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#1 |
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Random Account
Aug 2009
32·7·31 Posts |
Is there a formula for calculating the number of digits in a large number?
An example would be 218477
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#2 |
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Jul 2003
So Cal
2×34×13 Posts |
Yes, take the base-10 log, and round up.
log (2^18477) = 18477 log 2 = 18477 (0.30103) = 5562.1 so 2^18477 has 5563 digits. |
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#3 |
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Random Account
Aug 2009
32×7×31 Posts |
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#4 | |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts |
Quote:
The "LOG" button on your calculator is (almost certainly, anyway - just try log(1000) or something to check) the base 10 logarithm. I doubt you'll be able to replicate the first step on your calculator, since if it's anything like my TI, it has a limit of 100 digits. You can, however, replicate the rest. Note that 0.30103 is not exactly log_10(2), but is close enough for this example. If you did need the base x logarithm of y, you could use log(y)/log(x). (with log being the logarithm of any base, surprising as that may sound )
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#5 |
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Random Account
Aug 2009
111101000012 Posts |
On my calculator if I enter 1000 and press Log, it returns 3, meaning 103
Entering 2 and pressing Log gives me .301029996. This is the rounded number in his example. Multiply that times 18477 and i get 5562.1. It seems to work. On the GIMPS home page is 242643801. If I do this like above, then I get 12,837,063.2 |
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#6 |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts |
Yep, you've got it.
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#7 |
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Aug 2004
Melbourne, Australia
23×19 Posts |
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#8 |
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Aug 2006
3·1,993 Posts |
Yes, the 'right' way is to round down and add one. But usually you'd know if you were dealing with a power of 10.
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#9 |
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Aug 2004
Melbourne, Australia
15210 Posts |
Hmmm... this should be on the Mersenne Wiki.
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#10 |
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
224058 Posts |
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#11 |
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Random Account
Aug 2009
195310 Posts |
http://www.mersenne.org/bench.htm
This link is broken. It is in the last post of the thread above. Understandable. All of this is from 2004. |
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