I think I see a pattern after a few days, just not sure how to give an answer yet.

edit: removed spoiler for bonus, because Glycereine beat me to it.

edit:spelled Glycereine wrong....twice

Actually I spelled Glycerine wrong. But it was intentional .

I'm struggling with where to go on this one too after a couple days I have some reasonable ideas but not to go all the way to 100 yet...

God speaking to Lucifer?

No chance. 6 and 9 were eliminated

Edit:

I misunderstood. Thought no 6's or 9's would be used. You meant the number as a whole being flippable. I need more time now and to do the first part :-p.

Birds....

cats

Fighter Jet

Ya, I knew it was pretty easy. Okay, not one of my best ones, back to the drawing board for me...

Actually it was pretty good, I just would have changed the word apple in the riddle to something else. Small change really but would have made it quite a bit harder.

A Kiln?

an apple tree...but you said apple in the riddle, so probably not.

I was thinking Apple as well, but it's in the riddle.

Ante/Auntie

Oh of course!

A propulsion system on a train burning fossil fuels but yearning for a clean energy source that won't leave it covered in debris.

I'll only be disappointed because I didn't solve it .

Quick question, is it scarred or scared?

A Beehive.

A Pile of Leaves?

I have a couple things that fit most of it but none that I feel like fit it all .

A Pinata?

Starting with ROSE we know that the square must be 4 digits, so the squared number must be between 32 and 99 (31 is 3 digits, 100 is 5).

Since OR must be prime, R cannot be an even number.

This elimates the squares of: 45-54, 64-70, 78-83 and 90-94.

With the remaining squares the following have numbers that repeat themselves: 34,35,38,40,56,58,60,62,63,75,76,77,85,88 and 97.

Further of the remaining squares only the following have the first two digits reversed (OR) as prime.

32 - 1024

33 - 1089

37 - 1369

41 - 1681

42 - 1764

55 - 3025

57 - 3249

59 - 3481

61 - 3721

71 - 5041

84 - 7056

86 - 7396

89 - 7921

96 - 9216

99 - 9801

Now having reduced the possibilities of ROSE to 15, we move on to INNES, noting that the last digit or ROSE must be the 2nd to last digit in INNES, and 0,2,3,7 and 8 are not possibilities.

INNES must be the square of a number between 100 and 316 in order to be 5 digits.

Starting by eliminating any squares that do not have the 2nd and 3rd digits the same or have other digits that are the same, we are left with only:

197 - 38809

231 - 53361

245 - 60025

We note here that in the E place 1 number has a 0 and another a 2, neither of which is allowed based on previously established narrowing of ROSE. Thus INNES must be the square of 231, or 53361.

Knowing now that:

E = 6

I = 5

N = 3

S = 1

We can see that the only solution of the 15 remaining for ROSE that fits with the last two digits 16 is the square of 96, or 9216.

Thus R = 9 and O = 2.

So ROSE-INNES is

9216-53361, the squares of 96 and 231

I haven't had a chance to dig into PIGEOLET much and it hurts my brain anyways...

I should have said "non-negative" integer since 0 is included.

Sinners

1. Pride

2. Gluttony

3. Wrath

4. Lust

5. Greed

6. Sloth

7. Envy

Nice result.

There is a known proof for 2+2ⁿ that would verify your first answer.

I don't know the proof for 1+2ⁿ.

I imagine working out the answer for say n=5 would suggest the proof, and the odd answer would follow.

Just a hunch.

For the reasoning of both why the values of n that work, do so as well as which player in those values are the winner:

With some quick deduction you can see that no matter how many marbles two players have, 1 of those players will always take 1 marble and give away 2. This makes it so that if any value of n reduces to 2 players remaining, the result will mean there is a winner.

Alternatively if any value of n results in an odd numbers of players remaining (after any complete round) then the solution is no longer solvable (regardless of the number of marbles). This is so because from then on the number of marbles that each player has will vary only by 1, alternating between taking 1 and giving away 2, and taking 2 and giving away 1. Stalemate.

So for any number of working solutions, n must not reduce to an odd number at any time. For this to be the case, the number (after the first round) must always be a multiple of 2. The only way for us to ensure that the number be a multiple of 2 after each round is for us to start backwards from 2 and go up multiplying by 2. This can go on indefinitely, but remember that in the very first round of the game player number 1 will always be eliminated.

Other than player 1, the only players that are eliminated in the first round are the even players.

So if there are an even number of players, then half the players + 1 (number 1 is the plus 1) are gone after round one and the result ((n-2)/2) must be 2^x.

And for an odd number of players the results after round one are a reduction to (n-1)/2 players.

This is why there are two different formulas for values of n.

The two different formulas however 2ⁿ +1 and 2ⁿ +2 have two different winners. This is because adding the extra even number at the end of a working odd "n" still results in a working solution, but moves the winner one player to the right. This is because at the beginning of round 2 instead of starting by giving away 2 marbles, player 3 will start by giving away 1.

In all odd solutions the player that never gives away 2 marbles will be the nth player.

In the even solution, the winner is moved 1 player to the right (discounting even players because they cannot win) to player 3.

so, for a positive integer "x" the following is true:

For 2 players it is obvious player 2 wins.

n = 2^x +1 is a solution, who's winner is n

n = 2^x +2 is a solution, who's winner is 3

That may or may not have helped anyone confused...

I have two questions..

1. What do you mean by same value?

2. Is it even possible for them to be the same?

Same value meaning that for this given length of a side, The numeric value of Surface Area and Volume are the same. Units of course won't be.

You can take the two equations, for the surface area of a regular dodecahedron and the volume of that dodecahedron to be equal to each other and solve for the common variable in both equations, the length of a side.

Used calc and it gave me this: 2.6941667104173468292341373231142

(formulas from wiki, i did this calculation myself)

This is accurate from my calculations also but no real derivations on my own, same method as this poster.

napkin not good for keeping good decimal values.

I'm too tired for this

One thing that drove me nuts is that I'm solving for a unit-less side length.... lol.

I'll give someone else a chance at the winners . I have a theory though...

The solution is not too easy. any body els.

Can you help me clarify the clues?

I understand it's a 5 digit number, where the digits are labeled in order from left to right: I, II, III, IV, V

Then II + III = 10

III + V = 14

II is the 4th highest of the 5 digits

I is smaller than 2 times it's digit + II

and I + II + III + IV + V = 30

?

Also I'm assuming I can use each digit only once. Did I misunderstand something?

Thanks!! I encourage others to do it though without looking, it doesn't take too long but you can make it harder than it should be (I almost did)