John and Fred have a dilemma. They need to get back to town as quickly as possible, but they only have one bicycle between them.

John says, "Here's what we'll do. I'll ride the bike down the road, chain it to a lamp post, and continue on. Then, when you get to the bike, you ride it until you pass me, then go a ways further and chain it to a lamp post, and we'll continue leapfrogging that way until we get to town. Our average speed will be higher and we'll get there sooner."

Fred says, "No, that won't do any good, because even though we alternate walking and riding, every inch of the way will still be traveled on foot by one or the other of us, so we can't get there any faster than if we both walked all the way."

Who is right? Why?

tanstaafl.

John's right. One goes faster, then the other goes faster, but at a miniumum they're both progressing as fast as they would walking... It doesn't matter that one of them has to be walking at all times...

Matthew

Exactly. What's the trick? I don't see it.

Or is this really just a math problem?

Damn you, Doug.

I had a reply all typed in, then saw a fallacy in its logic and had to cancel it.

GRRR.

Greetings!

Some assumptions that need validation:

1) The bike is working and faster than either person walking.
2) Does the amount of time chaining and unchaining the bike delay the pair?
3) Do they walk at the same pace, or does one walk faster than the other?
4) What is the phase of the moon at the time, and is it in the sky?

okay, maybe not number 4...

john is right because even if they have to walk the whole way more/less, they won't be as tired as they would if they simply walked. if they aren't as tired they will indeed walk faster and get to town quicker so they can buy a car before the sale is over.

I think the trick is to see why Fred is wrong.

The point of the puzzle is that both Fred and John make plausible sounding explanations, but they can't both be right.

So, if you had to explain to Fred, you might tell him that even though every inch has to be walked, with the bicycle you are able to do it in parallel, so it doesn't take as long.

Richard.

Okay. Let's assume that they both walk at the same rate and they both ride at the same rate. Let's also assume that they ride three times faster than they walk (makes my example easier, and it shouldn't make much difference). Assume that there are a number of equidistant points labelled O (for Origin), A, B, C, D, etc.

For our origin, let's assume that the first ``leg'' has already been undertaken. That is, traveller 1 has already ridden the bike ahead, and let's assume that he got to point A while traveller 2 is back at the Origin. Their pattern looks like this:

        *               *               *

O 1 A 3 D 1 E 1 F 1 G 3 J 1 K 1 L 1 M 3 P 1 Q ...

A 1 B 1 C 1 D 3 G 1 H 1 I 1 J 3 M 1 N 1 O 1 P ...

*               *               *

The letters are the ``milestones'' and the numbers are the distances they travelled. The asterisks are where they left the bike. If they were to walk, then this would be the pattern:

O 1 A 1 B 1 C 1 D 1 E 1 F 1 G 1 H 1 I 1 J 1 K ...

You can see that they're already way behind in the same amount of time. I can't think of a reason for the pattern to not continue (<- split infinitive ).

Edit: Try to make it a little more legible.

Also, we're not taking into account all the hoodlums that will be stealing the bike's wheel.

Bike is faster. Assume walk speed = 1mph Assume bike speed = 2mph Assume lamp posts are 10 miles apart At 11 hours, it is clear that if both walk, they will be 11 miles away. Let's take the bike case. Assume Fred rides first. At 5 hours, Fred will be 10 miles away at the first post, John will be 5 miles away. John walks for 5 more hours (5 more miles) and gets the bike. During the 5 hours of walking, Fred also was walking for 5 miles. So now we're at 10 hours. Fred is 15 miles away and John is 10 miles away at the first post. John gets the bike and starts riding (@ 2mph). In one hour, Fred is still walking and has walked one mile. John has now ridden 2 miles. So Fred is at 16 miles and John is at 12 miles. This is now hour 11. Remember, at 11 hours walking they were only 11 miles. However, riding one is 16 miles away and the other is 12 miles away. Given that they will NEVER go SLOWER by walking, they are already ahead and will only get further ahead.

rjlov said:
> with the bicycle you are able to do it in parallel

That is they key right there I think. One or the other still has to walk every inch of the way as Fred said, but there are times when they are each covering a different part of the road at the same time. Hence, it won't take as long.

John's right, by the amount of time it takes the person riding to catch up to the person walking. From there the speed is only as fast as the person walking takes to reach the bike and the cycle begins.

