The worlds first riddle!

Lights : Probably my wording was bad.

The light is controlled by only one of the three switches, but you don't know which one. When that particular switch is ON, the light is on. The other two are just duds.

I hope this makes it clear.

Relative

I have an answer to the lights problem. However it is too good a puzzle to waste, so I'll PM the answer to you unless you say otherwise.

So you did it in one try! Nice work!
I remember it gave me quite a headache when I heard it the first time.

You guys, crack me up. I have not laughed so hard for ages.

We have Relative, apologising for only having 5 answers in an hour!

Then Lacomus, quotes," While you might have difficulty in using your bottle-lifting technique, I agree that, IMHO, your answer fits all of the conditions of the question."
(Now let me tell you about boards over acid baths . . . Mwuhahahahaha!)

Which can only be described as ?'you vote for me and I will vote for you'

Then Frank, "…they've got bars in Slovenia with ice machines that go down to -10 C!!!"

Come on Frank, in the States they are called a ?'Sauna' the outside temperature is -30c
In Slovenia. You go into the freezer to warm up.

The Barrels question, was very good.

Was there no answer to this question?
In the far, far north grows a lonely tree. Legend has it that when this tree is cut down, there will never be another Christmas. Ten years ago, a woodcutter drove a marker into the trunk of the fifty-foot tree at a point half his own height.
Not knowing of the legend, he promised himself he would finally get around to felling this lonely tree when the marker was as high as he. This woodsman was a bit of a legend himself- he was twenty-five feet tall ten years ago (Legends are rife in the northern reaches where it's winter year round).

Ten years ago, the woodsman grew at a rate of a foot a year. Today he grows at only eleven inches per year. Every ten years (on the dot) his rate of growth shrinks by an inch. Someday, he will stop growing altogether. The tree, on the other hand, will grow until it is cut down at a rate of one inch per year.

If legends live forever (and were true), in how many years will the last Christmas be

This one is true. (and clever)

During WWII, the Royal Air Force called upon a mathematician to improve their bombers. Specifically, they wanted to reduce the chances of them being shot down. Bombers are relatively slow moving targets escorted by smaller fighters to protect them from air attacks. But antiaircraft fire would occasionally knock them down. Therefore, a reasonable solution was to put metal armour on the bottoms.
However, putting metal armour all over the bottom would add too much weight.

Therefore, a mathematician examines the bottoms of all the bombers which had been in combat. Most had at least some damage to their undersides, clustered about seemingly at random over the bottoms. After examining all the locations the bombers had received damage, he made a recommendation.
What was it

I am still thinking about the lights

Try

To return to your problem of the Northern Woodman:

Frank's answer was that the spike would never reach the woodman's height - I think I report that correctly - and after a lot of thought I agree with him.

The main reason is that trees do not 'stretch' in growth, they grow from the top. So if you stick a spike into a tree around twelve foot from the ground and the tree grows to twice its height, the spike will remain at near enough twelve foot from the ground. I have often seen this when barbed wire is nailed to a tree. Long after the outward growth of the trunk has entirely engulfed the barbed wire, the scar the wire made on the tree is still the right height for fence wire.

So if the spike is at half his height then after he has grown it will always be less than half his height and never be at his full height.

Try : Last Christmas in 3720 years. Under additional rules : tree grows as a rubber band stretches, the day the mark was made is 10 years ago after christmas, the tree will be cut immediately and most importantly 1 foot =12 inches, right ?

Under who's bottoms ?

Ah, I know : the places that weren't damaged on the real planes

Planes --

He recommended that the RAF reinforce the armor on the areas that had NO holes.

In view of the ?'Christmas Tree' answers, I have asked the review board to look into the answer and their findings completely exonerate me to the extent that they only called me an ?'ignoramus' that's good isn't it?

Well this is a question of simple arithmetic.
The woodcutter grows 120" in the first ten years, 110" in the second ten years, and so on. This simple series adds up to 780 inches, when added to his original height of 300 inches is 1080 inches, or an even 90 feet. The woodcutter will never get any taller.

Clearly the marker will never overtake the woodcutter while he is growing, so the problem becomes when will the tree be ninety feet tall. And that is also simple arithmetic. The marker started at 150", so has another 930 inches to go, or 930 years. But that was ten years ago, so the last Christmas will be in 920 years. Oh no!

But as it happens, this isn't a question of simple arithmetic. It's a question of simple biology. Trees grow upward at their extremities; only the top of the tree is getting higher, not the bottom, which just gets wider.
So never, fear. Christmas will always be here!

Apologies to all I should have read my own notes more carefully.

Try this:
Walking home one day, you take a short cut along the train tracks. The tracks cross a narrow bridge over a deep gorge. At the point you are 3/8 of the way across the bridge, you hear the train whistle somewhere behind you. You charge across the bridge, and, just like in "Stand By Me" jump off the track as the train is about to run you down.
As it happens, if you had gone the other way, you would have reached safety just before being run over as well.
If you can run ten miles per hour, how fast is the train moving

Frank, and Relative are right about the plane. I thought that was very clever

40

The tree growth is a good one : just complex enough to obscure biology.
Though, i am not completely sure about very young trees - they seem to stretch a bit (or is it just my illusion)?

And BTW, if the tree stretches as a rubber band, the arithmetic is just a bit different : the part under the mark is 1/4 the size of the tree, so it grows just 1/4 inch per year, and thus 4* the number of years.

Try

"Vote for me and I'll vote for you"?

I prefer to call it being 'open to all eventualities' (But just as long as I get the votes, who cares what it is called??!!) :wink:

Relative, you are too good for me!

