It's said that milking stools have exactly three legs so they won't wobble when set on an uneven barn floor.

Not so for tables, since home floors are assumed to be planar.

Which gives rise to an interesting puzzle.

Assume you have a table whose four legs are the same length.

But the floor is warped. Three legs touch the floor; the 4th does not.

Is it always the case that for any mildly warped floor your table can be positioned so that all four legs contact the floor?

Prove your answer.

so what you're after is a proof of whether or not there always exists 4 coplanar points at set distances from one another on an irregular surface? Further, the points could be anywhere within the surface and at any angle relative to the assumed axis. I'd say it depends upon the degree of irregularity and the size of the surface.

Example: assume the floor is exactly the dimensions of the table legs. I think we'd all agree that for this special case, you can certainly create a mildly warped surface such that 3 legs contact on the same plane and the fourth does not. With no additional room to maneuver the table, a solution with 4 contacting legs is impossible.

The question is, can you improve on this special case by enlarging the floor?

I'd say yes, you can improve the likelihood of having all four table legs contact the floor, but you can't guarantee it. It's disproven in the one special case I mentioned, therefore it cannot happen in all cases.

Well, its easy to figure out the minimum number of people the professor's uncle could have given the money to by maximizing the number of high dollar gifts:

7^0 = $1 * 1 = $1

7^1 = $7 * 1 = $7

7^2 = $49 * 3 = $147

7^3 = $343 *3 = $1029

7^4 = $2401 * 3 = $2401

7^5 = $16807 * 3 = $50421

7^6 = $117649 * 1 = $117649

7^7 = $823543 * 1 = $823543

So, the professor's uncle could have given away the million to as few as 16 people.

The only way to give the money to more people would be to remove one of the high dollar gifts and split it up among the lower dollar gifts. The problem is that each high-dollar gift is exactly 7 of the next lower dollar gift, so I'm not sure how the professor's uncle could abide within his rule about not giving the same gift to more than 7 people if he chooses anything but the minimum 16. For example, suppose he wants to give out one fewer $343 gift. The minimum number of gifts to add would be 7 $7 gifts, bring the total number of $7 gifts to 8, which exceeds the maximum. You can try it for any of the amounts, and you'll find it doesn't work.

Perhaps the surest proof is the following:

Try giving 7 gifts each of $1 through $117649 (the maximum in each category). The total is only $960799. You can't give out any more gifts than that, because all categories are exhausted and an $823543 gift would far exceed the one million available.

So I say he has only one option - 16 people in the configuration above.

The type of liquid doesn't matter much, but the amount of pressure behind it at the faucet could make a huge difference. If the pressure "pushes" the liquid out of the faucet faster than it can drain by gravity alone, you'll fill up the sink and cup. Of course, if it was a compressible fluid (which would expand after leaving the faucet), even better!

Is your cup affixed to the bottom of the sink but on top of the drain? If so, it will plug the drain, and water will fill both the sink and the cup. Or, if it's a bathroom sink, you could pull the little lever that plugs the drain to cause the same effect.

6 miles between the two points

First pick he has a 26/53 (~49%) chance at grabbing a black.

If he chooses to swap, he now has a 26/51 (~51%)(at best) chance.

If he did draw black first time around, he still has a 25/51 (~49%) chance of drawing black if he decides to swap.

I would keep the first pick. Seems to be too simple of an answer to be right tho.

Same situation as Let's Make a Deal Riddle.

Instead of opening a door w/a donkey, he's removing a joker.

Not sure what the best site is for explanation, but here's one.

http://www.increasebrainpower.com/lets-make-a-deal.html

Always switch.

(Alternatively 26/53 < 25/51 < 26/51, so the switching makes sense.)

The devil is in the details.

As several have pointed out, the chances of drawing a winning card on the first pick is 26/53, and the worst case chance on the second pick is 25/51. However, 25/51 (0.4901) is not greater than 26/53 (0.4905), so there is a small decrease in your chances of selecting a new black card if you already hold one. This is different than the typical Monty Hall/Let's Make a Deal kind of game, where your chances do improve.

That being said, I think the difference in those probabilities is neglible in comparison with the chance for improvement if you do not hold a winning card to begin with (0.4905 vs 0.5098). So, yes, switch. Unless you draw the joker to begin with...

If Cody picks the joker to begin with, does the host tell him when she goes to remove it from the deck?

