The worlds first riddle!

That is interesting mark, don't know exactly what it means but, it still is interesting.

TTH:

7,7,3,3=24

I get 3/7= .4286 + 3 = 3.4286 x 7 = 24.00

Yup! 7(3/7 + 3) = 24


Mark:

5, 5, 5, 1

5 * (5 - 1/5)

Looking good!


RESISTANCE
I'll take a shot at the simplest case. I get 1/2 for the tetrahedron.

RESISTANCE (2)
I did a little Googling and found this:

tet: 1/2
cube: 7/12
oct: 5/12
dod: 19/30
ico: 11/30

The tet is its own dual. The cube and oct are duals. The dod and ico are duals.

Interestingly, if you add the resistances of duals, you get 1.


You are right; at first glance the question seems near impossible. However, when taken a step at a time…not so difficult!

They can be derived as follows. For any graph the resistance between two points, S and T, can be computed as follows. Set the potential of S to 1, T to 0. Solve a set of linear equations to find the potential of the remaining vertices. Compute the current flow out of S (which will necessarily equal the current flow into T). Invert the current flow to determine the resistance.

For our case for each graph let n be the number of vertices and d be the degree of each vertex. Specify adjacent vertices S and T. Setting the potential of S to 1 and T to 0 will determine the potential values on the other vertices. Consider rotating the solid around S (so T coincides with each of the neighbors of S).

Average the rotated potential values at each vertex. The average at S is clearly 1 (since S remains fixed). The average potential value at each neighbor of S is clearly the same by symmetry. Let this value be x. We claim the average potential at every remaining vertex is also x. This follows because there is no current flow into or out of any of the remaining vertices in any of the rotated states.

So what is x? By symmetry the average potential over all the vertices is .5. So since the potential at S is 1 we have x=(n/2 - 1)/(n-1). So the average current out S (which will be the same for each rotation) is d*(1-x)=d*(n/2)/(n-1). So the total resistance between S and T is the inverse of this or (2*n-2)/(d*n) = (2/d)*(1-1/n). Using the values of n and d for each solid we find the answers above.

This method should work for any sufficiently symmetric graph.


TTH thought, "That is interesting".

It is not known what Miss mi thought! My own thought is centripetal.



In a certain sequence, the first term is 2 and each succeeding term is 1 less than the square of the previous term. What is the forth term in this sequence



1000=W that a P is W

1001 = A N

1000=W that a P is W = 1000 words that a picture is worth

1001 = A N = 1001 Arabian Nights

When using decimal notation is STEPHENS< STEVENS

I don't even know if I understand the question

If the first term is......
2 then the next is (2x2) - 1 = 4-1=3
3 then the next is (3x3) - 1 = 9-1=8
8 then the last is (8x8) - 1 = 64-1=63
63 is the fourth term?

i was thinking
3+3+7+7=20 which in base 8 equals 24
5+5+5+1=16 which in base 6 equals 24

Stormy:

1000=W that a P is W = 1000 words that a picture is worth

1001 = A N = 1001 Arabian Nights


TTH:

If the first term is......
2 then the next is (2x2) - 1 = 4-1=3
3 then the next is (3x3) - 1 = 9-1=8
8 then the last is (8x8) - 1 = 64-1=63
63 is the fourth term

Confidence is all you need!


Thoh:

I was thinking
3+3+7+7=20 which in base 8 equals 24
5+5+5+1=16 which in base 6 equals 24

That is good thinking. Why didn't I think of that?



Lost in the city, you finally happen upon a rickety old bridge that leads to a shortcut. You need to know if the bridge is safe. As luck would have it, you recognize that on the other side of the bridge is the A2K office.

Everyone who works there either ALWAYS tells the truth or ALWAYS lies. You see three men on the other side of the bridge; Mark , Try and Thoh .


You shout out: "Are you truth tellers?"

Mark says something, but you can't hear what he said.
Try shouts, "He said he was a truth-teller."
Thoh shouts out, "No, he didn't. He said he was a liar."

Which man do you ask to find out whether or not the bridge is safe




Walk on the living, they don't even mumble
Walk on the dead they mutter and grumble

I am

I think that:

Thoh is a liar! This is because Mark would never call himself a liar!
Try is telling the truth if Thoh is lying!
Mark needs to speak up!

I would ask Try



Leaves!

(Tolkein )

OK - I'll speak up...

THOH: ARE YOU GOING TO LET TEHMEH CALL YOU A LIAR?

That was convenient of Try to make himself out to be the good guy...
And you're right; I would never call myself a liar.

no i am not

Well, that sure showed him
mark since you know computers and all, this one was for you:

TehMeh:


Thoh is a liar! This is because Mark would never call himself a liar!
Try is telling the truth if Thoh is lying!
Mark needs to speak up!

I would ask Try

You rock at teh Intarweb.


The first man would always say he is a truth-teller:
If he is a truth-teller, of course he would say so.
If he is a liar he would lie and still say he is a truth-teller.

This means that the second man told the truth.
(Of course, you could always ask Thoh; as long as you realize he is lying.)

Leaves!

(Tolkein)

Omg u b0rk t3h intarw3b!!1!!11!shift+one !1!!


Mark:

That was convenient of Try to make himself out to be the good guy...
And you're right; I would never call myself a liar.

Which means that you too could be a truth teller.




If 3x+4y=10 and 2x+7y=11, then x is




10001000 = O in a M =

31536000 = S in a Y =

8549176320 = D in AO =

9192631770 = OPS in a CA =

No one is even going to try and attempt this one?

When using decimal notation is STEPHENS< STEVENS

I didn't do the math so I will guess....seconds in a year?

Edit to add:
2?
One in a million?

31536000 = S in a Y = seconds in a year

digits in alphabetical order?No idea, not even a guess

Oscillations Per Second in a Cesium Atom

Atomic clock

I thought that was the answer to something....how embarrassing!

Even Stephens? - shot in the dark once again... :wink:

Sorry mismi40
Good guess, but not what I was looking for.
I thought mark might get it. It has to do with computer programs.

TehMeh
I would never have gotten that answer.

Oh...no way would I get that - I only know how to turn my computer on and off! Just gave it a shot for you!