i would argue that anyone in a gravitational field does not neccesarily have a tub.

For the nails and tha painting, put the nails in very very close together (so they basicaaly touch). Tie a not near the center of the string and wedge it between the nails. Either gets removed and it falls.

2^n (for n =2) is 4, a 4x4 board fill that however you like (left as an exercise for the reader :P)

for the next board size up all you need is 4 thick L shape to surround (sufficient but not neccesary) This shape can definitely be filled by L shaped blocks. (but i don't have the mathematical vocabulary to prove that.)

slowly begin to tip it on its side.
If you reach a point (45 degrees) where the level of liquid is exactly touches two opposite edges of the cylinder, then its exactly half full.

@ Draugnar, I'm 15 years old but an good at maths and have an interest in riddles (and in the question of free will)

this cylinder is transparent/open? that makes it easy...
I was trying to use an opaque cylinder.

You can look in the top, presumably, DC. That works perfectly, dave bishop. You are quite good and think outside the box. Did you like my answer for the picture frame riddle?

Oh, but it may be more or less than 45 degrees depending on the height of the cylinder. The key is if it reaches the front lip before (more than half), after (less than half), or at the same time as (exactly half) the liquid reaches the back edge.

Yes I did... the one with the knot? Ingenious.
And yeah... you're right about the 45 degrees thing.
I think I'll now post a similar riddle to my 9 ball riddle earlier

You have ten bags, each containing 10 balls.
9 bags contain 10 balls weighing 1KG each, while the tenth bag contains 10 balls weighing 0.9KG each.
You have to find which bag contains the heavier balls.

To do so you can make one weighing on a set of electrical scales which will give an exact answer for the total weight of the balls it carries.
You can put any number of balls from any number of bags on the scales to make this measurement.

How do you find, for sure, the bag containing the heavier balls?

I'm not understanding the question. Do you mean the bag that contains the lighter balls?

Yes, its meant to be this:

You have ten bags, each containing 10 balls.
9 bags contain 10 balls weighing 1KG each, while the tenth bag contains 10 balls weighing 0.9KG each.
You have to find which bag contains the lighter balls.

To do so you can make one weighing on a set of electrical scales which will give an exact answer for the total weight of the balls it carries.
You can put any number of balls from any number of bags on the scales to make this measurement.

How do you find, for sure, the bag containing the lighter balls?

Because your statement says that 9 of the bags contain 10 balls weighing 1K each, while your question asks how to find the bag containing the heavier balls. Wouldn't you be trying to find the bag that contains the lighter balls?

disregard that last post

hmmm. I'm not sure if this is right but I will give it a shot. What you could do is take one ball from this first bag, two balls from the second, three balls from the third etc... Then you weigh the bag that you created and depending on the total, you would be able to find out how many .9 KG balls were in the weighed bag, and thus the number of the bag that carries the .9KG balls.

Take one ball from bag one, two balls from bag two, etc.

You'll have 55 balls. The total weight will be 55 kg minus a number from 0.1 Kg to 1.0 kg.

If the number is 0.1 kg (total weight 49.9 Kg) bag 1 has the lighter balls.
If the number is 0.2 kg, bag 2 has the lighter balls, etc.

yep, well done guys

Back to the original puzzle.

You have 12 coins, one is fake. But you don't know whether it is heavier or lighter than the other coins.

You have a balance and you may use it 3 times. How do you find the fake coin?

Lucas

Oh, it was not the first riddle. I mean weighing scales.

I could find which out of 27 were heavier if I had 3 weighings

weighing any six coins, three on one side and three on the other, if they are different then you know you've got the fake, otherwise you know you've got 6 real coins.

if you've got the fake, take of 3 and put them aside, place 3 real coins and see if they match (putting them on the heavier side also reveals whether the fake is ligther or heavier) if you've still got a tipping scale you know the fake is ligther and which 3 coins it was in, if they balance you know it's heavier and it is in the three you put aside.

(once you have it down to three plus whether the fake is heavier or ligther you put three coins on for your last weighing)

If the first six weigh the same you've got six real coins, and again pick three plus thee of the ones you haven't weighed yet...

is this the goat v the car on the game show riddle? it was referenced in "curious incident of a dog in the nighttime." it's an awesome problem. it's called the monty hall problem, isn't it?

