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Can't solve this riddle

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Septimius said:
OK, so let me explain this one:

As Feep pointed out, the caterpillar will remain in its relative position when the bungee is expanded. As such, the caterpillar will traverse 1/100 of the bungee with its first centimetre. The second centimetre will give it 1/200 of the length.
As it's pointed out, this is a divergent series. It means that 1/(1 * 100) + 1/(2 * 100) + 1 / (3 * 100) ... + 1 / (n * 100) = 1.

Let's expand it further!

Given that the caterpillar moves 1 cm per second, and the expansion happens instantaneously, how long will it take for it to reach the other side?
That's a math-heavy question, and my high-school math teacher applauded me and basically gave me the rest of the year off when I managed to solve it, so it might be a bit esoteric.
Click to expand...
Looks like the harmonic series is hypergeometric and has no known equation that can precisely give the value after x terms. However, the harmonic series after n steps *converges* on the natural logarithm of n + something called the Euler-Mascheroni constant (γ), which is 0.57721.

So, we're solving γ + ln(n) = 100, ln(n) = 100 - γ, so n = e^(100 - γ). This value, according to Wolfram Alpha, is around 1.509 x 10^43 seconds, or 4.783 x 10^32 millennia. So, awhile. ^^ Any imprecision due to the technical inaccuracy of ln(n) + γ is easily destroyed by me not listing out like 30 significant digits.
 
VegiHam said:
Okay, the two person version has the numbers 1 and 2, the three person 1, 2 and 3, and the ten person 1 through 10. It seems odd that zero is constantly excluded. Would the puzzle become unsolvable if the possibility of zero were added?
Click to expand...
Yes, if there were more possibilities than the number of prisoners the puzzle would be unsolvable.
 
VegiHam said:
Okay, the two person version has the numbers 1 and 2, the three person 1, 2 and 3, and the ten person 1 through 10. It seems odd that zero is constantly excluded. Would the puzzle become unsolvable if the possibility of zero were added?
Click to expand...

if zero were added there'd be more numbers than people.
 
Feep said:
Okay, I got it. It's pretty easy if you think about it the right way.

The series is divergent, and therefore he will reach it.

We think about what fraction of the bungee the caterpillar will cross at each step. First is 1/100, of course. Then an additional 1/200 on the next bit. Then 1/300. And so on. These terms must be summed.

Factoring out a 1/100, we get the ever-familiar harmonic series: 1/1 + 1/2 + 1/3 + 1/4...

That series is divergent, therefore, the caterpillar can cross the rope no matter how long it is. Hooray!
Click to expand...

However, most caterpillars only live 2 weeks to a month.
 
Feep said:
Looks like the harmonic series is hypergeometric and has no known equation that can precisely give the value after x terms. However, the harmonic series after n steps *converges* on the natural logarithm of n + something called the Euler-Mascheroni constant (γ), which is 0.57721.

So, we're solving γ + ln(x) = 100, ln(x) = 100 - γ, so x = e^(100 - γ). This value, according to Wolfram Alpha, is around 1.509 x 10^43 seconds, or 4.783 x 10^32 millennia. So, awhile. ^^ Any imprecision due to the technical inaccuracy of ln(n) + γ is easily destroyed by me not listing out like 30 significant digits.
Click to expand...

How the hell did you end up at the Euler constant that fast?

I was tipped in the right direction by the chapter of math we were at, that there was a correlation between the natural logarithm and sum of the harmonic series. I started working out what it was, and eventually came to like 0.577, then found out it was Euler's Constant. It was pretty neat, calculating a constant one of the great mathematicians has named after him.

At any rate, I'm hugely impressed!
 
Septimius said:
How the hell did you end up at the Euler constant that fast?

I was tipped in the right direction by the chapter of math we were at, that there was a correlation between the natural logarithm and sum of the harmonic series. I started working out what it was, and eventually came to like 0.577, then found out it was Euler's Constant. It was pretty neat, calculating a constant one of the great mathematicians has named after him.

At any rate, I'm hugely impressed!
Click to expand...
I knew that the harmonic series and the natural logarithm were related from way back when, too. I did a quick test with n = 10 and found I was about 0.6 off, then with n = 11 and found I was about 0.59 off, and I figured this value was converging. Wrote a VERY easy script to just check at n = 10000 what the difference was, got 0.5772, typed that into Google, found what the constant was called, did the math. ^^

i.e. cheated with computers

efyu_lemonardo said:
Turns out that the exact value of n is 15092688622113788323693563264538101449859497.
Click to expand...
Welp
 
Septimius said:
How the hell did you end up at the Euler constant that fast?

I was tipped in the right direction by the chapter of math we were at, that there was a correlation between the natural logarithm and sum of the harmonic series. I started working out what it was, and eventually came to like 0.577, then found out it was Euler's Constant. It was pretty neat, calculating a constant one of the great mathematicians has named after him.

At any rate, I'm hugely impressed!
Click to expand...

by calculating sum(1/x) from 1 to N - ln(N) for large values of N, I imagine...

Feep said:
I knew that the harmonic series and the natural logarithm were related from way back when, too. I did a quick test with n = 10 and found I was about 0.6 off, then with n = 11 and found I was about 0.59 off, and I figured this value was converging. Wrote a VERY easy script to just check at n = 10000 what the difference was, got 0.5772, typed that into Google, found what the constant was called, did the math. ^^

i.e. cheated with computers


Welp
Click to expand...

