I’ll post one.
You have a 1 km track.
You are going to run 2 laps
You can run the first lap as fast or as slow as you want, call the speed on that lap Vone
You must run the second lap so that your average speed for the full 2km is twice as fast Vone.
i.e.
Speed lap 1 = Vone
Speed lap 2 = Vtwo
Average speed = Vave = 2*Vone
How fast do you have to run the second lap?
Nice one, another intuition buster
Spoiler!
Let's assume speed is in laps/second.
Time for first lap T1 = 1/Vone
Time for second lap T2 = 1/Vtwo
Avg speed = 2/(T1+T2)
We want this to be twice the speed of Vone, which is 2/T1
=> 2/(T1+T2) = 2/T1
=> the second lap must be done in 0 seconds.
You are one of 100 passengers boarding a 100 passenger airplane. Each passenger has a unique seat on the plane.
When a passenger boards the plane, they go directly to their seat and occupy it.
Exactly one of the other 99 passengers can't read their ticket. When it's their turn to board, they will grab a random seat that is unoccupied.
This, unfortunately, can have a cascading effect. If a passenger boards and finds their seat occupied, they will also grab a random seat that is unoccupied.
The random selection of seats is uniform - every unoccupied seat has the same chance of being selected. This includes, for example, the illiterate passenger who could choose their assigned seat with the same probability as any other unoccupied seat.
You happen to be last in line to board. What's the probability that you will be able to sit in your own seat?
You are incorrect that everyone after ends up in the wrong seat after the dummy. The bad passenger has a chance they randomly select their correct seat. But that person only bumps the person who seat they occupy, so it could be they select the last passenger's seat, and everyone is in the correct seat except them and the one person they bumped.
I presume the solution will involve factorials, and it's been a long time since I did those. Or it's a dirty trick and the plane us upside down with no seats and full of tigers.
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So it would just be the same chance that the wrong seat selector picks the correct seat. Since the minute they pick a wrong seat you have a zero chance to get yours as everyone after them ends up in the wrong seat
So if the wrong seat selector goes first, your chance is 1/100
If he gets on 2nd, its 1/99, etc
But I have been removed from Math too long to put that into a formula
Or I am way off (Most Likely)
EDIT: Thinking a bit more each of the scenarios has the same probability of happening 1/99 - The idiot getting on the plane first or 99th. So its that probability of thr event occuring (getting on plane) vs chance they select correct seat
You are incorrect that everyone after ends up in the wrong seat after the dummy. The bad passenger has a chance they randomly select their correct seat. But that person only bumps the person who seat they occupy, so it could be they select the last passenger's seat, and everyone is in the correct seat except them and the one person they bumped.
I presume the solution will involve factorials, and it's been a long time since I did those. Or it's a dirty trick and the plane us upside down with no seats and full of tigers.
Well it doesn't matter if some passengers end up in the correct seat, it only matter that once the dummy gets into a wrong seat you CAN NOT end up in correct seat going last. So the premise stands
Well it doesn't matter if some passengers end up in the correct seat, it only matter that once the dummy gets into a wrong seat you CAN NOT end up in correct seat going last. So the premise stands
I'm saying you can, because of the random seat selection factor. Once a passenger can't use their own seat, they'll take another one. Lets say dummy is supposed to be in seat 15, but they take seat 28. The person in 28 randomly takes 36. The person in 36 randomly takes 15. Well now everyone after that can still use their correct seat since those 3 mistakes cancel out and no more random selections happen.
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I'm saying you can, because of the random seat selection factor. Once a passenger can't use their own seat, they'll take another one. Lets say dummy is supposed to be in seat 15, but they take seat 28. The person in 28 randomly takes 36. The person in 36 randomly takes 15. Well now everyone after that can still use their correct seat since those 3 mistakes cancel out and no more random selections happen.
