mersenneforum.org Shortcuts on large computations
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 2005-08-28, 22:18 #1 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 11011111102 Posts Shortcuts on large computations Alright, quick question everyone. Is there a way to find the last few digits and the first few digits of a large number in exponent form, i.e. 2^2345343? I know that finding the last few digits would be easier, but finding more than the last 3 (as far as my mathematical knowledge goes, would take a very long time). And as for finding the first, say, 10 digits of the number...I have no idea how to do that. Is there an easy, fast way to find the first and last 10 digits of such a number, with any base? Thanks.
2005-08-28, 22:26   #2
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by Primeinator Alright, quick question everyone. Is there a way to find the last few digits and the first few digits of a large number in exponent form, i.e. 2^2345343? I know that finding the last few digits would be easier, but finding more than the last 3 (as far as my mathematical knowledge goes, would take a very long time). And as for finding the first, say, 10 digits of the number...I have no idea how to do that. Is there an easy, fast way to find the first and last 10 digits of such a number, with any base? Thanks.
Finding the last k digits is easy, but requires knowing some elementary
number theory. Just compute your number mod 10^k for your chosen k.

Finding the first few digits is also easy and requires nothing more than
second year high school algebra. Hint: "think logarithms". Of course, one
needed to compute the log to sufficient accuracy, but that is easy.
[but it does require some knowledge of infinite series or equivalent]

 2005-08-29, 20:19 #3 ewmayer ∂2ω=0     Sep 2002 República de California 2·13·443 Posts For the least-significant D decimal digits, compute your desired number using modular binary exponentiation, with modulus 10^D.
 2005-09-03, 18:56 #4 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 11011111102 Posts Thanks :)

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