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zakruti.com » Knowledge, science, education » Numberphile
The Archimedes Number - Numberphile

The Archimedes Number - Numberphile

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Rating: 4.0; Vote: 1
The Archimedes Number -mer: -Bigger than the number of atoms in the universe- is an understatement. (I'll be referring to the observable universe when I say universe) If you were to add an atom into a new universe (universe 2) every time you counted the number of atoms in our universe (universe 1, and then added an atom to universe 3 whenever universe 2 filled up, and keep doing that for as many universes as you need, you would still need several thousand universes to count to this number. This does not mean the number of atoms in several thousand universes, just to be clear; it means that, to quote another person who used this reasoning, -We're writing it in base universe. -
Date: 2022-04-09

Comments and reviews: 9


yes, sure that's the number, but for Archimedes to design such a problem and know it has a solution and that nobody else could solve it then Archimedes might know what it is about the problem that makes it so difficult to solve, or rather Archimedes knows what the number is made of, perhaps with a short 1 line formula to describe it, but by knowing what the number is made of, Archimedes ultimately may know more about the number than we do despite modern technology having printed out the actual number.
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Solving these equations (and checking solutions that are out there, I get the first part: Herd = 50, 389, 082.
Adding the requirement that WhiteBulls+BlackBulls be a square, I find the first solution when this answer is multiplied by 3x11x29x4657 = 4, 456, 749.
In other words, when the herd is 224, 571, 490, 814, 418. The WB+BB = 79, 450, 446, 596, 004 which is 8, 913, 498 squared.
Am I misinterpreting something?

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The number at 4: 52 is wrong. This isnt the total number of cows, but the number of dappled and yellow bulls at the solution to the first eight conditions (which isnt a triangular number, and thus this solution doesnt satisfy all nine conditions. The smallest number of cows satisfying the first eight conditions is is 224, 571, 490, 814, 418.
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I cannot reproduce the intermediate result 51, 285, 802, 909, 803 pieces of cattle. The solution of the first 7 equations gives for the total sum 50, 389, 082 - k. And with the next condition, at least k = 3-11-29-4657, which gives a total sum of 224, 571, 490, 814, 418. What do I miss?
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During the whole video I thought -this is an easy problem, just solve the linear equation, insert the parametric solution on the other statements and done! - then I realized the real problem here is that people was tackling the problem with no algorithmic theory or computers.
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I write a Python code (more than 200, 000 lines) that convert numbers in words (in portuguese) for numbers at maximum 3, 000, 003 digits! By the way, the Archimedes Number (the 1st one with 206, 545 digits) is more than 600 pages in A4 sheets using Arial font 11.
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A stack of 51trillion dollar bills is -3. 5million miles high.
If the average cow is 14, 000(-5ft)dollar bills in height, and covers -10sq. ft, and
Sicily is -10k square miles, with -2. 8m sq. ft(280k cows)per sq. mile, then.
That's a lot of cow shit.

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Not that impressive.
How about the total amount of dappled cows, white bulls and yellow bulls have to all be divisible by 2-57, 885, 161-1 (one of largest prime numbers we know of today? Bet that would add another Centillion digits or so.

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I wish I understood why anyone would want to solve this. It seems a bit like counting the grains of sand on a beach. When you have finished you have no new insights, just a large number of no use to anyone.
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