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Given the googolplex number and that 1 gigabyte = 1 073 741 824 bytes. How would I calculate the amount of disk space required to store the googolplex number?

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4 Answers 4

Googolplex is $10^{10^{100}}$. Now, the number of bits required to store the number $n$ is $\lg n$ (that is, $\log_2 n$). We have:

$\lg 10^{10^{100}}=10^{100}\cdot \lg 10$

A byte contains $8=2^3$ bits, and as you said, a gigabyte contains $2^{30}$ bytes. So a gigabyte contains $2^{33}$ bits. Hence you need $\frac{10^{100}}{2^{33}}\lg 10$ gigabytes overall.

This is around $3.867 \times 10^{90}$ gigabytes.

Now, of course, there is a much easier way to store googolplex and similar numbers: You can represent them as $a^{b^c}$ with $a,b,c$ small, and only need the bits to store $a,b,c$ and the information of your specific storage method (it's something around 14 bits). You can also represent googolplex by a single bit if you'd like, but then your number-representation algorithm will be rather specific...

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I think it's amusing that we claim we need so many gigabytes to store this number, but we reference it so casually in the same sentence! Taken literally, to store the information contained in the word "googleplex" we need enough space to store the expression $10^{10^{100}}$. Of course, if we wanted to store a number on the order of one googleplex that had random digits with full accuracy, this question becomes more relevant. –  barf Jun 2 '11 at 16:26

In the sense of information theory, you just need to store any program that will calculate a googolplex, which can be written in a few hundred bytes. I once saw one on the web, along with a suggestion that it not be run for 567 years. The justification was that by Moore's Law, you shouldn't run any program that takes more than four years to run-you should wait two years for speeds to double and then run it then. The program needed a factor of $2^{283.5}$ to meet that.

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I think that was referring to a googol only. :-) –  ThudanBlunder Jun 2 '11 at 13:19
    
@ThudanBlunder: As far as I recall, it was "calculating" a googolplex, which it really did by "outputting" a googol of zeros (which it then cautioned should not be sent to a display as they are much slower). –  Ross Millikan Jun 2 '11 at 16:29

If you use the trivial approach to store the number in its binary representation then you have to calculate the length of the number in base 2 to get the amount of bits you need to store it. Then you can convert the number of bits into bytes (1 byte = 8 bit). The amount of bits is equivalent to the number of digits in base 2 which is given by the $D_2$ function I referenced here.

So in total:

Number of bytes to store $x = D_2(x)/8$

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The best thing to due is attack where the system is more relaxed and handle's more simple computations.In the actual final configuration of a googolplex so unknown it is not easily recognized, and as for software programs capable of rendering a number so larger was shipped with the first version of Windows operating system MS Paint and MS Word pad, which used in continual contrast in conjunction with nano spec pint font's (so even the fact of a googolplex being such a large number it still has to be magnified of visual recognition) carried upon ten to the power of 10 nano spec's of a half pint font (so definitely going to need cyber space to store such large quantities) because both wordpad and paint were built to handle poster collages on a co-lasso scale because in a single 72 pint font character it can be stored one trillion billion nano spec's of a 0.01 pint font thus significantly reducing the size off the known universe with the accumulation of a googolplex units in quantity which is the equivalent of trying to find a needle in a hay stack. Another great googolplex units retrieval method know to have worked reflective imagery which tackles the high unification of numbers to the power simply by reflecting it'self, https://plus.google.com/photos/104235935005231527262/albums/5773073029403960097

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