Comments by "disgruntledtoons" (@disgruntledtoons) on "Numberphile" channel.

  1. Fun anecdote: Sam Lloyd was not able to patent the puzzle because he could not submit a solution to the challenge.
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  2. And then it shows him in bed with Lewis Carroll...
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  3. To win the game you must attack on as many turns as possible. You get a card at the end of each turn during which you take a territory, and turning in sets of cards earns the player doing so progressively higher numbers of armies each time this is done. After a while the armies gained for doing this are ridiculously high, and in a long-running game this provides the majority of the armies raised during the game.
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  4. I'll stick with 355/113 for approximations of pi, thanks.
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  5. In the 70's my father built a calculator from scratch. The case was made of plexiglas and was about the side of a desk-top telephone set of the same period. It had a power cable for plugging into a power outlet. It could add, subtract, multiply, and divide. He was working as an engineer at General Motors (in their tech center north of Detroit), and one day he came to work and it was gone. He went to his boss to report it stolen and found his boss using it. The boss begged him to let him have it for just that one day, and Dad agreed. The next day he went to the boss' office to retrieve it, and it was gone. Dad and his boss went to the boss' boss' office, and found that fellow using it. Dad passed on in 2003, but Mom still has the calculator and it still works.
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  6. I don't think that taking two integers, finding their logarithms, adding those logarithms, and then raising that logarithm to an exponent gives us any time savings, and if you need to know the exact value of the product, you are barking up the wrong tree with that process.
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  7.  @mrjava66  And don't forget the taxes that are passed along by the producers of everything you buy. Ultimately, all taxes are paid by working people.
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  8. The divisibility check for 8: If the hundreds place is even, then the last two digits must be divisible by 8. If the hundreds place is odd, then the last two digits must be divisible by 4 but not by 8.
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  9. However, the digits past the 40th one to the right of the decimal point have no practical value.
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  10. There also seem to be lots of prime quads, consisting of sets of values 30k + 11, 30k + 13, 30k + 17, and 30k + 19. The gaps are, as is to be expected, larger than the gaps between prime pairs, but the JavaScript that I am currently running to find them has reached 170 million and is still finding them.
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  11. Why didn't you mention that in graph theory, a tree's definition precludes it from containing loops?
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  12. A gap of n-1 primes appears much earlier than n!+2. There will also be a gap starting at the least common multiple of 2 through n, plus two, a much smaller number than n!+2.
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  13. 2^6 * 3 * 17. Seems like a pretty quick journey to me.
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  14. I'm not sure that the diagonalization proof is valid. From every finite set of reals you can construct a real that is not in the set. The same goes for any set of primes; from them you can construct a prime that is not in the set. So you've proven that the reals are not a finite set, but this particular proof does not establish a different cardinality from the integers. Perhaps there is a proof that establishes this, but the diagonalization proof is not it.
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  15. You can color from 1 to 8 in two colors without any number being the sum of two other numbers of the same color. There is only one way to do it: 1, 2, 4, and 8 are one color, and 3, 5, 6, 7 are the other color.
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  16. A minor correction. In the first example you referred to 7 as the dividend. The term for the number being divided into the other is the *divisor*; the number into which it is divided is termed the dividend.
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  17. But if you take the natural logarithm of that integer, divide by the square root of 163, you get a number that is 2.2373515 x 10^-31 away from pi.
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  18. "We call ourselves the '6th Panzer Army', because we've only got 6 Panzers left." -- SS Commander Sepp Dietrich
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  19. I suspect that Fermat's proof has been discovered at some point and was found to have an error in it.
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  20. If p is composite, then (p-1)! will be divisible by p (because the factorial will include one or more pairs of numbers whose product is p), and therefore (p-1)! + 1 cannot be divisible by p. The other half of the theorem is trickier.
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  21. According to Wikipedia, the number that Graham actually cited in the proof was F(F(F(F(F(F(F(12))))))), where F(n) is 2(^n)3. This is a much smaller number than the number described here, which Graham had used in an unpublished work and which was later made known under that name by Martin Gardner.
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  22. Even number of divisors = light is off. Odd number of divisors = light is on.
