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This question already has an answer here:

I'm trying to run this

N[
  Sum[2/(10^(Mod[-(10^(10^8 - 1) - n^2), n]*3810217)*(10^(n*3810217) - 1)), 
    {n, 1, Floor[Sqrt[10^10^8 + 45708]]}], 
  Floor[Sqrt[10^10^8 - 10^(10^8 - 1) + 45708]]*3810217]; 
AbsoluteTiming[Flatten[Position[Partition[RealDigits[%][[1]], 3810217, 3810217, -1], {(0) .., 2}]]]

and I'm getting error messages that I have exceeded my max extra precision = 50

I've read through the documentation, and I can't figure out how to find my machine's maximum precision. I've tried:

N[MachinePrecision]

15.9546

and I know that can't be right because I've already ran toy problems and I've gotten 1000 precision easy.

I know there are ways to set precision goals , but I'd like to know what the limits are first cause I plan on running my computer to the max. If I knew the max precision, I could maybe modify my expression to come within range.

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marked as duplicate by Oleksandr R., bbgodfrey, Mr.Wizard Feb 16 '15 at 14:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ $MaxExtraPrecision is not a fixed quantity. If you want it higher (or even infinite), you only have to ask. $\endgroup$ – Oleksandr R. Feb 16 '15 at 4:48
  • $\begingroup$ I'll quote myself from a prior comment to you. "At the same time, perhaps read the Mathematica documentation, guides, tutorials, and the guides at this very site. The basic operations you've queried about are all covered within." This is not a mechanical-turk manual site... $\endgroup$ – ciao Feb 16 '15 at 7:17
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    $\begingroup$ Please read: (7564), (10624), (55292) $\endgroup$ – Mr.Wizard Feb 16 '15 at 14:28
  • $\begingroup$ @Mr.Wizard , Thanks! That straightened out a few things. Better late than never. You are the best thing on this site, and all of stackexchange. The Great and Powerful OZ . $\endgroup$ – user24719 Sep 11 '15 at 23:30
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First you must understand the difference between machine precision arithmetic (CPU floating point) and Mathematica's own arbitrary precision arithmetic.

You get the first by calling N with one argument and the second by calling N with two arguments, the second being the precision you want to maintain. When using Mathematica's own arbitrary precision arithmetic, the limit to the precision you can achieve is the size of your computer's memory.

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