Certainly if you had all the odd currencies from 1 cent to 49 cents as well as a 50-cent piece, you could do it. This would require 26 coins. But perhaps there's a better solution.

North-East 3 steps land on 3

North-East 3 steps land on 5

North-West 5 steps land one step north of border

(you do travel through a 'blank' square however)

?blank square?

yeah, but I think that's cheating... need to stay inside the grid (although most puzzles need you to think outside the box, this one does not ). The blank square is the one just northeast of the 7 & 4 at the top. Famous puzzle, famous answer, so I'll refrain from posting. But I finally remembered where I'd seen Prof. Templeton's avatar before...

1 horse @ 50 = 50

39 cattle @ 10 = 390

60 sheep @ 10 = 60

1 + 39 + 60 = 100

50 + 390 + 60 = 500

It's an interesting problem because you have 2 equations, but 3 unknowns. Logic tells you that you have to have an even multiple of 10 sheep, or you won't get a 0 in the ones place of the amount spent. From there, I employed a third method you didn't mention: brute force

Here's what I've never understood about the gossip ring in these type of puzzles:

They gossip about the cheating nature of each individual husband, but never seem to share their thoughts on the total number of cheating husbands. So while they have useful information about what eveyone else's husband is doing, they can't even deduce what their own husband is doing until you add the extra information, which is really a start time. When you add the specific start time, each wife can count how many days it's been since the announcement and therefore they can confirm whether there are 8, 9, or 10 cheating husbands in their midst (wives of cheaters know of 8 and suspect there could be 9, wives of the faithful know of 9 and suspect there could be 10). I guess that these wives are just too sensitive to even that kind of information, although I'm sure there are ways to get it without a direct wife-to-wife confrontation. What they need - whether it's a simple poll, or a visiting missionary's announcement - is to be coordinated in their efforts to expose the unfaithful. To me, the gossip poll is more efficient, but produces a less interesting puzzle.

And here's the other thing I don't get:

What about the double standard here? Presumably the wives know of the cheating husbands because they are the ones the husbands are cheating with! It's because Sally has been cheating with Harry that she can issue the first gossip rumor about Harry - she doesn't have to expose her own guilt, but condemns him. Perhaps the husbands should get together and share what they know. A woman with a shaved head is probably much more ostracized than a man...

Bushindo, as long as the prisoners are together the previous night, why don't they alphabetize all 100 of them, each learn his/her ordinal number, and THEN divide into two groups. As you say, have each prisoner memorize the names AND ORDINALS of the 50 people in his/her group. (I haven't considered how they should proceed on the day of the big test, but it feels like a more valuable bit of information...)

Looks like bushindo was saying the prisoners were limited to 49 bits of information. A name and an ordinal would be one bit each. I agree that it makes sense for them to alphabetize themselves prior to dividing, but I'm not sure how to divide or how beneficial it would be to only know part of the equation. The prisoners could only memorize their own ordinal plus the names and ordinals of 24 others. So, when they got to the jar room, how do they open the jars? Do they go 1-50, but if they come across a name they know that's higher than 50, they open 1-49 plus that jar? Or do they open all the jars matching ordinals they memorized, plus the first 25 others in line? I have no idea how to calculate the odds that names would get put where they belong. Say they know 3 and 6 from memory, and happen to come across 4 & 5 in the initial 50 jars. They could sort 3-6 properly in that case. But if they only came across 4, where would you put it? Presumably you'd want it in 4 or 5 (because it's between 3 & 6), but which? And what do you put in the other jar (4 or 5)? It would have to be something out of order, and that would be a big problem. It almost seems like the extra information is enough to confuse, but not to help. The more I think about it, the more I think you're limited to a sort and search procedure unless you know every name.

Bushindo, that is an improvement, and I agree that HH's solution offers a slightly smaller expectation that 0.95 (I doubt it is as small as 0.9945, but I can't do the math to be sure). The worst case: the first prisoner goes through 50 jars without finding his number, then he is only able to correct 49 jars (he'll wind up putting some garbage number in his own jar). That means it is possible for the second prisoner to have the bad luck of chasing another chain of 50 jars without success. For this to happen, the remaining 50 have to form a chain, with P2's number at the end AND it is the number written on the slip that P1 had to put into his own jar.

The probabilities, as we've seen before, are low: what is the probability of there being two permutation chains of length 50? That's why I think the expected value is still only a tiny bit less than 0.95.

