Prime

You could win a fortune trading options on that stock with some well timed transactions.

Here are few corollaries and further conclusions on the subject.

True, it does depend on common as well as uncommon factors. Here is how.

So that just leaves question 2... That is interesting! The ubiquitous e. I suppose, whenever you get a series with a mixture of factorials and exponents, you can always look how to convert the series into f(ex). A correction to my post with the birthday (question 2) equation.

I can calculate this probability as a series, but I haven't found a polynomial solution yet. But I suppose, EH is looking for a polynomial solution. Is that so? Or is the problem to calculate probability for two specific cards? But that would be too simple, not EH-like problem.

Here the first statement by S "I know, you don't" gives me problems. (I assume, by that he meant that he knew that P couldn't tell the numbers without him saying so.) Otherwise the answer could be:

Sorry, my mistake. I didn't pay due attention to the statement of the problem. I just saw the names P and S, and assumed this problem was the same. (The other puzzle had two two-digit numbers.) So scratch my previous posts. However, I don't see, why product couldn't be a 4th power of a prime. Say, 81 could be 9*9, or it could be 3*27.

As for this variation the answer could be: To find whether this is the only possible solution, you'd still save a lot of time by writing a computer program. I suspect, there are other solutions. As far as puzzles go, this variation is more fair than the one posted before. Because solving the other one without a computer program was not a feasible task. I would change the question of the puzzle "What could be the numbers?"

A more difficult variation of same puzzle had been posted awhile ago.

That is the formula for a key position! That simple operation has a name in Boolean Algebra and Computer Science. And so T-Roe has found the magic formula. The derivation and proof are yet to be found. It could be a serious math/logic problem. I wouldn't spend all my free time on it. If someone finds it, I'll appreciate if they publish it here. If I stumble upon the proof again, I'll publish it then.

That's the one! A true Alternator. One side note. Here we redefine the print command to print without skipping to a new line.

Ah, but what is "self-aware"? What is "desire"? I say those are the terms to describe certain tendencies, the mechanism of which we cannot explain. So, we have a new problem! Build an Alternator. That is program A, which outputs a different program B, which, in turn, begets program A. I can see a solution.

See my previous post.

Is that right Prime? That's a very strong hint. Indeed, the formula has to do with exactly that. Let me change the rules of the game a little. Instead of leaving the last coin for your opponent, you must take the last coin to win. Interestingly, the key positions do not change. E.g., for the new rules {1,2,3} position is still lost for the player who has to move from it. Only those key positions are reversed, where there is not more than one row with more than one coin remaining. For example, {1,1,1} becomes a position from which you can force a win if it is your turn to move. The reason for the rule change is to keep the formula consistent. With the old rules (win if your opponent takes the last coin), the formula for the key position works only while there are more than one row with more than one coin left. After that you just have to leave odd number of rows with one coin each to your opponent. With the new rules (one who takes the last coin -- wins), the formula for the key position works all the way to the end. I suppose, this could be a hint too. As I said earlier, I remember the formula, but forgot the derivation. So don't count on my help too much. I grew too lazy (perhaps, incapable) to duplicate the derivation/proof I did twenty some years ago. I would like someone else to do it and publish the proof here.

Did I miss any? That's all of them! Part 1 is now officially solved! Also, partial credit to T-Roe and to bonanova. Part 2 is the most fun. And it is difficult.

I see, CR's concern was not unfounded... The alternator will go into oblivion after 2 runs, same as the reducer. I did not get the "attack". What is it? Whreas "new" seems to be a command relying on some Operating System services for creating another copy of self. That clearly goes outside the scope of this puzzle. In general, as EH has already pointed out, a good programming language has many possibilities for creating quines. This puzzle relied on limiting programming language capabilities.

Actually, I messed up my difference table a little. Still the same answer is possible.

It would be easier, if there were less contestants. Nice puzzle, though.

Reusing what Bonanova has already given away,

So I see, you have figured out some of the key positions. Namely... I suppose, you can see all positions below that one. So, if you figure out all key positions above that one -- the Part 1 of the puzzle is solved.

can someone confirm this? The first part of the puzzle is to figure out the {3,5,7} position (the winning path of moves). Then we can move on to figuring out the formula. Your second point seems to imply that you can take coins from more than one row on a single move. That would be against the rules. Exactly! A valid move is taking any greater than zero number of coins from one and only one row. To take coins from another row, a player must wait for the next turn.

Just a thought...

I am game. My first move is taking 1 coin from the row of 5, leaving you with 3, 4, and 7. Your move.

I have highlighted the relevant parts of the OP (Original Post). Clearly, it implies that you don't know whether counterfeit coin lighter or heavier. Here is a detailed explanation of how your method could fail: 1. Weigh 9 against 9 and one side is lighter. You took the lighter group of 9 and 2. Weigh 3 vs. 3 with 3 on the side. They weigh the same. So you take 3 on the side and 3. Weigh 1 vs. 1 with 1 on the side. They weigh the same, so you assume (wrongly) that the coin on the side is the counterfeit. Whereas, it was the one the 9 in the heavy group, which you have abandoned. The answer has been posted throughout this topic many times over. And I have already noted that you can figure out counterfeit out of 39 coins with 4 weighings. But you are not guaranteed to get it out of 27 in less than 4. Three weighings is enough for 12 coins, where you don't know whether counterfeit is heavier, or lighter.

Good proof, but it lacks the final touch. (Same as several other proofs here.) a=3 Then a is a prime and it is divisible by 3. This case must be tested separately.