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  23. VanSickle's Conjecture is that every discovery in mathematics will eventually have an application outside of mathematics. I'm wondering what Van Eck's sequence will be used for. (Proof of VanSickle's Conjecture is left as an exercise for the student.)
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  24. And then in later centuries comes alternative circuit theory, where the complex numbers allow us to replace calculus with algebra and make engineers' lives much, much easier.
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  25. There are either an infinity of Lychrel numbers, or there are none at all, for each Lychrel number, when reversed and added, must yield another Lychrel number, or else it's not a Lychrel number.
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  26. For the next one you can show that the series 1/2 + 1/3 + 1/5 + 1/7 + ... +1/p + ... also diverges.
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  27. The US Navy probably presented the problem in its academy because the goat is the mascot of the US Naval Academy.
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  28. Did the observed numbers match your model?
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  29. I'm an American and I've never been confused by any of the Anglicisms described here.
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  30. There is a more radical form of this game where you take anu number, and express it as a sum of powers of two, and then express each exponent in that sum as a sum of powers of two, and so on, until no number greater than two is found in the entire express. Then for each step, subtract one from the expression, and increase each remaining two to three. Subtract one from this expression, and replace every remaining three with four. Continue with the process of subtracting one, and then replacing each remaining n with n+1. The number will soon grow beyond calculability, even with low starting values, but it is easily proven that the game will eventually end.
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  31. It was a bit comical watching all of these guys whack each other.
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  32. Whole numbers, natural numbers (the whole numbers, and zero), integers, rational numbers, quadratic numbers, constructable numbers, algebraic numbers, computable numbers, and then the reals.
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  33. Also, for each prime, the repeat period is either p-1 or some number that divides into p-1.
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  34. If nobody had used exponentation up until that point, just write a 9 down and to the left of what the previous entrant left on the board.
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  35. "I feel like that might not be enough paper." Since you said that you would label all of the primes, your feelings are correct.
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  36. Question from the back: Every real number has an efficient sequence of approximations. However, does every constantly-increasing sequence constitute the denominators of an efficient sequence of approximations for a real number?
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  37. The number i allowed us to use algebra to solve problems with AC circuit theory, instead of having to use calculus. This makes engineering power systems much easier. The modern world really does depend on this invention.
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  38. Also, there is a practical application of finding the repeat period. You can convert some text into an integer (by concatenating the codes into a massive integer), and divide that integer by a smaller number. The remainder becomes a checksum for that text. The best divisor for this is a prime with the longest repeat period when dividing binary numbers.
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  39. How does the sequence grow if inventories are not self-referencing?
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  40. It's a shame that you can't get funding for infinity.
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  41. The reason that you can start with any two positive integers and get a sequence with ratios approaching phi is because for any two positive integers a and b such that a < b, the absolute value of phi - (a+b)/b is less than the absolute value of phi - b/a.
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  42. Sigh. Giordano was not martyred for his cosmological views. His cosmological views were unorthodox, but it was his specifically theological views that got him killed.
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  43. If you look at the successively higher powers of phi, you notice that the difference between the value and the nearest integer grows smaller, and in fact the difference is a power of the inverse of phi. Lucas numbers follow the formula phi^n + (-phi^(-n)).
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  44. The reason the Hilbert hotel appears to be a paradox is because infinity is not an actual number and so any comparison of it to another number is an invalid operation.
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  45. Another way to approach the same issue is to take the prime factorization of (10^n)-1. Counting only the primes that were not in the factorization for lower values of n, you get 3 when n = 1, 11 when n = 2, 37 when n = 3, 101 when n = 4, and so on.
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  46. So now we need the sequence of even values which have a greater number of Goldbach representations than any smaller even number, and the sequence of even values which have fewer Goldbach representations than any greater even number.
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  47. How many people noticed that x^3 + 5x + anything has no local maximum or minimum as soon as he first sketched the graph?
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  48. While watching 3,714,289 being trotted, out, I first subtracted 35 from the thirty-seven, leaving 214289. The 21 and 42 are each divisible by 8, so those are dropped, leaving 89, which leaves a remainder of 5 when divided by 7, all before the camera cut back to James.
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  49. The most amazing thing is that graph theory, number theory, topology, set theory, and calculus are all interconnected.
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  50. Who does a secret Santa in an office of three?
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