it will burn twice as quickly. Ned waits patiently, and as soon as AA is completely gone, a half hour has elapsed. He then immediately touches a match to point B, starting BC burning, and igniting the unlit end of AB.
At the moment B ignites, Fuse AB has half an hour left. Now it will burn twice as fast, and finish in fifteen minutes. As soon as AB disappears, Ned extinguishes BC and knows what remains will burn for forty-five minutes.
Now to abscond with Pauline!


Help!

You are trapped in the Prison Tower of Contrivances. Fortunately, you have managed to pry open bars on the window to your cell. Unfortunately, the window is half way up the two hundred foot tall tower.

Conveniently, two fifty-foot ropes are tied to two iron rings in the ceiling of your fifty-foot tall cell. The iron rings (and consequently the ropes) are about a foot apart.

Your confinement has weakened your bones somewhat, so you cannot safely fall more than a few feet without risking serious injury. However, your upper-body strength and climbing skills are top-notch.

Armed only with a pocketknife, how much of the rope can you liberate to aid in your escape

Crossing the Amazon : I thought at first there must be some glitch in the text, or I don't get something, because Ed can be used for every trip:

HHHMMM
Ed and a monkey
HHHM M M
Ed and a human
HHM M HM
Ed and a human
HM M HHM
Ed and a monkey
H M HHMM
Ed and a human


But I was missing a point, the same one as Adrian:
On the second trip, when a human arrives with Ed, with one monkey already on the other side, making a lone human with 2 monkeys alone, dancing to the Bach's fugue in F-minor..

The only solution is as posted by Try