Thread: Post A Logic Puzzle

because a) Monty is being inificent, and b) he could be taking out more then one dollar/rand

what intreges me is the fact you said usually

I figured out what was confusing about the waiter one. The amount paid is the amount the cook have plus the waiter... you never count the amount the customers have..

I totally missed the waiter question. When it says each customer has effectively paid 4 dollars it doesn't make sense because you pay two different people. (10/3+1)x4+2=15 dollars.

As for the devil one, although it intrigues me why you'd want to pay for heating in hell, but if you think about it the key is in the infinite days given. The probability that Monty would take any bill out of the sack is 1 over the course of infinity, so you'd end up with no money. On the other hand if you have finite days, you'd end up with exactly the same amount no matter which banker you choose.

The only time havent read back dammit i wasted five minuate of my life.

Well i will post the hardest puzzle i have seen i wonder if you could do it.

Miscellaneous Problems

28. Suppose five dots are arranged in a three-dimensional space so that no more than three at a time can have a flat surface pass through them. If each set of three dots has a flat surface pass through them and extend an infinite distance in every direction, what is the maximum number of distinct straight lines at which these planes can intersect one another?
29. Suppose a diagonal line is drawn across each of the six sides of a cube from one corner to the other. How many distinct patterns are possible if one includes all six sides of the cube in each pattern and counts as one pattern any patterns that can be made to coincide by various rotations of the cube as one rigid object?
30. Suppose the thirty edges of a regular, i.e., perfectly symmetrical, dodecahedron are rods, two of which are painted white and the rest black. How many distinct patterns can thus be created, counting as one pattern any patterns that can be made to coincide by various rotations of the dodecahedron as one rigid object?
31. Suppose ten marbles are inserted into a box based on the tosses of an unbiased coin, a white marble being inserted when the coin turns up heads and a black one when the coin turns up tails. Suppose someone who knows how the marbles were selected but not what their colors are selects ten marbles from the box one at a time at random, returning each marble and mixing the marbles thoroughly before making another selection. If all ten examined marbles turn out to be white, what is the probability to the nearest percent that all ten marbles in the box are white?
http://www.eskimo.com/~miyaguch/power.html#power13

Well their more on the site and if you type in titan test in google then your get IQ of about 160-259 puzzles see if you can do it.

As for the devil one, although it intrigues me why you'd want to pay for heating in hell, but if you think about it the key is in the infinite days given. The probability that Monty would take any bill out of the sack is 1 over the course of infinity, so you'd end up with no money. On the other hand if you have finite days, you'd end up with exactly the same amount no matter which banker you choose.[/b]

DING DING DING DING DING. Correct.

Although the answer is debatable.

Ross-Littlewood paradox
Main article: Balls and vase problem
Suppose there is a jar capable of containing infinitely many marbles and an infinite collection of marbles labelled 1, 2, 3, and so on. At time t = 0, marbles 1 through 10 are placed in the jar. At t = 0.5, marbles 11 through 20 are placed in the jar and marble 1 is taken out; at t = 0.75, marbles 21 through 30 are put in the jar and marble 2 is taken out; and in general at time t = 1 − 0.5n, marbles 10n + 1 through 10n + 10 are placed in the jar and marble n is taken out. How many marbles are in the jar at time t = 1?[/b]

This is pointless both problems are paradoxes so we wouldnt be able to figure it out

solve this

This statement is false.
is this statement true.

R=1

R=T
R^2=RT
R^2-t^2=RT-T^2
(R-T)(R+T)=T(R-T)
<strike>(R-T</strike>)(R+T)=T<strike>(R-T)</strike>

2=1

I did this today in math class and didn&#39;t get the answer, anybody know what I did wrong, or did I prove 1=2

R=1

R=T
R^2=RT
R^2-t^2=RT-T^2
(R-T)(R+T)=T(R-T)
(R-T)(R+T)=T(R-T)

2=1[/b]

check fourth line again the right side is wrong.

whats wrong?

whats wrong?[/b]

Expand the brackets on the forth line then go backward. I would tell you were you went wrong but it wont do you any good just work backward and it will become clear.