Of course, BellaxPallus offers a marvelous option as well!

edit: once again, HH has typed quicker and thought deeper than I. Once again, I buy his argument. In fact, his probabilities are computed only for the worst case, and are not adjusted by the probability of that case coming to pass (ie, there are two chains of length 50 or one chain of length 100), so I think the actual expectation is a tad higher still.

I think you're right about the expectation being slightly higher. I've made the simplistic assumption that all chain lengths have equal probabilities of occuring (i.e., 1, 10, & 100 jar chains each occur 1/100 times, etc.), but I doubt that's the case. I would expect that chains at the high end of the spectrum (80-100 jars, for example), are relatively rare, and chains at the lower end of the spectrum are the most common.

And, really, you're looking at two coincidental probabilities: first, that such a chain exists at all in the random ordering of jars; second, that the specific number you're looking for is part of that chain. Combining the two gives the probability that the slip you want is in a particular length chain. And, while long chains are rare, the probability of being in a long chain, given the existence of a long chain, is relatively high. Conversely, while short chains are common, the probability of being in a particular short chain is relatively low.

There are way too many combinations of additive factors of 100 for me to compute how probable a 100-jar chain is. I'll leave that to the statistical geniuses out there (you know who you are). But I'll say this: the probability of a 100-jar chain is very close to 0. And the probability of prisoner #2 being in the latter half of a 100-jar chain is 50% of very close to 0. So, the probability of prisoner #2 living is halfway between very close to 1 and 1.

Going back to prisoner #1, he lives unless there's a chain longer than 50 and his number is in it. Like I said above, I'm not going to compute the additive factors for 100, but I did for 10. And if I counted right, chains longer than 5 occurred only in only 12 of 37 possibilities, or about 32% of the time. So, if there were 10 prisoners and they got to look in 5 jars, prisoner #1 is saved 68% of the time right off the bat because there's no chain longer than 5 jars for his name to be in. The other 32% is further reduced when you consider that, even if a chain longer than 5 exists, prisoner #1 might not be in it. I get that, in the case of only 10 prisoners, prisoner #1 has a chance of surviving that is almost 75%. Prisoner #2 has an expected survival rate of 98.6%. Those numbers should improve substantially as the number of prisoners increases.

HoustonHokie's approach is one that I didn't consider. Very nice. However, I think the expected number for his approach will be a tiny wee bit smaller, though. My approach is along the line of CaptainEd's, though with a slight difference.

We can modify's CaptainEd's approach a bit. Let the first guy sort the first 50. Let the second prisoner spend 6 searches on the first sorted 50 to find his name. He then can spend the remaining searches to sort jar 51-94. That way, we can bump his chance of living to about 94%. The 3rd guy will need to do a binary search on jar 1-50, another on 51-94, and a simple search on the last 6, with guaranteed 100% survival. So the expected number of prisoners saved is (.51 + .94 + 98 ) = 99.45

But we can do better...

First, I want to revise the probability that prisoner 1 will live. He gets to open and shuffle 50 jars and he might find his name in those 50. But, even if he doesn't, he still has to tell the warden which jar his name is in at the end of his sorting. He picks a 51st jar at random, and could be right. So his chance is actually 51%.

[i realize that in my first post I was talking about numbers instead of names, but it is easier to deal with numbers and I assume that prisoners with perfect memory would also know everyone else's name and order alphabetically, which they could translate into numbers for purposes of sorting. An interesting twist on this problem would be if the prisoners did not know each others' names, but I won't go there now.]

Second, I wanted to clarify my assertion that prisoners 2-100 will live. The simple beauty of this solution is that prisoner 1 will put 49 slips of paper into their correct jars (the 50th slip of paper will be put into whatever jar is empty at the time, and is likely to be incorrect). The worst case situation for prisoner #2 is this: first, he finds that his name is not in jar #2 (51/100 chance). Second, following the search pattern I described before, he opens 49 more jars and does not find his name in those, which is possible, but not very likely (1/51 chance). Third, he then gets a chance to guess at one of the jars that he did not open and does not find his name there (49/50 chance). So prisoner #2's probability of finding his own name in jar #2 is (49/100) based on what prisoner #1 did. His probability of finding his name in a jar other than #2 is (50/51). And the probabilty of him guessing the right jar if he does not find his own name in all of that is (1/50). So prisoner #2's chances of living are (49/100)*1 + (51/100)*(50/51) + (1/100)*(1/50) = 0.9902.

