Math puzzle for those who enjoy them... - Page 3 - Science, Math and Philosophy Forum

Lol bro, your "number theory challenge for fun". Did you run out of space in the margin for all your solutions?

It was malformed with the condition on zeros anyway. "Interesting" is not a mathematical condition. I tried to solve it without zeros, and yeah, it's not solvable other than by brute force.

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07-18-2020 , 12:27 AM

#53

masque de Z

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How is it malformed when i reedited before seeing your line. You post too quickly. People edit their work. I had already neutralized the bs solutions by removing them on re-edit. But even initially i had said the numbers are not all zero hinting of course they are not 1 also since its the same problem. But i went a step further and gave you also a^(2*3*4*5...) or smaller same power logic even. Then i saw your post as it was produced while i was writing my better version which took sometime in order to produce decent examples. Not that i expect good faith people to already do that for you (ie remove the obvious trivial crap as an option) because of course its not the idea in mind clearly to have a trivial worthless answer. Of course not. That happens only when people are not hostile dickheads to others and spend their time instead on building something better out of the ideas of others for all to win instead of ridiculing or rejecting them.

How do you know brute force is the only way by the way? Did you prove that?

But lets return to current more solvable project at present.

For a solution to your problem to be broadly worthy one must come to it naturally and not as an application to some combinatorics or abstract algebra group theory class only.

So here it is then that way (now that i got some more free time after tax week) that way, my way, the way a true researcher that is unafraid of the darkness finds the solution even if they do not know the math theorems behind it yet...maybe even for that reason developing them later.

Spoiler:

The first guy can only be helped by the actions of the captain but cannot know anything besides that or what the captain will try to do for them. We have yet to realize what that assist can be and may not waste it until we are certain it kills everything that leads to death for all if that is possible or maximizes their chances if not.

The captain can put the number of the first guy in the right box or do something else that will secure he get to that number before all 50 moves even if it is exactly on the 50th. If someone other than the first guy needs help because there is a way for the first guy to get to its number before 51 moves then the help will go there etc.

So given what i stated above in earlier posts regarding partial derangements its possible there are a number of fixed points in the permutation that goes like e^-1/k! in probability approximately. We cannot ignore their help.

We need a way to choose what we select next after a choice is made that proves not successful, without being able to have anything known about the sequences to us other than that the captain might have found a way to make our worse nightmare not possible to happen or to happen with minimum probability.

If we are #1 and open a box eg 56 and see 67 there, then we know 67 doesnt have 67 in it and it can therefore have 1 in it. The same cannot be said about any other number box because any other box can have still some chance to be a fixed point with not 1 in it.

So it becomes best option then to check 67 after 56 because we remove the fixed point wasted trials. Then at 67 if 1 is not there we must check the number found there for exactly the same reason because the contents of that box are not that number and can be our number still with higher chance than any of the other boxes (conditional probability argument).

Speaking of which what if 1 is already on 1? Then the above logic will never take us to our box. To remove that possibility we must start from our box first.

So we start at 1 and see say 59. Then we go to 59 and see 6. Then we go to 6 and see 9. Then we go to 9 and see 17. Then we go to 17 and see 1? Will it be like that? Yes. Always like that but different number of steps. We will have 1 in our horizon that way because we will never get that way to a fixed point. Why do we get to our number for sure? Because this process never finishes unless it takes us to it. It always takes us to a different number. Because every time we open something and it is not our number it cannot be the end of the process because it will keep changing into something not seen before until we hit our number if we started from its original box eg 1. Basically this process has an end that takes it back to the first box (the box that has 1 in it takes you back to look to box 1 right?). We will be able to get back to our number eventually but the question is with what chance this is in less than 51 moves?

So we must start from our box to see if we got lucky. If we get lucky great. The others must do the same starting from their boxes. If its not there then follow the sequence emerging.

A permutation therefore has what one recognizes as a number of cycles by looking in my link earlier.

The cycles must be disjoint because they start from a number and end to the same number or they would literally take all numbers if 1 went to 2, 2 to 3, 3 to 4 etc. The cycles since disjoint can have at most a period of 100. They are like islands. If we have a period that is longer than 50 then that is the only period longer than 50. So all people would be able starting from their number to either get it right away or after a while that is less than 50 moves.

So the captain can intervene and break that period over 50 (if any) to a smaller one and provided both pieces resulting are less than 50 we now have a solvable situation.

