On Tue, Mar 13, 2012 at 6:19 AM, Stuart Anderson <stuart.errol.anderson@gmail.com> wrote:
Is the Library of Babel a sphere whose consummate center is any hexagon, and whose circumference is inaccessible.?
Using mathematica (and continued fractions - could have done it without them) I got a remainder of 385, that is there is at least one hexagon with only 385 books - 12 shelves of 32 books and one book left over. Put that book on its own shelf and there are seven empty shelves left. Nice numbers
The remainder can be calculated with continued fractions. In mathematica ' ContinuedFraction[(5^2623999)/128] The result was a 182165 digit number followed by 5 terms of the continued fraction expansion of the remainder ie {305631237.... ..... 339715957641, 1, 1, 1, 1, 25}
then convert the continued fraction terms into a fraction; FromContinuedFraction[{0,1, 1, 1, 1, 25}] = 77/128 the result; 77/128 (of a hexagon) 640 * 77/128 = 385 books
I wrote to Bloch, William Goldbloom author of The Unimaginable Mathematics of Borges’ Library of Babel (2008) and he did the same calculation using mod arithmetic in Maple; irem(25^1312000,640); or 25&^1312000 mod 640;
Both of them quickly return 385.
Maple? Mathematica? Continued Fractions? This is something you can calculate in your head in under 2 minutes. Use the Chinese Remainder Theorem. 25^1312000 = 0 mod 5. 25^1312000 = 1 mod 128, since 1312000 is a multiple of 32. (you might think that the exponent of (Z/Z128)* was 64, since it has 64 elements. But since (Z/8Z)* has exponent 2 (3^2 = 9 = 1 mod 8 is the only case you need to check, since the other elements are 1, -1, and -3), (Z/128Z) has exponent 32. (To see that 1312000 is a multiple of 32, you just need to verify that 1312 is a multiple of 4, which follows from the fact that 12 and 100 are multiples of 4). So to compute 25^1312000 mod 640, we just need to find a number mod 640 that is 0 mod 5 and 1 mod 128. 128 = 3 mod 5, and 3 * 3 + 1 = 10 = 0 mod 5, so the solution is (128 * 3) + 1 = 385. Andy