Why don't they double?

Yep. Every inch does have to be walked. But with John's method, part of the time they're walking at the same time in different areas, thus walking those inches twice as fast.

I've got one. Three guys go to a hotel and want to rent a room. They are charged $30 for the room, so they split it, $10 each. When they get in their room the hotel manager realizes he forgot to give them the special weekday rate of $25, so he gives the bell hop $5 and tells him to take it to the three guys. The bell hop, knowing that three guys cant split $5 evenly, pockets $2 himself, and gives $3 to the guys. So each guy has paid $9 for his share of the room, and the bell hop has $2.
9 X 3 = 27
27+2=29.
What happened to the other dollar?

The hotel manager has it. $25 divided by three is roughly $8.33.

I'd explain it better, but I'm kinda drunk again.

Bike is faster, Solution attached.

That's an easier one. Order of operations. It's 3*9+27 minus the 2 the bellhop kept to bring you the the proper answer of 25 total dollars.

hmm... think you are wrong... $25 arent to be splitted between the guys....

Hmm... a though nut to crack when i come to think of it. But then again i have been awake for about 28 hours now...
The possibility i can only half think of is that its a logical problem. Somewhere you use an even number or the zero as a non value... and on another place you are using it as a value... but as i said.... i am very uncertain of this... too tired...
*dreams away to the bed, then wakes up by the knowledge that I have to go to work now*

damn.... perhaps i´ll come back and think of it some more after ive slept a bit...

Positive I'm right. $30 initially for the room, $5 off for the mistake = a $25 room. No matter how you split it.

damn.... i´m too tired....
Now that ive thought some more I may have been too fast to judge... You are right... $25 divided by 3 is 8,33 and not $9. Allthough HOW the manager got a hold of them is another question.... perhaps I´m partially right... my dignity and pride certainly hopes so.

Note: The person I referred my discussion too was not to lectric but to bxgurl. But then again... your arguments are the same so i´m wrong either way.

hehe.... *sigh* What i meant before was that 25$ wasnt suppsed to be splitted between the guys since in reality they have paid 25+2(tip?)=$27 so the split is 27/3=$9
Damn.... how am I wrong.... I can feel it somewhere... but i cant put my finger on it....

The hotel manager does not have it. He has $25.
The bellhop has $2.
The 3 guys have $3 refunded.
There is no "missing" dollar.

The guys think they paid $27.
The manager thinks they paid $25.
The difference is in the bellboys pocket.

And what exactly are those three guys doing in such a cheap room?

Hmmm. I've spent far too long thinking about this. The thing that we have to realize is that in the end, we're not trying to get to 30, we're trying to reach 25. The problem does a remarkable job of convincing us that we're trying to reach $30, which is a completly irelevant figure.

In the begining, they put in $30, the manager takes all of it, and we're at zero.

After that, they put in 27, the manager takes 25, the bell hop takes 2, and we're at zero.

In the arithmetic, they put in 9, the bell hop puts in 2, and we get to 29, but the question remains is how they managed to get the bell hop to put money in instead of taking money out. Must have something to do with the fact that there are three guys in a cheap hotel room.

That's a damn good mind trick. I still want to beleive it, even though I understand why it's false.

Matthew

It's an old one, google finds lots of instances.

...although there's some other interesting problems at some of these sites, too.

Yeah, It's easier if you just think of it as the men having $30, 25 of which goes to the manager and 2 to the bellhop with 3 leftover. But the way it is worded confuses you.

My Dad used this trick to exersize good thinking about the way problems are worded once when I was a kid. We actually had the whole family sit around the table with real money and did the work. Of course, now I realize that it was a trick of wording, but I was sort of young then.

I might be picking the wrong crowd for this one, but here goes:

You have a mile long track and you drive a car around it at 30mph. How fast do you have to take a second lap to average 60mph for the two laps?

Infinity mph!

You'd have to travel the second lap at infinite speed, or, rather, take zero seconds to complete the lap.

That would be the only way in which you could cover the two miles in a thirtieth of a second...

Bryan.

See, I knew this was the wrong crowd. The only person who has ever gotten this one when I asked was a math guy and understood the problem immediately. Other people just usually argue with me that "the average of 30 and 90 was 60 when I was in school." Never mind that speed is a function of distance and time.

