Digits in the answer
The number of seconds until the ant finishes takes 43,430 digits to write. Printed in small type, it fills a dozen pages or so. The age of the universe in seconds has 18 digits.
A rubber rope 1 km long. An ant sits at one end and crawls toward the other: 1 centimetre a second. And every second the rope is stretched evenly by another kilometre, carrying the ant along with it. The end runs away a hundred thousand times faster than the ant crawls. Will it ever reach the end?
The stretch carries the ant along with the rope, so it never takes away the share of the way already covered. Every second the ant adds a crumb: 1 cm of a rope that keeps getting longer. The crumbs shrink, but their sum grows without limit, like the harmonic series 1 + ½ + ⅓ + … So the ant gets there, only after ≈ 1.6 × 1043,429 seconds. The universe is about 4 × 1017 seconds old.
A classic puzzle popularised by Martin Gardner. It is about how a sum of ever smaller pieces does not have to stop at some limit.
Made of rubber, 1 km long at the start. One end is tied to a post, the other is being pulled. The rubber stretches evenly: every bit of it gets longer by the same factor, like a drawing on a balloon as it is blown up.
Starts at the tied end and crawls toward the other at a steady 1 cm a second relative to the rubber under its feet. It holds on to the rope, so when the rubber stretches, the ant rides along with the bit it sits on.
First the ant crawls its 1 cm. Then the rope is instantly stretched by another kilometre. After the first second it is 2 km long, after the second 3 km, after the hundredth 101 km. The end runs away at 1 km a second, 100,000 times faster than the ant crawls.
Will the ant ever reach the far end of the rope? If so, when?
"The end runs away a hundred thousand times faster. The ant will never catch it"
It will catch it. The stretch pulls the ant along too: everything it has covered stretches with the rope. So count not metres but the share of the rope. And the share grows every second, if only a little, and those gains together exceed any number.
A 1 km rope, an ant at 1 cm/s, the rope gets 1 km longer every second. Pick an answer.
A breakdown of your answer will appear here.
Hard to decide? Open the hints one at a time: the easy one first, then stronger ones.
Do not count the metres to the end, count the share of the rope the ant has covered. What does an even stretch do to that share?
In second number k the rope is k km long, so 1 cm is 1/(100,000 · k) of it. The ant arrives when these shares add up to 1.
The sum 1 + 1/2 + 1/3 + … + 1/n: is there a number it never passes? Group the terms: 1/3 + 1/4, then 1/5 + … + 1/8.
First the sandbox: choose the rope length, the ant's step and the stretch, and see whether you can stop it. Then the game: six ropes, and for each you guess at which step the ant gets there, or that it never does. Some ropes are pulled differently: by a few percent each time.
It gives the answer away, so make your own guess first.
At the top the rope always spans the full width, and the marks show metres: when it is pulled, the marks slide apart. The ant stands at the share of the rope it has covered. At the bottom the same share on a chart: across is the step number on a log scale, each mark is 10 times more steps.
The steps are counted by a simulation: the ant's metres and the rope's, step by step. Past 20 million steps the formula from round 4 takes over, and the chart will say so.
At which step does the ant arrive?
The simulation knows no formulas: it keeps the ant's metre and the rope's length, adds the ant's step and then stretches everything proportionally. The formula from round 4 counts only the share: the sum v/L₁ + v/L₂ + … On the chart the line is the formula, the dots are the simulation, the dashes are the continuous version where the rope is pulled smoothly.
The whole trick is to count not metres but the share of the rope. The metres to the end really do grow. But the share the ant has covered only grows, and the only question is whether it reaches one.
“In plain words” is hidden on all puzzles.
The rubber stretches evenly, so an ant that sat at one third of the rope is at one third again after the stretch: that third is simply longer now. The end runs away, but the ant rides forward with the rubber too. So the thing to watch is the share. Every second the ant adds 1 cm to a rope that is k km long, that is 1 / (100,000 · k) of the rope. This card does not explain why such crumbs ever add up to the whole rope. The next card and the strict layer do.
A 4 m rope, the ant covers 1 m per step, after each step the rope gets 1 m longer. The shares: ¼, then ⅕, ⅙, 1/7… Each is smaller than the one before, yet their sum passes 1 at the seventh step. After the sixth step the ant was only 4 cm short. Our puzzle works the same way, only each share is 100,000 times smaller, and the wait is unbelievably long.
“Rigorous” is hidden on all puzzles.
1/100 000 · (1 + 1/2 + 1/3 + … + 1/n) = Hn / 100 000
In second k the rope is k km long, and 1 cm is 1/(100,000 · k) of its length. Add them up: you get the harmonic series divided by 100,000. The ant arrives when the sum 1 + ½ + ⅓ + … passes 100,000. And it passes any number, because the terms can be grouped: ½, then ⅓ + ¼ > ½, then ⅕ + ⅙ + 1/7 + ⅛ > ½, and so on. Each next group is twice as long and gives more than ½. There are infinitely many groups, so the sum is infinite. Nicole Oresme found this proof around 1350.
Across is time in seconds, each mark 10 times more. The sum 1 + ½ + … + 1/n is almost a straight line here: it is about ln n + 0.5772. To add one to the sum you must multiply the time by e ≈ 2.718. That is why after a year the sum is only 17.8, after the age of the universe 41.2, and it reaches 100,000 only at n ≈ e100,000 − 0.5772 ≈ 1.6 × 1043,429.
Rope L, the ant covers v per step, after each step the rope gets longer by s. During step k it is L + (k − 1)s. The share after n steps:
ψ is the digamma function, "the harmonic number for fractional arguments". In our puzzle L = s = 1 km, v = 1 cm, so the share is just Hn / 100,000, and Hn ≈ ln n + γ, γ ≈ 0.5772.