@Lucas

It is a question of working out all the possibilities, finding which coin is counterfeit, and whether it is heavier or lighter. Let us begin...
Let me call the coins A through L...
First weighing I weigh A, B, C, D vs E, F, G, H
Two possibilities - they don't balance or they do
& Let's say they don't - two possibilities again A, B, C, D is heavier than E, F, G, H or A, B, C, D is lighter than E, F, G, H.
# Let's say A, B, C, D are heavier so either one of A, B, C, D is heavier OR one of E, F, G, H is lighter. Then I will weigh A, B, H vs D, E, L leaving C, F, G aside.
Three possibilities - either they balance, A, B, H, is heavier than D, E, L, or A, B, H is lighter than D, E, L.
* Suppose they balance. Then C, F or G is the fake. So I weigh F against G.
Two possibilities - they balance meaning F, G are real and C is fake and heavier, or they don't balance meaning whichever was lighter in that last weighing (F, G) was the fake, and lighter.
* Suppose A, B, H was heavier than D, E, L. So, either A or B is heavier or E is lighter. Weigh A against B. If they balance, E was fake and lighter. If they don't balance, whichever one of A, B was heavier is the fake and heavier.
* Suppose A, B, H was lighter than D, E, L. So, either H was lighter or D was heavier. Weigh H against L. If they balance, D was fake and heavier and if they don't balance, H is fake and lighter.
# Supposing A, B, C, D was lighter than E, F, G, H, take everything in the asterisk section and switch the letters A, B, C, D with E, F, G, H and vice-versa.

& Suppose A, B, C, D balance against E, F, G, H, then they're all real. Then we weigh I J against K, A
Two possibilities - they balance or they don't.
% Say they balance, then L is fake and we weigh it against any of the real coins to see whether it is lighter or heavier.
% Say they don't, then either I, J is heavier than K, A or I, J is lighter than K, A.
@ Say I, J is lighter than K, A, then, either one of I, J is fake and lighter, or K is fake and heavier. You weigh I against J. If they balance, K is fake and heavier; if they don't whichever is lighter of I vs. J is fake, and lighter.
@ Say I, J is heavier than K, A, then, either one of I, J is fake and heavier, or K is fake and lighter. You weigh I against J. If they balance, K is fake and lighter; if they don't whichever is heavier of I vs. J is fake, and heavier.

All that for a fake penny. I would have just taken the hit.

@orathaic
Your solution requires more than three weighings. Check out the different options and see that you usually need four.

This was on my math homework last week.

Let a1 , . . . , an be distinct positive integers and let M be a set of n − 1 positive integers not containing a1 + ⋅ ⋅ ⋅ + an . A grasshopper is to jump along the real axis, starting at the point 0 and making n jumps to the right with lengths a1 , . . . , an in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in M.

e.g. in terms of numbers

123=3, 456=4

First step eliminates 6 of the possible solutions. Gives you two possible weights or a definite weight of a regular penny. The weight of one penny may equal 1 or 1_1/3.

14=2, 25=3

By dividing the first weighings by three and comparing them to these measurements, you can deduce which one has the fake penny. 14's average penny is 1. 25's is 1.5. If they equal the same, you know it's one of the two you left out. It must be one because it corresponds with the first measurement.

The last measurement is between two pennies.

Simpler version. I don't know if this is what you said LeoDaVinci.

Okay, this one has nothing to do with math:

One day, while looking through the archives of early twentieth century explorers, you come across a scrap of paper with this diary entry.

“June 24

It was pitch dark when I got up this morning. As I cooked my breakfast I vowed that this would be the day I started to observe the local inhabitants.

I set out due south, and within half a mile of the camp I came across the tracks of a mature specimen. Although I could see the outline of his bare feet clearly, there was something odd about them. It took me a moment to realize that the right foot was taking a larger step than the left. By now I had already visualized the individual I was tracking, and decided to refer to him as “George.” I was able to follow the tracks another three miles south, but then lost them completely. I decided to make a sweep directly west, since I had planned to explore that way anyways, and it seemed to be as good a direction as any for spotting George again. As I trudged on, I marveled at the desolate landscape, and realized I was the first European to see these sights.

I continued on my westward trek until supper time, when I paused to heat up some stew. After another half hour, I decided to give up, and turned to head north. I was enjoying the peaceful evening but after a couple of miles of northward journey I noticed my lantern was running quite low on fuel. I turned it down to the barest glimmer, and started to look for the camp. I knew it would be only a mile or so north of where I stood.

Finally, I was able to make out the tents. I had just picked up my pace when suddenly a strange figure loomed in front of me. Startled, I realized it was in fact the object of my research. He seemed a hearty specimen, but considerably shorter than me. Looking closer, I knew I had found George: there was something funny about his left foot – perhaps it had been injured in the past. He didn’t seem surprised at running into me, but turned and looked over his shoulder as he walked away.

I watched as he vanished into the dark, and then rushed into camp to detail my first encounter.”

The question: could George swim?

"Let t be a real number, and let floor(t) be the greatest integer less than or equal to t, i.e. floor(t) is t, rounded down to the nearest whole number. Define x=floor(t) and y=floor(2t). Write a polynomial in the variables x, y which is identically zero."

Isn't this just (2x-y)(2x-y+1)? One of those has to be zero.

and for yours, Plastic Hussar, I think the point is that the only way the west thing would work out is if you were at one of the Earth's poles. So I assume that the answer is yes, he can swim, since he is a polar bear?