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.

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.

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..

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