They're all liars. That's the only way they could all disagree about P, but agree about Q. So, P has 3, Q has 1, and R has 2.

I've never seen why

Everyone has to try

So hard to deny.

For many with a brain

Heard creation's refrain:

The man-God came!

I fear it may be too late

For us to debate.

Are you just going to wait?

He gave us His all

And is now standing tall.

Maybe you hear His call.

But if you decide

To just run and hide

I won't deride.

Why don't you think twice

And to newbies be nice?

Here's my advice:

Had this riddle instead

About another god read

Would you have said

"Away with him now!"

Or would you solve it somehow

And then take a bow?

Why do you choose

Such limited views

And then come to bruise?

The Den isn't the place

For such a disgrace

Or for hate to embrace.

Please open your mind.

I think you will find

'Tis better to be kind.

It has to be the apostle Paul.

His name used to be Saul and was changed following his visit to Damascus.

He was imprisoned several times in his ministry.

There is a famous shipwreck story on the way to Rome.

His jailer was converted in Phillippi.

He was bitten by a viper in Malta (following the shipwreck) with no ill effects.

Welcome to the Den, DPT!

First post....hopefully I'm thinking about this right.

If you letter the quarters from the top row across each row you get...

abc

de fg

h i j k l

mnop

qrs

There are 15 possible equations such as a+b+c=38,

a+d+h=38...etc.

If you put those equations together as much as possible like b+c=d+h from two equations listed you end up with 19 equations. 19 equations, 19 unknowns, algebra should be enough to solve it. I'm feeling a bit lazy so I'm not solving them right now, but I'll list them all if someone else is inclined to solve them.

1. b+c=d+h

2. a+b=g+l

3. a+d=m+q

4. c+g=s+p

5. h+m=r+s

6. q+r=l+p

7. d+f+g=m+i+b

8. d+e+f=k+o+r

9. e+f+g=i+n+r

10. n+o+p=i+e+b

11. m+n+p=g+k+r

12. m+n+o=b+f+k

13. m+o+p=d+i+r

14. m+i+e=f+k+p

15. m+e+b=d+n+r

16. g+o+r=b+f+p

17. g+k+o=d+i+n

18. h+i+k+l=a+e+o+s

19. h+i+k+l=c+f+n+q

Good thought, but...

By having 19 equations with 19 unknowns, you assume that a unique solution exists. I doubt that one unique solution exists - there are likely many solutions (if there are any at all), similar to a magic square. And, because your equations have no way to constrain the sums to be 38 or for the individual variables to be integers from 1 to 19, that opens the door to even more solutions. Finally, I think that if you carry out the substitution process that one would normally do to solve these types of equations, you'd find that several are redundant and you'd end up with something like a = a or 0 = 0 at the end. So, I don't think algebra helps much. At least not here.

television

Four of a kind:

There are 13 combinations (A-K) and each can be obtained with 48 different kickers, so the number of Four of a Kinds is 13 * 48 = 624.

Straight Flush:

There are 10 Straight Flushes in each suit (A high to 5 high), so the number of Straight Flushes is 10 * 4 = 40.

Therefore the Straight Flush beats a Four of a Kind, because it is more rare.

nevermind, didnt think this one quite through

You're in the ballpark - just keep in mind that the hour hand moves, too...

It happens every 32:43 minutes starting at 12:16:22. So, exact times are:

12:16:22

12:49:05

1:21:49

1:54:33

2:27:16

3:00:00

3:32:44

4:05:27

4:38:11

5:10:55

5:43:38

6:16:22

6:49:05

7:21:49

7:54:33

8:27:16

9:00:00

9:32:44

10:05:27

10:38:11

11:10:55

11:43:38

Well i'm not sure that I have arrived at an answer but here my initial thoughts

The person picking second does have the advantage statistically speaking. The second chooser should always chose the same as the first player except switching the last flip. So if Davey chose HTH for example, Alex would then want to chose HTT. This is because Davey has a 1 in 8 chance of being right for any 3 flips. While Alex is really only guessing on the outcome of the 1 coin (the last and different one) and so has a 1 in 2 chance.

So on the surface If I am Davey I would want 4 to 1 odds (or better) from Alex (note I haven't really thought through this last part, but i'm guessing it is correct).

My first thoughts were the similar, but then I thought about it again, and I don't know.