How do you break that? Go at the start of the biggest over 50 cycle eg its smallest number and count how many moves it takes to get back to the same number. Then before its its 50 or exactly there put there that number in that box and the contents of that box to the other box creating 2 cycles that are shorter than 51.

eg 1 5 79 11 23 89 ... 45 67 ...98 1 . Go at 45 and put 1 there (if position of original 1 is very far over 50 steps away).

Thanks for an interesting problem.

Last edited by masque de Z; 07-18-2020 at 12:32 AM.

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07-18-2020 , 01:09 AM

#54

masque de Z

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Quote:

Originally Posted by d2_e4

Interesting is not a mathematical condition? What does that mean? Interesting is whatever still remains not answered that has the possibility for something funny happening like actually a limit existing in that streak that comes to an end or more things to learn about how things are connected.

https://en.wikipedia.org/wiki/Waring%27s_problem

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07-18-2020 , 01:11 AM

#55

PairTheBoard

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Quote:

Originally Posted by masque de Z

Spoiler:

PairTheBoard

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07-18-2020 , 09:57 AM

#56

d2_e4

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Quote:

Originally Posted by masque de Z

I'm in a bit of a rush so I'll come back to the rest of your post later, but just wanted to respond to this quickly. I mean that you said "some" zeroes are allowed, as long it makes the solution "interesting". In one of the examples you gave, you had nonzero terms except one. Obviously, I understand intuitively what you meant, but it's not a rigorous condition. For example, what if I came up with a number that is the sum of 2 squares, 2 cubes, 2 4th powers, 2 5th powers, 2 6th powers etc. and pad the rest with zeroes? Is that sufficiently "interesting"?

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07-19-2020 , 02:11 AM

#57

masque de Z

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Quote:

Originally Posted by d2_e4

The answer to this of course is that which only a wise thinker chooses ie to research the damn this thoroughly. After the trivial a^(2*3*4*5*...) single nonzero element each time solution all is possibly interesting. So if you do other powers with multiple elements and one was with only a few or one i dont mind because some number will be 7th powers of course. That opens also the door to asking can a 7th power be described by say 7 7th power terms or less that are not the same as original(ie not the stupid thing) (Fermat generalization may be known even) . Take that to mean not 1 7th but 2 and higher.

But if you got 2 2 3 3 4 4 5 5 6 6 7 7 etc is a separate problem requiring all to be nonzero. You can choose that too. I want to make it easier to have some room for some zeroes. That makes it less strict.

When someone proposes something the correct thing is to try to generalize, improve it reframe it etc. Thats how good will people with genuine curiosity behave. This is not a pissing contest. It is an opportunity to learn.

Here is a conjecture i have since teenager.

Every number with the exception of a small set of less than 25 members (i think if recall correct search have to look programs or say no larger than 10000 i think and just reran them now to find the first exceptions are 15, 23, 55, 62, 71, 471, 510, 646, 806, 839, 879, 939, 1023, 1063, 1287, 2127, 5135, 6811, 7499, 9191, 26471) can be written as a^2+b^3+c^4+d^5 with a b c d non negative integers. I searched random very big ones and it worked every time and i think its clean above a big enough number. We know of course why this matters because we have Lagrange's theorem that says every number can be written as sum of 4 squares and the Waring problem relates to it. Going 2345 is far more impressive in difficulty and interesting if it has only a finite number of exceptions. Rerunning after many years programs with faster cpus now to check better as we speak so that the above is more accurate lol.

Last edited by masque de Z; 07-19-2020 at 02:38 AM.

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07-20-2020 , 12:36 AM

#58

masque de Z

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Updating exceptions list for 2345 conjecture {15, 23, 55, 62, 71, 471, 510, 646, 806, 839, 879, 939, 1023, 1063, 1287, 2127, 5135, 6811, 7499, 9191, 26471} They get progressively harder to find and going very high randomly always worked on any trial of 8-9 digit numbers.

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07-20-2020 , 06:20 AM

#59

masque de Z

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My favorite small prime number representation

9876543211=611^2+737^3+312^4+1^5 or 675^2+977^3+269^4+82^5 among the probably 200 or more other ones for the same number. With so many backups its hard to miss isnt it?

Exception list above remains the same up to 200000.

Some degeneracy heuristic argument probably can be made for a finite list.

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07-24-2020 , 05:19 AM

#60

d2_e4

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You're like the Ramanujan of SMP, bro.

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07-24-2020 , 05:50 AM

#61

d2_e4

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I've been working on a pet project myself lately. I posit that for all n>2, the following relationship holds:

ABS(log(lcm(1,2,…,n))−n) < √n * ln(n)^2

Where lcm is the least common multiple. I've verified it experimentally with n up to 1,000,000. I'm struggling to prove it definitively, though. Any help would be appreciated.