IF anybody else noes where you went wrong dont tell give him hints you dont want to do his homework.

if you still cant get the anwser have you consider taking stimultacs
http://youtube.com/watch?v=OvcDmPcGVG0

This isn&#39;t my homework, the teacher gave the whole class the answer I had to leave before I got it though.

try doing the actual numbers next to it as well

is it that R^2 is equal to R, because R is 1? I have no idea

try my idea

for example:

R=1 [no need to do anything here]

R=T [1=1]
R^2=RT [1^2=1x1] [1=1]
R^2-t^2=RT-T^2 [1^2-1=1x1-1] [0=0] <--that is the trick
(R-T)(R+T)=T(R-T) [(1-1)(1+1) {x by zero} = 1(1-1) {x by zero}
(R+T)=T [(1+1)=1]

2=1

Oh ya I was close, when I said you couldn&#39;t factor 0.

A man has 9 bags filled with gold coins all of the exact same size and wieght. His friend the magician wishes to play a trick on the man so he takes one of the bags, empties it and fills it with iron coins of the same size. He then magically makes them appear gold instead of grey. The man finds out his friend did this and demands that he undoes it. The friend promises to return everything to normal if the man can figure out which bag contains the fake coins. The iron coins weigh half as much as the gold coins and the man may only weigh any arrangement of coin from any bags, once. (one total) How does he do it? =P

A man has 9 bags filled with gold coins all of the exact same size and wieght. His friend the magician wishes to play a trick on the man so he takes one of the bags, empties it and fills it with iron coins of the same size. He then magically makes them appear gold instead of grey. The man finds out his friend did this and demands that he undoes it. The friend promises to return everything to normal if the man can figure out which bag contains the fake coins. The iron coins weigh half as much as the gold coins and the man may only weigh any arrangement of coin from any bags, once. (one total) How does he do it? =P[/b]

What this seem retarded is this right. Well it simple just weigh the bags individualy intill you find the bag that weights half as much then the rest whitch would be iron. I really think their something missing

Originally Posted by becomingagodo

What this seem retarded is this right. Well it simple just weigh the bags individualy intill you find the bag that weights half as much then the rest whitch would be iron. I really think their something missing
[/b]

he can only weigh them once.

he can only weigh them once.[/b]

I thought all you need to do i weight all the individual nine bags once to find the lightest.

you could, but he wont, because he knows he can do it with only one, so why wast the effort?

you could, but he wont, because he knows he can do it with only one, so why wast the effort?[/b]

Number of coins in the bag their might be infinite many coins in each of the nine bags. I normally think outside the box and that came to me.

Originally Posted by becomingagodo

Number of coins in the bag their might be infinite many coins in each of the nine bags. I normally think outside the box and that came to me.
[/b]

That might make it a little hard to pick up the bag XD

Plus, the merchant would be rich.

But to make everything easier lets say theres 50 coins in each bag and each of the gold coins weighs 100g. No infinites for you >

That might make it a little hard to pick up the bag XD

Plus, the merchant would be rich.

But to make everything easier lets say theres 50 coins in each bag and each of the gold coins weighs 100g. No infinites for you >[/b]

Finally some assumption can be thrown away. Their two more how many bags can you weight if it all nine then it easy and can you take the coins out of the bag and what do you mean arrangment. If the coin were infintly small then it wouldnt be that hard to pick up. Also does the scale measure exaculy what condition is it in strength of gravity does the scale suffer from some sort of uncertainty princple i.e. is it moving and below plank constant.

Originally Posted by becomingagodo

Finally some assumption can be thrown away. Their two more how many bags can you weight if it all nine then it easy and can you take the coins out of the bag and what do you mean arrangment. If the coin were infintly small then it wouldnt be that hard to pick up. Also does the scale measure exaculy what condition is it in strength of gravity does the scale suffer from some sort of uncertainty princple i.e. is it moving and below plank constant.
[/b]

You are trying to make this way too complex 0.o

You can take any amount of coins from any or all of the nine bags and weigh them. Lets pretend this scale is a heavenly scale that takes perfect mesurements mmkay?

Here is a simple one you may have heard.

Two men, each with his son, sit down to dinner. There are only three plates. Each person has his own plate. How can you explain this?