Comparing Davey's 1/8 chance on any 3 flips with Alex's 1/2 on the last one is comparing apples and oranges, I think. They both have the same probability of the first 2 flips matching their chosen combinations, and then they both have a 1/2 chance on the last coin. I'm beginning to think the odds are even, no matter who chooses first.

they first meet after 1.36 seconds, and then they meet every 2.72 seconds after that. So, at the end of 60 seconds, they will have met 22 times and end up in the same positions they started in.

I think I could stack them in layers of 4, 1, & 4 octahedrons, where the apexes of the top layer touch the apexes of the bottom layer. The single octahedron in the middle layer could fit in the gap in between the other 8. It doesn't come out exactly cubic, so I'm guessing there's some more optimization to be done, but I get a side length of 1 / (2 * sqrt(2)) = 0.3536. My overall volume used only comes out to 9/48, so I imagine there's a more compact arrangement possible.

menial

penpal

bengal

melted

they'll be realigned in the exact same position at 700 days, which is the least common multiple of 100 and 350. In that amount of time, Mercury will have circled the sun 7 times and Mars twice, which means that Mercury will have "lapped" Mars 5 times. So, the first time that they align in any position is 700 / 5 = 140 days.

Not proven, that is too pat an answer.

The assumption is that a simple cubic packing is the best way to get 64 spheres into the box, but it hasn't been shown mathmatically (sic).

http://www.tiem.utk.edu/~gross/bioed/webmo...herepacking.htm

It is clear that for very small particles they will assume a hexagonal packing, and will occupy about 74% of the space. The post about porosity is interesting, but natural materials would naturally tend towards the most efficient packing - hexagonal.

So the question is, given that we are constrained by the number of spheres (they are relatively large compared to the box) does the space wasted by offsetting the cubes outweigh the increase in packing density?

HH had an interesting approach, but I think the calculation for the first layer is d = a/3.5 rather than a/4. Even if you show that you can't quite fit those spheres, you can always shrink them down and still be bigger than a/4.

For example, what if you did 6 rows of packing that way, and then jumbled the remaining 10 spheres in on top?

I will say that box 1 and 2 are the lightest, that box 3 likely weighs the same, but that box 4 likely weights more. Unless the density of the cubes is less than that of the surrounding meduim (which wasn't specified!).

how tall does 6 layers of spheres get? I would say that the 6 layers are at least 7*radius tall. If d = a/3.5, then r = a/7, and at 6 layers, the spheres have already reached the top of the box. You won't be able to stuff any more in. In fact, if d = a/4, then after 6 layers, the high of the packed spheres would be 7/8 a, and you still have more than a full layer of 9 spheres to dump in the top of the box. Hexagonal packing is efficient in an infinite sense (or if you happen to have a hexagonal or spherical container). In fact, if the OP had specified 63 spheres instead of 64, the answer might be completely different, because then you could use an offset pattern that is 7 layers tall. But then again, it might not. The problem is the large outside gaps that bound two sides of each layer, which greatly reduce the efficiency of the packing.

euchre

I've been wrong before, and I guess I'll be wrong again. I actually get a ratio of about 0.80 with dimes and quarters if I place them in a rectangular grid. Guess I should have drawn it out first

A similar arrangement of pennies and dimes yields 0.786 (small improvement over just pennies or just dimes).

The first two are the same: 0.785, assuming a "unit" table is square. If it were instead a regular hexagon, the ratio could increase to 0.906, but that sounds too much like another brain teaser I looked at recently... The 0.785 comes from the ratio of the area of a circle with diameter d to a square with sides d, which is (pi*d²/4) / d².

Actually, the answer is the same no matter what you're using. The problem is that you haven't defined the ratio of quarters to dimes (for example). So, to maximize the ratio, I would use quarters to cover the majority of the table and place one dime in place of a quarter somewhere. My ratio would be very close to 0.785. Now, if I had to use the same number of each coin, that would be a different story. But I don't see that requirement, yet

Incidentally, my answer would change if a dime had a diameter smaller than 9.9 mm, because then I could fit it in the spaces between quarters.

The first two are the same: 0.785, assuming a "unit" table is square. If it were instead a regular hexagon, the ratio could increase to 0.906, but that sounds too much like another brain teaser I looked at recently... The 0.785 comes from the ratio of the area of a circle with diameter d to a square with sides d, which is (pi*d²/4) / d².

while it would not surprise me if the answer came out as all are equal, I wonder is it possible to have a closer packing than a simple 4x4x4 grid. If that is the case then we might be able to fit slightly larger spheres than diameter a/4 in the box.