I'm pretty sure there's a fundamental theorem in calculus that guarantees that the discrete sum and the integral converge up to a constant.
 
The Friendly Monster said:
Yes, if there were more possibilities than the number of prisoners the puzzle would be unsolvable.
Click to expand...

efyu_lemonardo said:
if zero were added there'd be more numbers than people.
Click to expand...

So the number of possibilities has to equal the number of people? Oh, I see how it works! Value 1, 2, etcetera have to be every possible answer right? That way, no matter which value the property takes, the guy championing that value has to have written his own number correctly in order to make the value and that's why they get released.

So in the ten person version, the group of all the numbers has a property that can take ten values and is formed out of some combination of the integers from 1-10. Each prisoner must champion on possible value and thus whichever it takes they go free. Presumably this property has to be something easy to calculate from the visible numbers. It's got to be one operation on the numbers, and I'd guess it needs to be commutative, so each prisoner can do the same thing without it mattering. Multiplying the numbers seems difficult with ten of them, so is it addition? Is it ten possible values of the total of all the numbers?
 
VegiHam said:
So the number of possibilities has to equal the number of people? Oh, I see how it works! Value 1, 2, etcetera have to be every possible answer right? That way, no matter which value the property takes, the guy championing that value has to have written his own number correctly in order to make the value and that's why they get released.

So in the ten person version, the group of all the numbers has a property that can take ten values and is formed out of some combination of the integers from 1-10. Each prisoner must champion on possible value and thus whichever it takes they go free. Presumably this property has to be something easy to calculate from the visible numbers. It's got to be one operation on the numbers, and I'd guess it needs to be commutative, so each prisoner can do the same thing without it mattering. Multiplying the numbers seems difficult with ten of them, so is it addition? Is it ten possible values of the total of all the numbers?
Click to expand...
keep going!
 
efyu_lemonardo said:
keep going!
Click to expand...

I think I've got it. Opposite guy always makes the total of the two numbers three, and agree guy makes it two or four. So you can call them 2n+1 guy and 2n guy. For the ten prisoners, each one is assigned a role, 10n guy, 10n+1 guy, through to 10n+9 guy. Then all they do is add up the total of all the other guys numbers and write the number down that needs to be over their cell in order to make the grand total fit their assigned rule. Since these ten groups cover all integers, one of them will be right and they all go free. The strategy can't fail, as long as they all get their sums right.

Thinking about it, I guess they're also, 'ends in 0 guy', 'ends in 1 guy' and so on, since we're dealing with 10s. That probably makes this easier for them then it would be if there were nine of them. "Okay Bob, just add up all our numbers, then write down the one you'd need to have to make the total 5 more than a multiple of nine."
 
VegiHam said:
I think I've got it. Opposite guy always makes the total of the two numbers three, and agree guy makes it two or four. So you can call them 2n+1 guy and 2n guy. For the ten prisoners, each one is assigned a role, 10n guy, 10n+1 guy, through to 10n+9 guy. Then all they do is add up the total of all the other guys numbers and write the number down that needs to be over their cell in order to make the grand total fit their assigned rule. Since these ten groups cover all integers, one of them will be right and they all go free. The strategy can't fail, as long as they all get their sums right.

Thinking about it, I guess they're also, 'ends in 0 guy', 'ends in 1 guy' and so on, since we're dealing with 10s. That probably makes this easier for them then it would be if there were nine of them. "Okay Bob, just add up all our numbers, then write down the one you'd need to have to make the total 5 more than a multiple of nine."
Click to expand...

WELL DONE
indeed the choice of ten was made to simplify finding the remainder
 
Two Words said:
I have two riddles. One I have figured out, the other I can't figure out. I'll give them both. If you can figure it out, answer in a spoiler. No cheating!

Solved Riddle:
You walk down a road until you hit a fork. One direction will lead you to your destination, the other will leave you lost. There are two men at the fork in the road. One always lies, the other always tells the truth. You can only ask one question and only one is allowed to answer. What do you ask and who do you ask it to in order to determine the correct way to go. Both know the correct way.


Unsolved Riddle:
It's the same riddle, but add a third person who sometimes lies and sometimes tells the truth. You can ask three questions. There is some dispute on how many questions you may ask here.
Click to expand...

1) "Which would the other say leads me to my destination?" Take the other path.

2) Damn. I got nothing.
 
So the options are big bowtie and extra big bowtie?

Would it be any different if it were bowtie and nothing?

Edit: you erased it! Does that mean it was a dud?
 
Mista Koo said:
I still have no idea how the prisoner riddle works. I shouldn't have read the solution.
Click to expand...

Here's a detailed explanation:
Say I'm one of the prisoners. If I knew the last digit of the sum of all ten numbers, then I could figure out my number because I could add up the 9 other numbers and based on the last digit of that number there could be only one possible number to add in order to get the correct last digit for the sum of all ten. Try a few examples and you'll see this is true.

Now since I don't know the last digit of the sum of all ten, I can't do this. But what I do know is that this digit has to be in the range 0-9. So each prisoner decides to guess a different last digit between 0 and 9, and that guarantees one, and only one, will guess correctly.
 
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