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- Xref: sparky sci.crypt:5787 alt.security.pgp:236
- Path: sparky!uunet!cs.utexas.edu!news
- From: orb@cs.utexas.edu (Norman Richards)
- Newsgroups: sci.crypt,alt.security.pgp
- Subject: Re: RSA Question (was Re: PKP/RSA comments on PGP legality)
- Date: 17 Dec 1992 10:59:55 -0600
- Organization: CS Dept, University of Texas at Austin
- Lines: 20
- Message-ID: <lj1cgbINNemn@kara-kum.cs.utexas.edu>
- References: <1992Dec14.190615.13954@macc.wisc.edu> <PHR.92Dec16022546@napa.telebit.com> <r09LrAXKBh107h@nadir.uucp>
- NNTP-Posting-Host: kara-kum.cs.utexas.edu
-
- In article <r09LrAXKBh107h@nadir.uucp> crespin@nadir.uucp (Rudy Crespin) writes:
- >Same number X^3(mod, 391) = 133
- >(phi(N) is not divisible by 3) What is my number?
- >
- >I tried to figure out t, but got stumped.
- >Thanks again,
- >Rudy Crespin
-
- since n=391 we can find n=p*q=17*23
- (p-1)*(q-1)=352
- you gave e=3 so my guess is d=235 since 3*235 mod 352 = 1
-
- 133^235 mod 391 = 27 , so that is what I got
- to check, 27^3 mod 391 = 133 so all seems well.
-
- So, is X=27?
- _______________________________________________________________________________
- orb@ccwf.cc.utexas.edu "Two roads diverged in a wood, and I -
- I chose to climb the nearest tree.
- And that has made all the difference."
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