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# Help me understand a little logic problem.

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The bit behind the worm is also stretched, so the second minute figure is 3cm of 2m.

posted by pompomtom at 4:50 AM on September 5, 2012

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# Help me understand a little logic problem.

September 5, 2012 4:39 AM Subscribe

Help me understand this. Math/logic puzzle follows...

Here are the puzzle and answer, as given by Futility Closet:

This makes zero sense to me. Won't the worm always be 1/100 of the way along? First minute: 1 cm of 1 metre. Second minute: 2 cm of 2 metres. Third minute: 3cm of 3 metres. And so on forever.

Am I mistaken somehow?

Here are the puzzle and answer, as given by Futility Closet:

*A worm crawls along an elastic band that’s 1 meter long. It starts at one end and covers 1 centimeter per minute. Unfortunately, at the end of each minute the band is instantly and uniformly stretched by an additional meter. Heroically, the worm keeps its grip and continues crawling. Will it ever reach the far end?*

*Surprisingly, yes. At the end of the first minute the worm has crawled 1/100 of the way to the far end. At the end of the second minute, it’s crawled (1/100 + 1/200) of the way. So it will reach the far end in t minutes when 1/100 (1/1 + 1/2 + 1/3 + … + 1/t) equals or exceeds 1. The expression in parentheses is the harmonic series, which can be made as large as one desires, but it’ll take a while: In this example the worm will reach its goal in t = 1.509269 × 1043 minutes, or about 286,961,000,000,000,000,000,000,000,000,000,000 centuries.*

This makes zero sense to me. Won't the worm always be 1/100 of the way along? First minute: 1 cm of 1 metre. Second minute: 2 cm of 2 metres. Third minute: 3cm of 3 metres. And so on forever.

Am I mistaken somehow?

*First minute: 1 cm of 1 metre. Second minute: 2 cm of 2 metres. Third minute: 3cm of 3 metres. And so on forever.*

The bit behind the worm is also stretched, so the second minute figure is 3cm of 2m.

posted by pompomtom at 4:50 AM on September 5, 2012

Ah! Thank you very much.

On a side note, I read the words "stretch" and "uniform" and "elastic band", and yet visualized the worm on a ruler that is extended only from the far end. I find that a fascinating, self-created error -- my brain unconsciously simplified the problem to make it easier and quicker to "solve".

Thanks again.

posted by TheHollowSeasThatRoar at 5:17 AM on September 5, 2012

On a side note, I read the words "stretch" and "uniform" and "elastic band", and yet visualized the worm on a ruler that is extended only from the far end. I find that a fascinating, self-created error -- my brain unconsciously simplified the problem to make it easier and quicker to "solve".

Thanks again.

posted by TheHollowSeasThatRoar at 5:17 AM on September 5, 2012

As a side note, here's a way to calculate that value of t, since the quoted answer leaves that out.

What we need to do is solve the equation

100 = (1 + 1/2 + 1/3 + ... + 1/t)

for t. My first instinct here was to get an estimate by approximating the sum by the integral from 1 to t of dx/x, which is log(t). It's easy enough to argue that this will always underestimate the sum, so it will overestimate the time required to reach the end. Indeed, the solution we get this way is t = exp(100), or about 2.688x10^43, which is the right order of magnitude but too large.

Fortunately, the relationship between the sum and the integral is, in this particular case, very simple. The difference between the nth partial sum and log(n) converges to a known constant "gamma," called the Euler-Mascheroni constant. So, we can get a much better estimate by calculating exp(100 - gamma), and this produces the result in the answer above.

posted by Aquinas at 8:50 AM on September 5, 2012 [2 favorites]

What we need to do is solve the equation

100 = (1 + 1/2 + 1/3 + ... + 1/t)

for t. My first instinct here was to get an estimate by approximating the sum by the integral from 1 to t of dx/x, which is log(t). It's easy enough to argue that this will always underestimate the sum, so it will overestimate the time required to reach the end. Indeed, the solution we get this way is t = exp(100), or about 2.688x10^43, which is the right order of magnitude but too large.

Fortunately, the relationship between the sum and the integral is, in this particular case, very simple. The difference between the nth partial sum and log(n) converges to a known constant "gamma," called the Euler-Mascheroni constant. So, we can get a much better estimate by calculating exp(100 - gamma), and this produces the result in the answer above.

posted by Aquinas at 8:50 AM on September 5, 2012 [2 favorites]

Also note, you must assume it is an ideal worm. Real worms cannot walk for 286,961,000,000,000,000,000,000,000,000,000,000 centuries.

posted by iotic at 11:02 AM on September 5, 2012 [1 favorite]

posted by iotic at 11:02 AM on September 5, 2012 [1 favorite]

And an ideal elastic band! For a real elastic band, it will eventually snap, catapulting the worm off into space, where it will cover half the distance to the next math/logic puzzle a la zeno's paradox.

posted by yohko at 11:17 AM on September 5, 2012 [1 favorite]

posted by yohko at 11:17 AM on September 5, 2012 [1 favorite]

This thread is closed to new comments.

posted by A Thousand Baited Hooks at 4:48 AM on September 5, 2012 [2 favorites]