Well, I must admit that (several years ago) you could probably have classed me as a "math guy". Well, a physics guy, anyway...

Bryan.

Okay, here's one for people who are "math guys":

I have a sequence such that

x_1 = 1

and

x_{n+1} = (1 + sum_i=1^n{{x_i}^2}) / n

for n>=1.

So the sequence begins 1, 2, 3, 5, 8, ...

Hypothesis: all x_n are integers.

Puzzle: prove or disprove the hypothesis.

This one has been bugging me on and off for over 10 years, so if someone has a solution, I'll be happy all day.

I think that should be 1,2,3,5,10 right? Just want to make sure i parsed that ascii formula right before i keep going

Yep. Every inch does have to be walked. But with John's method, part of the time they're walking at the same time in different areas, thus walking those inches twice as fast.

Yes.

A very succinct explanation. Well done!

tanstaafl.

so if your sequence is correct, then I think you mean "/ x_n" rather than "/ n". If you mean "/ n" then I think the sequence is 10 not 8, and after about n = 8 or so the numbers just explode upwards -- so much so that I can't even check to see if they really are integers (i wanted to run a few rounds to make sure the hypothesis was right).

However, if you meant "/ x_n" then that's a whole different ballgame... And those provably are integers, and it's actually the fibonacci sequence ... here's why:

define sum(a(b), b, c, d) to be "the sum of a(b) where b goes from c to d" .. to make parsing easier.

i'm switching to definiing n in terms of n-1 rahter than n+1 in terms of n... it made my brain hurt the other way for some reason.

restatement:
x_n = ( 1 + sum( (x_i)^2, i, 1, n-1) ) / x_(n-1)

pull out the last value of the sum:
x_n = ( 1 + sum( (x_i)^2, i, 1, n - 2) + (x_(n-1))^2 ) / x_(n-1)

divide through:
x_n = x_(n-1) + (1 + sum( (x_i)^2, i, 1, n - 2)) / x_(n-1)

the 1 + sum(.., n - 2) by definition is actually x_(n-1)*x_(n-2), which makes it now:

x_n = x_(n-1) + (x_(n-1)*x_(n-2)) / x_(n-1)

simplify and you get:

x_n = x_(n-1) + x_(n-2)

Like I mentioned, though, if you meant "/ n", then I need to go back to the drawing board.

Mike

I hate you for doing this to me ... So I got the "/ n" version down to:

x_n = [ (x_(n-1)) / (n - 1) ] * [ x_(n-1) + n - 2)

if you set x_1=1 and x_2=2 (wonder why it doesn't give you x_2 itself ... weird)

But now I'm kind of stuck ... If (x_(n-1))^2 + (n-2)(x_(n-1)^) somehow can be factored so that there is an (n-1) term in the numerator, that would cancel out the n-1 in the denominator and prove it to be true. But my brain has stopped functioning now that it is 2:30.

You realize you've set jEmplode back an entire day with this question

Mike

If you continue to use algebra on the BBS, I will have to hunt you down and kill you. Slowly. With a very dull knife.

In reply to:

---

If you mean "/ n" then I think the sequence is 10 not 8, and after about n = 8 or so the numbers just explode upwards -- so much so that I can't even check to see if they really are integers

---

Yes, I do really mean "/n", and yes, the numbers do get very big very quickly - it doesn't take long to run out of screen space And sorry for the mistake in my mental arithmetic - x_5 is indeed 10.

It's a long time since I've actually looked at this one, but ISTR making some progress by extending the sequence to include x_0 = 1. Then the expression becomes

x_{n+1} = sum_i=0^n{{x_i}^2} / n

and

n.x_{n+1} = (n-1)x_n + {x_n}^2

n (x_{n+1} - x_n} = {x_n}^2 - x_n
= x_n (x_n - 1)

So

x_{n+1} = x_n + x_n(x_n - 1) / n

Which would be great if I could somehow show that x_n or x_n - 1 is a multiple of n ... Unfortunately, it appears to alternate, so that for odd n, x_n is a multiple of n, and for even n, x_n - 1 is a multiple of n.

