20170923, 10:59  #45 
Sep 2017
2^{3}·3 Posts 
dear sir if you don t mind can you pm your skype id than i give you more exmpel..

20170923, 13:26  #46 
Feb 2017
Nowhere
2×13×137 Posts 
Curiously, the OP's original question was posted to a Wolfram Stackexchange chat room the day before it was posted to this august Forum.
The user zhk had identified the trouble as an overflow problem. I also found an old thread in this Forum with exactly the same kind of calculation as in this one. Major difference: When asked, the OP in that thread described precisely what he was trying to do and why: It's possible to calculate an unknown RSA modulus? 
20170923, 13:43  #47 
Sep 2017
2^{3}·3 Posts 
yes DR sir my calculating string same like this:http://www.mersenneforum.org/showthread.php?t=13524

20170923, 14:36  #48  
"Robert Gerbicz"
Oct 2005
Hungary
2·3·5·47 Posts 
Quote:
Code:
m1^e==c1 mod N [1] m2^e==c2 mod N [2] And why would anybody solve this problem where the original message is KNOWN , these are m1 and m2. Or there are other c values, where m is unknown. 

20170923, 15:00  #49 
Feb 2017
Nowhere
2×13×137 Posts 
We've repeatedly told the OP "The numbers are too big"
Repeatedly ignored. 
20170923, 16:25  #50 
Sep 2009
2^{3}·239 Posts 
Could it be done by calculating everything mod some suitable prime larger than N, eg 2^5211 (assuming N is RSA512)? Or mod several smaller primes and applying the Chinese Remainder Theorem to recover N?
Chris 
20170925, 13:15  #51 
Sep 2017
2^{3}·3 Posts 
Dear all.
my string same like this N modul recover.. 
20170925, 13:20  #52 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 

20200813, 04:15  #53 
Sep 2017
2^{3}×3 Posts 
no body try to solve this issu overflow?

20200813, 06:50  #54 
Sep 2002
Database er0rr
2^{2}·3·7·41 Posts 
Arbitrary precision arithmetic is a solved problem.

20200813, 10:45  #55 
Sep 2017
2^{3}×3 Posts 
dear sir,
your send link not solve the probleam its all ready solve like old exp65537 5 digt but new exp12 digit can it possibel with exp calculating start? 
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