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TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
I'm too lazy to check 14 is the only ambiguous sum of possible ages.
That's what you have to do, otherwise you don't know if you've actually solved it.
Seems pretty easy

you just partition numbers between 10^n and 10^(n+1) into
the intervals:

[10^n, 2*10^n], [2*10^n+1, 4*10^n+2], and then [4*10^n+3,10^(n+1)]

then you just have to show 10^(n+1) > 8*10^n+6 which holds when n>0. and you have
your proof from Bertrand's postulate.
It is easy when you are given the hint to use Bertrand. You solved it correctly, although you used different intervals than I did (my solution is in reply 27).
 
T

tardis

Member
Sep 7, 2019
73
It is easy when you are given the hint to use Bertrand. You solved it correctly, although you used different intervals than I did (my solution is in reply 27).
I edited my post a bunch of times. You are right the intervals I use are redundant. Using simpler intervals it easily generalizes: k primes between b^n and b^(n+1) whenever b^(n+1) >= 2^k b^n.
 
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
Using simpler intervals it easily generalizes: k primes between b^n and b^(n+1) whenever b^(n+1) >= 2^k b^n.
Well spotted. The condition b>=2^k is equivalent.
 
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T

tardis

Member
Sep 7, 2019
73
The problem is this:

There are two villages A, and B, and a river R. Starting from village A, you have to go to the river to get water, and then go to village B.
The problem is to find the shortest path from A to P to B such that P is a point on the river.

You may use the fact that the shortest path between two points is a straight line.

If you know your calculus you don't need to use calculus to solve this.

The solution is to find a point C which is A reflected vertically about the river.
The idea is that for any point Q on R, the distance from C to Q is the same as
the distance from A to Q:

1640674725300

So the shortest path from A to P to B is the same as the one from C to P to B,
but that is just the line from C to B:

1640674806200

I didn't want this to be a trick question so I made the original image large enough to
visualize the point C below the line.

This is called Heron's problem if you are interested in more research into it. Not sure if this counts as a "brain teaser" but I hope it's interesting.
 
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TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
Yup! Too bad. 😅
This makes no sense. There is absolutely no connection whatsoever between the act of giving a quarter dollar to two people and the current time. If I were a moderator you'd be banned for this.

In order to catch the bus that will take you to the annual Sanctioned Suicide convention in Paris (Texas), you have to leave the house in 45 minutes. Unfortunately, you've lost your job due to the "violent assault in the women's restroom" incident (people are so touchy these days), and now you're only left with two pieces of string and a box of matches (and absolutely nothing else). One string takes an hour to burn down, the other takes 30 minutes. How do you know when to leave? You may neglect air resistance and friction in this problem.
 
Dr Iron Arc

Dr Iron Arc

Into the Unknown
Feb 10, 2020
18,937
This makes no sense. There is absolutely no connection whatsoever between the act of giving a quarter dollar to two people and the current time. If I were a moderator you'd be banned for this.
Obviously since the donor in question is a father, aka the kind of person who makes lame dad jokes, he was all but guaranteed to have fully intended on only giving out that exact amount of money at that exact time distributed randomly between his two offspring.
 
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TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
Obviously since the donor in question is a father, aka the kind of person who makes lame dad jokes, he was all but guaranteed to have fully intended on only giving out that exact amount of money at that exact time distributed randomly between his two offspring.
That may well be the case, but you haven't answered my question yet: How do you know when to leave?
 
GenesAndEnvironment

GenesAndEnvironment

Autistic loser
Jan 26, 2021
5,743
In order to catch the bus that will take you to the annual Sanctioned Suicide convention in Paris (Texas), you have to leave the house in 45 minutes. Unfortunately, you've lost your job due to the "violent assault in the women's restroom" incident (people are so touchy these days), and now you're only left with two pieces of string and a box of matches (and absolutely nothing else). One string takes an hour to burn down, the other takes 30 minutes. How do you know when to leave? You may neglect air resistance and friction in this problem.
Mark both strings' lengths with blood (bite the hand or something). Immediately light both on fire. After the 30min has been burnt, wait for the second string to reach 3/4 of the way and ctb.
 
T

tardis

Member
Sep 7, 2019
73
In order to catch the bus that will take you to the annual Sanctioned Suicide convention in Paris (Texas), you have to leave the house in 45 minutes. Unfortunately, you've lost your job due to the "violent assault in the women's restroom" incident (people are so touchy these days), and now you're only left with two pieces of string and a box of matches (and absolutely nothing else). One string takes an hour to burn down, the other takes 30 minutes. How do you know when to leave? You may neglect air resistance and friction in this problem.

can't you just leave immediately and wait for the bus? If not you can simply fold the thirty minute string in half and light the hour string. then wait until it is the same length as folded up 30 minute one..
 
GenesAndEnvironment

GenesAndEnvironment

Autistic loser
Jan 26, 2021
5,743
can't you just leave immediately and wait for the bus? If not you can simply fold the thirty minute string in half and light the hour string. then wait until it is the same length as folded up 30 minute one..
It's a condition that you have to leave in 45 min. You don't know which string is which time (could be different materials, etc). You also don't know the lengths.
 
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TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
Mark both strings' lengths with blood (bite the hand or something). Immediately light both on fire. After the 30min has been burnt, wait for the second string to reach 3/4 of the way and ctb.
This is what happens when you don't neglect air resistance. You're actually almost there. That last part about waiting for the second string to reach 3/4 of the way is a bit imprecise.
can't you just leave immediately and wait for the bus?
You live in a country with severe covid restrictions. If you wait at a bus stop for more than five minutes, special state forces will detain you and subject you to a lobotomy.
Ah. well then the obvious solution is to simply light both of them from both ends. 30+15 =45
This phrasing is too imprecise for my taste, but I think you mean the correct thing. Now let's say you have only three matches, and one match only burns long enough to light one end of a string.
 
GenesAndEnvironment

GenesAndEnvironment

Autistic loser
Jan 26, 2021
5,743
Now let's say you have only three matches, and one match only burns long enough to light one end of a string.
Use the already lit strings to light each other? Btw, I think my blood strats were good enough...
 
TheAmazingCriswell

TheAmazingCriswell

I predict...
Apr 28, 2021
1,354
Use the already lit strings to light each other? Btw, I think my blood strats were good enough...
My idea was this:
Light both strings at the same time. Once the 30-minute string has burned down, you light the remainder of the 60-minute string from the other end.
 
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