See you are much smarter then me. My mind did not make that connection. This is why I stay out of this thread! Been 25 years since doing math haha
Oh! This discussion maybe steered me to a solution?
Spoiler!
Because you are last, there is a 50% probalbilty you get your correct seat, because before you, there are 2 seats left. If the dumb passenger has caused an unresolved chain reaction, your seat is blocked. If the reaction is resolved, your seat is not blocked.
Oh! This discussion maybe steered me to a solution?
Spoiler!
Because you are last, there is a 50% probalbilty you get your correct seat, because before you, there are 2 seats left. If the dumb passenger has caused an unresolved chain reaction, your seat is blocked. If the reaction is resolved, your seat is not blocked.
Love the discussion! Jason14h, I hope you stay in the thread - the discussion and flow of ideas is how the problems get solved.
Fuzz, your reasoning is a little bit suspect.
Spoiler!
Yes, you are left with two possibilities - your seat is either blocked or it isn't. But do either possibility have the same probability (ie 50/50) or will your seat get blocked, say, 9 times out of 10?
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Oh! This discussion maybe steered me to a solution?
Spoiler!
Because you are last, there is a 50% probalbilty you get your correct seat, because before you, there are 2 seats left. If the dumb passenger has caused an unresolved chain reaction, your seat is blocked. If the reaction is resolved, your seat is not blocked.
Spoiler!
I think you are right with the wrong logic
The question is does the jack asses seat or your seat get filled first. Anyone who is sitting in their own seat doesn’t matter.
So person sits in own seat, doesn’t matter nothing changes. Jackass sits down equal prob of him having his seat or your seat. If neither occurs we created a new Jack ass. Eventually new Jack ass gets on and again is 50-50 of sitting in the Jack ass or your seat. If neither of those conditions a new jackass is created.
The question is does the jack asses seat or your seat get filled first. Anyone who is sitting in their own seat doesn’t matter.
So person sits in own seat, doesn’t matter nothing changes. Jackass sits down equal prob of him having his seat or your seat. If neither occurs we created a new Jack ass. Eventually new Jack ass gets on and again is 50-50 of sitting in the Jack ass or your seat. If neither of those conditions a new jackass is created.
We have a winner!
The formal proof might go something like:
Spoiler!
Let's call the troublemaker T (I like the evolution of T from illiterate->wrong seat selector->dumb passenger->Jackass!)
Proof by induction.
Let's say T is the 99th passenger or (for our numbering scheme) 1 ahead of your position. Then T can choose either his seat or your seat. Probability of you getting your seat when T is 1 ahead of you is:
P(1) = 1/2
We now assume that
P(k) = 1/2 for k<=n
What if T is n+1 ahead of you?
Then P(n+1) = T choosing his seat OR T choosing any other passenger's seat except for yours (making that person the new T).
=1/(n+2) + n/(n+2)*P(k) where k <= n
=1/(n+2) + n/(n+2)*(1/2) (P(k)=1/2 when k<=n by our induction assumption)
=2/(2n+4) + n/(2n+4)
=(n+2)/(2n+4)
=1/2
I like the formalism of the above proof, but honestly, I like GGG's descriptive proof better because it is more obvious what is going on.
Love the discussion! Jason14h, I hope you stay in the thread - the discussion and flow of ideas is how the problems get solved.
Fuzz, your reasoning is a little bit suspect.
Spoiler!
Yes, you are left with two possibilities - your seat is either blocked or it isn't. But do either possibility have the same probability (ie 50/50) or will your seat get blocked, say, 9 times out of 10?
And here I thought you would take issue with the inverted tiger plane.
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This problem was part of an IMO (International Math Olympiad) problem. It requires a bit more math than some of the others I've posted, but I hope you try it as it has a neat idea behind it.