Maybe, maybe not.

I think that mpapineau has the elegant answer for a cubical arrangement of spheres. Because one sphere in a cube requires a certain proportion of its volume (approx 50%), 8 spheres in 8 cubes (or 27 s in 27 c, etc) require the same proportion of the overall cube containing them. The only way to change this is to change the spacing, as armcie has suggested. To complete the proof, we need to show that a cubical arrangement is the best.

If, for the 64-sphere arrangement, your spheres have diameter > a/4, the best arrangement you can get for the bottom layer is 3 x 3. The next layer would also be 3 x 3, offset from the bottom layer so the centers of the spheres lined up above the centers of the gaps in the bottom row. So, in order to get the requisite 64 spheres, you'd have to have roundup(64/9) = 8 layers of spheres (7 layers of 9 + 1 layer of 1). For a cube with side a, you can make the following assertions about the spacing:

a = 3.5d + 2.5g_h

a = 4.5d + 3.5g_v

(2d)² = (d + g_h)² + (d + g_v)²

where:

d = sphere diameter

g_h = horizontal gap between spheres in the same layer

g_v = vertical gap between spheres in layers n and n + 2

The diameter of the sphere comes out to be .197a, which is smaller than a/4. Because the spheres are smaller then they could be with the cubical arrangement, they don't satisfy the requirements of the OP.

you'd like to say that no composers are theives, but that's not valid with the information we've got. Politicians are a subset of the overall group of theives, and, even though there is no overlap between composers and politicians, it is entirely possible that composers are a subset of theives as well, either in whole or in part. Or, they might not be theives at all. I guess the most we can say is that some theives are politicians.

Well, I'm sure that the "steepest slope" method I proposed in my last post will eventually solve the puzzle. But it will not be efficient about it. After 72 moves, I'm giving up. The problem seems to be when the blank space is in the corner - you can't make any choice, and I kept getting into situations where the blank space was in a bottom corner and it would pull a low number down, forcing the inversion sum to rise. I got a sum as low as 8 by my 42nd move, but by my 72nd, I'm back up to 14, and I've been as high as 16. I've thought about removing the e term and seeing if that makes any difference (it might not put the extra weight on keeping the blank low on the board, which seems to be problematic). But I won't try again until later.

I got a list of 924 magic squares and another list of the 324 pan-magic squares from a couple web sites, popped them into Excel, and had it calculate the inversion sum for all of them (by the way, I decided e works better if the bottom row is 0 and top is 3 - that way, a complete square is 0 instead of 4). The lowest even number I got was 46 (which I guess would be the optimal starting point??), but it only showed up on the pan-magic ones, so I'm think the list of 924 is somehow incomplete, and it's hard to search for a particular square to know how incomplete it is.

My thought on efficient solving is, given a particular starting point and inversion sum, choose your moves so that the inversion sum trends lower. In other words: every move has at least two possibilities and up to four (two to four squares that you can slide into the blank spot). Compare the inversion sum for all possibilities, and choose the lowest. You ought to be able to keep the sum equal or lower almost all the time, except possibly in the corners, where you don't really have a choice. I don't know what to do in case of a tie, except look at the next generation possibilities. I'll try that with one of the N = 46 squares and see what I get.

I wasn't too sure if that square should be included. Wolfram seemed to imply that it wasn't. Can we figure out how many "magic squares" are possible? Surely some would be easier to get to from the original set-up then others.

I've seen different numbers for that. There are 324 pan-magic squares - of that I'm certain, but they are a subset of the total number of magic squares. For the total number, I've seen 924 and 4080, among other estimates.

I know. It was said "tongue in cheek".

Here's the numbers I get

13,2,6,3,5,2,0,1,2,2,2,2,2,0,0

Which equals 42. How did you come up with 53?

It is solvable - I miscalculated. Although I would have said 46 by adding the row of the blank space.

14 03 08 05

09 04 15 02

07 10 01 12

00 13 06 11

Hmmm, they all equal 30 so far.

A property of magic squares is that the sums will always be the same - all the sums will equal 30 if the range is 0-15. The permutation inversion sum for that board is 53, so that board isn't solvable anyway.