If prisoner #2 does not find his own name, he will not have opened a jar that prisoner #1 also opened, except that he might open jar #1. This is because each slip of paper that prisoner #2 finds will not be in order, leading him to the next jar. But every name that prisoner #1 found got put into the proper jar, except the last name, which was put in jar #1, and prisoner #2 might find prisoner #1's name. Likewise, every name that prisoner #2 found got put into the proper jar, except the last name, which was put in jar #2. So, if prisoners 1 & 2 are executed, 98 of the names are properly sorted in the process. It is still possible (very unlikely, but possible), that prisoner #3 would come in after 1 & 2 are dead, and find that his name is not in jar #3. But, with 98 names properly sorted, he will be guaranteed to find his own name, as will every other prisoner thereafter.

The analysis begs the question, can there be fewer than 98 names sorted after prisoner #2 is done, because that might lead to a situation where a future prisoner is in danger? Of course, but only if the prisoners find their own names. Say, for example, that prisoner #1 manages to sort names 1-49 into jars 1-49 during his turn. Then prisoner #2 would open jars 2-49, find the proper names there, and only be able to sort 1 or 2 more names. But you can see what's happening here - the early prisoners are living and sorting the later prisoner's names. So the worse the earlier prisoners fare, the better off the later prisoners are. And prisoners 3-100 are guaranteed to live, as noted above.

So, the total number of prisoners expected to live is:

prisoner 1: 51% chance

prisoner 2: 99.02% chance

prisoner 3-100: 100% chance

total = 99.5002 prisoners live (slightly better than I calculated before)

My guess is that you can always save 99 of the prisoners, and the other prisoner has a 50% chance of survival. Here's how it operates:

Each prisoner first opens the jar labeled with his number. If he finds the piece of paper with his number in it, he's spared. If not, he opens the jar labeled with the number of the piece of paper he holds. Again, if he finds his own number, he's spared. If not, he continues in the same fashion until he either finds his own number, or exhausts his 50 attempts. If he finds his own number, he places that slip of paper in the first jar he opened (labeled with his own number) and then opens the next highest jar that he has not yet opened. If jar and paper match, he moves on to the next highest jar. If they don't, he continues as before to move slips of paper to their corresponding jars until he completes the circuit or exausts his 50 attempts.

The first prisoner has a 50% chance of finding his own number. You can't change those odds much because he has no knowledge of the jars and there is no organization to any of them. But when he's done, 49 other prisoners have a 100% chance of finding theirs. If #1 finds the #2 piece of paper, then #2 will find his own number right off the bat. Otherwise, he opens jars and continues the sorting process begun by #1. Because #1 correctly sorted every slip he found, #2 won't open any of the jars #1 opened until he's found his own number, which he will be guaranteed to do in 50 attempts. After he finds his number, #2 uses the same procedure as #1 to sort higher jars.

Thus each prisoner #2 or higher will find his own number and live, and prison #1 has a 50% chance of finding his own number. So, on average, 99.5 prisoners are saved.

Postscript: a quick note about my assumptions. I assumed that when the prisoner chooses the jar at the end of his turn, the warden compares that jar with his number at that point in time to see if it is correct. If, instead, the prisoner gets to choose the jar where he thinks his number will end up, then they should all choose their own numbered jar, because they will all be correctly sorted at the end (and everyone will live).

The second question is very similar to the first, except you substitute 16 for 4 in the equations so it looks like this:

combin(N,2) * 2 = (N - 5) * 16 + 5 * [(N - 1) * 2 - 16]

there are two possible solutions for N here as well: N = 10 or N = 17

If N = 10, then the 5 friends each lost 8 of their 9 competitions and ended up with just 2 points apiece. That doesn't work, because the OP says the 5 friends advance. Instead, they lose 8 of their 16 competitions and wind up with 16 points apiece. Interestingly, all their opponents do the same, and all 17 tie with 16 points apiece. Everyone from that round advances!

The total number of points available is the number of competitions (C ) times 2. C is equal to the number of combinations of 2 opponents [C = combin(N,2) where N is number of contestants] We know that N-5 contestants scored 4 points each and the remaining 5 contestants scored (N-1)*2 - 4 points each. Putting it all together,

combin(N,2) * 2 = (N - 5) * 4 + 5 * [(N - 1) * 2 - 4]

solving for N, we get two possible answers: N = 5 or N = 10. The OP appears to assume the presence of more than 5 contestants, so I say there were 10.