Define a function f over all positive integers as:
Enter The Search for Planet X, which fills a very specific gap that also speaks directly to my own interests: It’s a deduction game, with a logic puzzle at its heart, and is competitive (rather than cooperative) with multiple ways to score, so that you can still stay in the game even if you aren’t the first to solve the ultimate mystery. It’s brilliant, one of my favorite games of 2020, and has me feeling nostalgic for the days of GAMES magazine and other logic puzzles (hell, even that one section on the LSATs, which I thought was hilarious because so many people do those puzzles for fun) while also making gameplay easier with a companion app.
The Search for Planet X is just what it sounds like: You’re astronomers searching for a potential ninth planet (sorry, Pluto), larger than Earth and with an orbit way farther from the sun than Neptune’s, the existence of which was suggested in a 2016 hypothesis by two CalTech researchers.
This problem was part of an IMO (International Math Olympiad) problem. It requires a bit more math than some of the others I've posted, but I hope you try it as it has a neat idea behind it.
Define a function f over all positive integers as:
We know from the third condition that only odd numbers will work.
It is also true that f(n) will always be odd. Equation 4 is an even minus an odd and equation 5 is an odd minus an even, both of which are always odd numbers, and equation 3 bounds all solutions for even values of n to 1, 3, and the solutions for equations 4/5.
That confirms my first statement that f(n)=n only for n=odd, as f(n) is always odd.
I subtracted the 4th equation from the fifth. That gives:
F(4n+3)-f(4n+1)=f(2n+1) - f(n)
Since the left side is 2 consecutive odd numbers we know that it can only work if the difference between them is 2. Although I suppose that is only true if both of those numbers satisfied that condition
However, it isn't true for all odd numbers. I built a table up to 32 and the following set satisfied the condition {1,3,5,7,9,15,17,27,31}
I suspect the answer could be forced out of analyzing that pattern, although it may not be long enough yet. I'm on vacation without my computer so I did the first 32 with pen/paper - I would have used excel and done a larger sample otherwise.
We know from the third condition that only odd numbers will work.
It is also true that f(n) will always be odd. Equation 4 is an even minus an odd and equation 5 is an odd minus an even, both of which are always odd numbers, and equation 3 bounds all solutions for even values of n to 1, 3, and the solutions for equations 4/5.
That confirms my first statement that f(n)=n only for n=odd, as f(n) is always odd.
I subtracted the 4th equation from the fifth. That gives:
F(4n+3)-f(4n+1)=f(2n+1) - f(n)
Since the left side is 2 consecutive odd numbers we know that it can only work if the difference between them is 2. Although I suppose that is only true if both of those numbers satisfied that condition
However, it isn't true for all odd numbers. I built a table up to 32 and the following set satisfied the condition {1,3,5,7,9,15,17,27,31}
I suspect the answer could be forced out of analyzing that pattern, although it may not be long enough yet. I'm on vacation without my computer so I did the first 32 with pen/paper - I would have used excel and done a larger sample otherwise.
Great start!
Spoiler!
I think you've approached the problem correctly. Trying to manipulate the equations is always the first start, but those equations are a bit ugly.
Computing f(n) for the first several n is absolutely the right step. Finding a pattern in the computed values is tricky though, and requires some insight.
I will say that your list where f(n)=n looks like it is missing one value.
This problem was part of an IMO (International Math Olympiad) problem. It requires a bit more math than some of the others I've posted, but I hope you try it as it has a neat idea behind it.
Define a function f over all positive integers as:
Tried replacing f(x) with x in the final 2 lines. The first one evaluates to n=1, so x=5 which implies n=5 as one of the solutions.
The 2nd one appears to be generally true.
So my guess for the solution is n=1, n=5, and n ≡ 3 mod 4
Edit: just realized the 2nd last line is generally true for powers of 2. So include all numbers that are 1 larger than powers of 2 in the solution.
n=1, n = 2^p + 1 for all integer p, and n ≡ 3 mod 4
Edit: that's probably wrong too because n=11 doesnt work in the final equation. Weird behaving function to say the least! Am I kind of on the right track at least?