40 it is. Now, answer me this:

You work at the local greasy spoon waiting on tables. At the end of a particularly long shift, another waiter challenges you.
He proposes the following game:

Each of you will take turns placing plates on a clean table. All the plates must be the same size, lie flat on the table, and no plate may overlap another. The first person who is unable to place another plate on the table without it falling off or moving another place must wash the dishes for the next week.

The table is a perfect circle, with a five-foot diameter. Each plate is also a perfect circle with a ten-inch diameter.
If you are given the choice of going first or second, which do you choose, and what will your strategy be

Please don't post the lights answer just yet, I am still working on it.

The snail and the rubber band:

I wanted to post this old puzzle but turns out I forgot the solution
I am struggling to solve it, , but decided to post it anyway.
If I don't solve it, we'll have to ask the Board to decide on the correct solution!

Anyway, we have a rubber band, 1 meter long.
At one end of the band is a snail, travelling at a speed of 1cm/second (a supersnail ) trying to get to the other side. But, alas, the rubber band is stretching, it's far end moving at a horrible velocity of 1 meter/second. Now this band is a superband also, and is infinitely stretchable.
The first question is : can the snail ever make it to the other side?
(in the first second the snail makes 1cm, but the distance to the far end just increased by almost a meter!)

The second question is, how long before (if ever) the snail reaches the end?

Good luck and bye for today!

Tryagain : I couldn't resist solving this one more, it's excellent!
Don't post the result..

Lacomus wrote, " What is a Star Trek?"
Can this be true? Is there a place on this planet that the ?'treckies' have not reached. I will vote for you.

The riddle grows longer to make it easier.

We live not on land, or sea.
We have two eyes, but cannot see.
We close at night, and open at dawn.
Forget the Birds, and bring on the Bees
It makes reproduction a breeze
The closer you are, the more you reflect,
on whether, your answer can be correct.
Part above and part below
Wet and dry, don't you know.
Claude, saw the clouds around us.
In a back garden in France.
But, at the end only lies.

We are


Two exceptionally quick-witted people were invited to a party in their honor. The party was to be held in a private house in the city of Mathematica.

Mathematica, predictably, is laid out in a perfect grid. Every house's address is composed of two positive numbers, indicating its relative position to the Town hall in the southwest corner. For example, House 3-8 is two houses east of the Town Hall and seven houses north.

Currently Mathematica is one hundred by hundred houses in area.
Our two famed puzzlers, let's call them Relative and Lacomus, were both given the address weeks in advance. And while both were quick-witted, they also were short memoried. Only a few hours before the party the following dialog occurred:

Lacomus: I'm afraid I've forgotten the address. I can only remember the product of the two numbers, and that the first number wasn't greater than the second.

Relative: I've forgotten it too, but can only remember the sum of the two numbers, and that neither number was 1.

Lacomus: I can't figure out where the party is.

Relative: I knew that.

Lacomus: OK, I know where the party is.

Relative: OK, so do I.

Lacomus and Relative were being perfectly truthful, and no other information was exchanged apart from this dialog.

What was the party's address

In a small town there are three houses: one white, one blue, one green. There are also three somewhat under utilized utilities: gas, electricity, and water. (they face each other. eg white opposite gas etc)

Is it possible to connect each house to all three utilities, so that no connection crosses another

Try

Before I head off to do some weekendly gadding about:

The three houses and the three utilities can not be connected as described and I can prove it (To be correct, Euler and I can prove it). There will always be one connection that cannot be made.

Also, and I make no other comment, Monet clinched it.

And now, being brilliant and also having a poor memory, I have to remember the address of that fantastic party I have been invited to.

<<Wanders off stage left muttering about sums and products and where the heck is Relative when you need him>>

Have a good one. :wink:

Relative, having stayed at home, has been thinking about the last time they went to the party with Iacomus.
"Wait a minute!" he slaps his forehead, "did that Mathematica really have 100x100 houses? I think Iacomus and I weren't really that brilliant finding that party, hmm, could say we were pretty lucky!"

And with that, decides to ask Tryagain forgiveness and 'shed a light' again on number of houses in Mathematica..

Lacomus is correct, all the connections can not be made.

The number of houses are not important, and only show that there are plenty of addresses to choose from. The champagne is beginning to go flat so I offer a few clues. :wink:

From Lacomus and Relative's first two statements, we know the product, let's call it P, isn't prime (because neither number was 1):

But much more informative is Lacomus second statement that he can't figure out where the party is with the information Relative supplied. If P was the multiple of exactly two prime numbers, Lacomus would know it by now (he would simply factor his number into its unique prime components). Therefore, he knows the address is not two prime numbers (although it could be one prime number and a composite number). By stating this, he is telling Relative as much.

Well, I was thinking along the same lines and came up with this:
p= a*b
s=a+b

1.) a and b are not prime
2.) s is not even: if it was even , there would be at least one s=p1+p2 where p1,p2 are prime (Goldbach conjecture), otherwise Relative can't know Iacomus can't solve the problem.
3.)All of the factors of p are less than half of M, where M ix max coordinate. Otherwise Iacomus can isolate this one factor.
4.) because of 2.), Iacomus knows he must put all factors 2 on one side, otherwise sum is even and then 2.) would follow. So he only needs to inspect combinations with odd sums.

From these I could easily eliminate most sums, and quickly found two solutions:
My initial two solutions were 3,8 and 4,13.
I knew there were many more, so I asked for some additional rule.

Besides, even without coordinates I found my party as well