Each contestant participated in 9 competitions. Our five friends won 7 of their 9, claiming a total of 7*2*5 = 70 points. The remaining 5 each won two contests, scoring 5*4 = 20 points among them. 70+20 = 90, which is equal to combin(10,2) * 2

they didn't know it was World War I yet - World War II hadn't happened

there are four liars

I wondered if there was a geometrical way to prove a maximum arrangement. Here are my thoughts on that subject.

The 288 number we got originally has a direct relationship to the radius of the circle and the size of a tile. Namely, 288 = (27 * 12) / 9 * 8. The 27 * 12 term translates 27 feet into inches and the 9 represents the size of an individual tile. The 8 comes from the fact that in every quadrant of the circle there is a horizontal and vertical change equivalent to the radius of the circle, and you need continuous tiles to make that change. 36 tiles are required in each direction, and 2 directions are required for each quadrant, and 36 * 2 * 4 = 288. So that's where 288 comes from. And as long as the tiles are in a tight rectangular grid with no offsets, I think 288 is the most you can do. It also appears to be the least.

Now, if you allow offseting of the tiles, which sparky suggested, you can get a few more. I've managed to get up to 291 tiles in my previous posts, but I thought it might be possible to get even more. The questions are, how to do it, and how many can you get?

Assume that the offset between rows isn't necessarily half a tile, but is set so that the number of tiles impacted in each row is maximized (after all, if I were a vandal, I would want to vandalize as many tiles as possible - not that the vandals have any control over how the tiles are laid out, but just go with me here - perhaps the contractor is in cahoots with the vandals, and just hasn't been caught yet). That would require at least 3 tiles per row because you can always set one side to intersect two tiles and the opposite may only intersect one tile, but it could be as many as 18 for the top or bottom row. You can't guarantee two tiles on both sides, because there shouldn't be any gaps or odd-shaped tiles in the rows.

If that's the pattern, the number of tiles can increase significantly. I did an optimization like that and ended up with 338 tiles. It may be possible to get a few more. I feel pretty certain that you can't go as high as 360 (which would be like adding another quadrant to the 288 solution), but it might be possible to get near 350.

BTW, 338 tiles results in $254.43 per vandal.

Good day all,

I got the same 288 tile answer based on a simple square grid pattern.

Question: What if the tiles were offset by 1/2 their dimension every other row? Would this change the number of tiles touched?

Thanks for pondering!

Yes, actually, it would.

I've got a scenario with 290 tiles now. So the price goes up to $218.08. Good thought.

EDIT: I just found another with 291 tiles. So it goes up to $220.88 per.

Okay, just the line this time

but this is one scenario:

Using the same circle as before, I now get 288 tiles with the grafiti line (I assume the line is infinitely thin, as it is in most geometric problems). So:

Materials cost = 288 / 10 => 29 boxes of tiles @ $11.20/box = $324.80

Labor cost = 288 / 4 => 72 15-minute periods @ $7.5/period = $540.00

Total cost = $864.80

Divided among 4 students = $216.20 per student

but this is one scenario:

I set a grid of tiles up so that the center of one tile was at the center of the circle, which appears to meet the requirements about not touching, etc. I got 4205 tiles that were at least partially covered with grafiti. So:

Materials cost = 4205 / 10 => 421 boxes of tiles @ $11.20/box = $4715.20

Labor cost = 4205 / 4 => 1052 15-minute periods @ $7.5/period = $7890.00

Total cost = $12605.20

Divided among 4 students = $3151.30 per student

N₁(n) = (n - 2)² for all n > 2

They are the interior blocks on the bottom layer.

N₂(n) = N(n) - N₀(n) - N₁(n) - N₃(n) - N₄(n) - N₅(n) for all n > 2

Basically, everything that's left over after doing all the others.

N₃(n) = {[(n - 1) / 2 - 1] + (n - 2)} * 4 for odd n > 1

N₃(n) = [n / 2 - 1 + (n - 2)] * 4 for even n > 2

These are the blocks on the four corners of the interior layers and the outer blocks that are not corners on the bottom layer.

N₄(n) = 4 for all n > 2

They are the blocks in the four corners of the bottom layer.

N₅(n) = 1 for odd n > 1

N₅(n) = 0 for even n

It's the solitary block on top of the pyramid.

if N is the total number of blocks,

N(n) = (1/6)n³ + (1/2)n³ + (1/3)n

works for both odd and even values of n

the number of blocks with 0 painted faces = N(n-2)

Basically, it's the pyramid completely contained inside the outer pyramid.