Pulled smoothly at speed c (length L + ct), the share grows by dθ/dt = v / (L + ct). Hence a logarithm and an exact answer:
In jerks and smoothly the answers differ only by a factor of eγ ≈ 1.78: 1.6 and 2.8 × 1043,429 seconds. Both numbers have 43,430 digits. The page checked the formula against the step-by-step simulation in round 3.
“Deeper” is hidden on all puzzles.
Martin Gardner spread the puzzle widely: it appeared in Scientific American, and in 1982 he included it in his book of paradoxes "aha! Gotcha: Paradoxes to Puzzle and Delight" (pp. 145–146). In some retellings a worm on a rubber band takes the ant's place. Who first came up with the puzzle, the sources we checked do not say, so we do not claim it.
The puzzle is still being extended: in 2013 Mark McCartney devoted a paper to it, "Extending the rubber rope: convergent series, divergent series and the integrating factor", in the International Journal of Mathematical Education in Science and Technology. As the title says, in the generalisations the series can diverge or converge. A converging series means the ant never gets there. That is round 5.
When people say the universe is expanding, they mean roughly our rope: distant galaxies are not flying through space, the space between them stretches, and distances grow proportionally. Light from a distant galaxy is the ant: it moves at the same speed relative to the space it travels through, while that space stretches beneath it.
If the expansion goes like our rope, without acceleration, light from any galaxy reaches us eventually, if only after a very long time. But if the expansion accelerates, and observations say our universe does, there is a limit for light emitted today: the event horizon. What lies beyond it will never reach us. Round 5 shows this on a model.
Mathematics: if the rope is pulled with acceleration, the ant may never get there, and the boundary is exact. Logic: a sum of ever smaller pieces can be infinite, and you can build it from books. Practice: the same calculation decides which galaxies we will never see.
Mathematics
A 10 m rope, the ant covers a percentage of those 10 m per step. The share is the sum v/L₁ + v/L₂ + … If the rope grows by the same amount each time, the denominators grow like k and the sum is the harmonic series: infinite, the ant gets there whatever its step. If the length grows like k², the sum 1 + 1/2 + 1/5 + 1/10 + … is finite: with L = 10 + 10(k − 1)² it equals (1 + π coth π)/2 ≈ 2.077 ant steps per 10 m, so the ant gets there only if its step is at least 48.2% of the rope. If the rope grows by 10% each time, the sum is a geometric series with limit 11 · v/L, and the threshold is 9.1%.
The general rule for a length growing like kp: for p ≤ 1 the ant always gets there, for p > 1 only if it is fast enough. The boundary runs exactly between the harmonic series and anything that grows even slightly faster.
Logic
Stack identical books so the top one sticks out over the next by ½ a length, that one over the third by ¼, then ⅙, ⅛… Each stack holds, because the centre of gravity of everything above sits exactly over the edge of the book below. The total overhang is half the harmonic series: (1 + ½ + ⅓ + … + 1/n) / 2. Just 4 books carry the top one fully past the table edge, 31 by two lengths, 227 by three.
The logic is the ant's: the terms shrink to zero, but that is not enough for the sum to be finite. They must shrink fast enough. Compare the fly between two trains: each leg is five times shorter than the one before, and infinitely many legs add up to a finite 75 km.
Practice
Two toy models with today's expansion rate H₀ ≈ 70 km/s per megaparsec. Uniform, like our rope: the distance to a galaxy grows by the same number of kilometres every second. Then light gets through from any distance, and the time is the ant's formula: (1/H₀)(eH₀D/c − 1). Accelerating: the distance is multiplied by the same factor each time. Then the share of the way light covers has a limit, and from galaxies farther than c/H₀ ≈ 14 billion light years, light emitted today never arrives.
The real universe is more complex: the expansion first slowed down and now speeds up because of dark energy. By Lineweaver and Davis's estimate (Scientific American, 2005), the event horizon is now about 16 billion light years away. Light that galaxies beyond it emit today will never reach us, although we do see their old light. Galaxies held together by gravity, like our Milky Way, do not stretch.
The number of seconds until the ant finishes takes 43,430 digits to write. Printed in small type, it fills a dozen pages or so. The age of the universe in seconds has 18 digits.
In a day the ant itself crawls only 864 m. But the stretch carries it along with the rubber, so at the end of the day it is already 10.3 km from the post. By then the rope is 86,401 km long, and 0.012% of it is done.
The French philosopher Nicole Oresme proved around 1350 that the harmonic series has no limit: he grouped the terms so that each group exceeds ½. It is one of the oldest proofs about series and is still taught from textbooks.
The sum 1 + ½ + ⅓ + … exceeds 10 only at the 12,367th term. To reach 100 takes about 1.5 × 1043 terms. The series is endless, but it crawls slower than the ant.
Martin Gardner included this puzzle in his book of paradoxes "aha! Gotcha" (W. H. Freeman, 1982). That is the source the Wikipedia article on the puzzle cites.
The event horizon of our universe is now about 16 billion light years away (Lineweaver and Davis, Scientific American, 2005). Light that galaxies beyond it emit today will never reach us. An ant on a rope pulled ever faster is in the same position.
Share your own explanation, an objection or a variant of the puzzle. You need a GitHub account to comment.
Is “Advanced” the right level for this puzzle?
Thank you! This helps set the levels right.
I keep adding new puzzles, new ways to explain them and write-ups of the discussions they start. Leave your email and I'll write when something new is out. No spam, one-click unsubscribe.
The sign-up form will appear here soon.