
Re: 120^288 and how to get it nearer to 10^604 Chapt13.40085 Maxwell Equations placing demands on mathematics #651 New Physics #771 ATOM TOTALITY 5th ed
Posted:
Jun 19, 2012 5:45 PM


On Jun 19, 3:19 am, Archimedes Plutonium <plutonium.archime...@gmail.com> wrote: > On Jun 19, 3:47 am, "io_x" <a...@b.c.invalid> wrote: > > here this is ok > > (16) > c:=(120 ^ 291)/ 81 > > (16) > > 135911930413328771927856248088227273330584837917347940523287916343518691899 9_ > > 690585504819382216526236402082584790197171041378399859102382461453415865321 _ > > 289342578352265056153572709886088953858778381318346248947192203901956277713 _ > > 062733148570612814040537186033993596502486544301179402174908486464784319131 _ > > 505496424448000000000000000000000000000000000000000000000000000000000000000 _ > > 000000000000000000000000000000000000000000000000000000000000000000000000000 _ > > 000000000000000000000000000000000000000000000000000000000000000000000000000 _ > > 000000000000000000000000000000000000000000000000000000000000000000000000000 _ > > 000 > Io, division will not work because it leads to fractions
Normally, division will indeed lead to fractions, but not in this case, because we can take advantage of powers.
Notice that 120 factors as 2^3*3*5. This means that we have:
120^291 = 2^(3*291) * 3^291 * 5^291 = 2^873 * 3^291 * 5^291
So there are lots of factors of three around, making division very simple indeed. Of course, AP subsequently corrected his mistake, and so we want to divide by 27, not 81 here:
120^291/27 = 2^873 * 3^288 * 5^291
This number is a perfect cube and is exactly three times the number that Io produced above.
> There maybe a 9.999.. x 10^603 type of perfect cube factorable by 120.
Indeed there is. Notice that if a number is divisible by 2*3*5 (which equals thirty), its cube will have 2^3 * 3^3 * 5^3 as a factor, and therefore 2^3 * 3 * 5 (which is 120) as well. This gives us a recipe for finding a cube of the form 9.999...*10^603 that has 120 as a factor:
 Begin with the number 10^604.  Take its cube root and discard everything after the decimal.  Divide by thirty and discard everything after the decimal.  Multiply it by thirty.  Cube it.
The resulting number will be the largest cube less than 10^603 with 120 as a factor. As it turns out, this number is approx. 10^604  2.111213954...*10^404, so there are 199 nines before the first digit other than nine. (This is expected since the d/dx(x^3) is 3x^2, so we expect the largest cube less than 10^604 to be on the order of 10^402 away. The extra two orders taking us to 10^404 is due to the additional constraint that the cube have 120 as a factor.)
Of course, most of AP's cubes weren't divisible merely by 120, but by large powers of 120. It turns out that the number that we just found happens to have 120^2 as a factor, but no more factors of 120. (Due to excess powers of two, finding cubes divisible by 120^2 or 120^3 are not uncommon by this method, but we must be especially lucky to find 120^4.)
Most likely, the cube that we just found won't help AP much, but it does provide an answer to his question.
Meanwhile, I noticed that Io started a thread looking for factors of floor(pi*10^603) beyond the ones that he found (I presume by trial division). Such factors will be difficult, since the number to be factored is only a few orders shy of the RSA numbers. (This is why the link that I provided earlier stops at 250, with about half of those incompletely factored.)