(Hmm, I'd forgotten how ugly TeX notation can be when you're not used to it)

Try a runsable spoon. not as sharp but quite effective.

x_{n+1} = x_n + x_n(x_n - 1) / n

I think you have to define x_2 explicitly .. My brain can't comprehend why, but if you use this to find x_2, it returns 1 -- mine did this too (since x_n - 1 = 0 for x_1) ... no idea why.

I wonder since we have two different equations, is it possible to set them equal to eachother and find out anything more interesting.

ms

"Why a spoon, cousin? Why not an axe?"

"Cause it's dull, you twit, it'll hurt more!"

Actually John never says that the other guy is walking. Maybe the other guy would hitchhike. Its would be faster. Maybe the other guy has a skateboard, a pod racer, a teleportation booth?

In reply to:

---

I think you have to define x_2 explicitly .. My brain can't comprehend why,

---

Yeah, I don't understand it either, as the original equation defines x_2 correctly. Perhaps there's a hidden divide-by-zero in the derivation (a bit like the one in the standard "proof" that 1=2).

Anyway, back to my DSSSL hacking - I've almost got something that interprets a user tag "print" to decide whether or not to output each subtree.

In reply to:

---

Try a runsable spoon. not as sharp but quite effective.

---

runcible?

I was thinking you could maybe get a little further because you know x_n was an integer and you could define:

since x_n was an integer, x_n = y / (n - 1) where y is an integer (1 + sum...). I haven't tried it yet, but I wonder if you sub that in to the x_{n+1}, does it get you any further (does that (n-1) cancel anything out). The next step has got to be that you take advantage of the fact that you know the previous number was an integer ........

ms

If you assume:
1. John and Fred walk the same speed
2. John and Fred ride the same speed
3. Lamposts are evenly spaced
4. Each section of travel is bounded by start point, town, or a lamp post.
5. There is a lamp post at the start and stop points, or the distance from the start or stop point to the nearest lamp post is the same as from lamp post to lamp post.
6. DELTA=(section walk time)-(section ride time)
7. They Ride faster than they walk.

John is correct.

For an even number of sections John and Fred will both arrive at town sooner (and at the same time). For each pair of sections, they will reduce there travel time by DELTA.

If the number of sections is odd, the scenario is the same as above except that John will arrive at town first and shave an additional DELTA off of his travel time (Fred gets no additional travel time reduction).

Since John concocted the proposition, perhaps he does not intend to chain up the bike to the first lamp post at all...

Since John concocted the proposition, perhaps he does not intend to chain up the bike to the first lamp post at all...

ROFL, best comment made so far.

"Since John concocted the proposition, perhaps he does not intend to chain up the bike to the first lamp post at all... "

damn, you made me crack me up

I have the proof, but it is too large to be contained within this box...

If you continue to use algebra on the BBS, I will have to hunt you down and kill you. Slowly. With a very dull knife.

So MathML in posts is enabled only for moderators?

Peter

Sorry runcible spoon. I'm sure that's the spelling I had before.

So it's a spork, right?? Oh, it has a cutting edge....

A knorkoon?

You are correct. I woke up at 1:30 in the morning going "wait... that isn't right!" I was just too lazy to post and point out my error. Oh well... It was caught.

Oh wait... you weren't talking to me.. I was refering to my equasion earlier on the walking thingie... the 2 on the bottom is incorrect, it should be 2X. Since all the trials I had tried were using 1 mph as x, it always worked, but it would NOT have worked if the speeds were 2mph and 5mph, respectively. anyway, blech.

Sorry runcible spoon.

Ah, nifty. In your first post, I thought you'd just horribly butchered the spelling of re-usable.

I guess that answers the Chunky Cambells Soup question...

A bump to this very old thread. Seems Click and Clack liked Doug's little puzzler.

Reading of question:
http://play.rbn.com/?url=cartalk/cartalk/demand/CT0417-07.ra

Reading of answer:
http://play.rbn.com/?url=cartalk/cartalk/demand/CT0418-04.ra

I don't know how long these links will last, they might take them down after a week or so.

Congrats, Doug! Do you get anything, like a mug or a Shameless Commerce gift certificate?

Ah, this brings back memories.

Do you get anything, like a mug or a Shameless Commerce gift certificate?

Naaah... they only give that stuff to the people who get the correct answer to the puzzler.

tanstaafl.