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07-24-2020 , 09:55 AM

#62

PairTheBoard

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Quote:

Originally Posted by d2_e4

Are there numbers up to 1,000,000 that are relatively close? Does there appear to be some kind of intermittent asymptoting going on? How big of a fudge factor would work to improve the rhs for numbers up to 1 million?

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07-24-2020 , 10:25 AM

#63

d2_e4

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I'm trolling. That's one of the equivalent reformulations of the Riemann Hypothesis. Didn't think anyone was going to seriously attempt it lol.

Feel free to prove it of course. It'll make you a millionaire

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07-24-2020 , 10:45 AM

#64

d2_e4

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But just looking at it, lcm(1...n) will increase every time n has a prime power in its factorisation that hasn't been seen previously, so for example for n 1-20, lcm(1...n) will increase at 1,2,3,4,5,7,9,11,13,15,16,17,18,19 I think. If these numbers were "too dense", e.g. we had a bunch of twin primes in a row, then the inequality would presumably fail.

Last edited by d2_e4; 07-24-2020 at 10:50 AM.

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07-24-2020 , 12:17 PM

#65

d2_e4

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18 doesn't belong there^^

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07-25-2020 , 03:35 PM

#66

PairTheBoard

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Originally Posted by d2_e4

18 doesn't belong there^^

I happen to know 18 and she goes pretty much wherever she likes.

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07-25-2020 , 11:18 PM

#67

masque de Z

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https://en.wikipedia.org/wiki/Chebyshev_function

https://en.wikipedia.org/wiki/Von_Mangoldt_function

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07-26-2020 , 09:27 AM

#68

d2_e4

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Quote:

Originally Posted by masque de Z

https://en.wikipedia.org/wiki/Chebyshev_function

https://en.wikipedia.org/wiki/Von_Mangoldt_function

Very good. I was hoping you'd have a go at it

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07-26-2020 , 09:53 AM

#69

d2_e4

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Quote:

Originally Posted by masque de Z

Spoiler:

This is correct, although a roundabout way of getting there. I prefer the following wording of the solution:

The naive solution has us losing almost all (1-2^-100) of the time, because we are operating with insufficient information. The key to solving the problem relies on the concept that we can "create" information, seemingly out of nowhere, by turning the lockers into a directed cyclic graph with no "tails" as PTB put it. A linked list in programming terms, one that is guaranteed to cycle.

Your solution is correct though. Ok, now figure out the probability we win without the captain, as number of prisoners --> infinity and they each open n/2 lockers.

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07-26-2020 , 11:04 AM

#70

lastcardcharlie

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Quote:

Originally Posted by d2_e4

The key to solving the problem relies on the concept that we can "create" information, seemingly out of nowhere, by turning the lockers into a directed cyclic graph with no "tails" as PTB put it.

That seems to be a strongly connected graph, but each vertex must also be contained in a unique cycle. "Cyclic graph" is ambiguous. /nit

https://en.wikipedia.org/wiki/Cyclic_graph

https://en.wikipedia.org/wiki/Strong...cted_component

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07-26-2020 , 11:12 AM

#71

d2_e4

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Quote:

Originally Posted by lastcardcharlie

I didn't know the term, thank you. I knew the term "directed acyclic graph", so I just took out the "a" on a wing and a prayer. I was hoping someone would correct me if I got it wrong

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08-23-2020 , 03:50 AM

#72

d2_e4

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Masque,

Can you solve the sum of the reciprocals for the squares?

1/2++1/4+1/9+1/16... = what?

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08-23-2020 , 05:31 AM

#73

masque de Z

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What is Integrate[x/(Exp[x] - 1), {x, 0, Infinity}]

Can you solve the sum of the 1/n^2k?

What about proving Stefan Boltzmann law (same thing) or playing around with Fourier series?

But lets try sum[ 1/prime^2 ] or derive Sum 1/n^2 in a way not seen on the internet anywhere?

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08-23-2020 , 09:47 AM

#74

d2_e4

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Quote:

Originally Posted by masque de Z

1. -((x+1)/e^x) + C
2. Not sure what you mean by solve? It's the zeta function evaluated at the even numbers. It's always some rational multiple of pi^2k, given by the Bernoulli numbers.
3.No idea about that one
4. Sum 1/prime diverges I think. 1/prime^2 is an interesting one, will have to think about that.
5. Integrate ln(2cosx) from 0 to pi/2. Blackpenredpen did a video on it.