[sf-lug] Ex 11.7 Exponentiation (understandable) & RSA algorithm (not so!)
jim at well.com
Mon Jan 12 10:13:21 PST 2009
i'm sorry too: i still don't get it. for their
3 * x = 1 (mod 11)
their solution is 4.
i read this as x = 4, and therefore
3 * 4 = 1 (mod 11)
assuming i'm reading correctly, the mystery is
how to understand 1(mod 11) == 12
i read 1(mod 11) as 1%11 == the remainder after
11 is divided into 1: integer arithmetic 1/11 == 0
and 1%11 yields 1.
if my assumption is incorrect, then the mystery is
3 * x = 1(mod 11)
On Mon, 2009-01-12 at 09:09 -0800, Asheesh Laroia wrote:
> On Mon, 12 Jan 2009, jim wrote:
> > i'd like verification of my understanding of this:
> > "Let d be the reciprocal of e mod r."
> > my arithmetic:
> > if e is 35 and r is 9111,
> > then d = 1/(35%9111) ## == 1/35 == 0.028571429
> No, sorry! Reciprocal isn't defined that way in modulo arithmetic. I'm
> glad you asked.
> Quoth Wikipedia: "In modular arithmetic, the modular multiplicative
> inverse of x is also defined: it is the number a such that a*x ≡ 1 (mod
> n). This multiplicative inverse exists if and only if a and n are coprime.
> For example, the inverse of 3 modulo 11 is 4 because it is the solution to
> 3*x ≡ 1 (mod 11). The extended Euclidean algorithm may be used to compute
> In modular arithmetic, 1/a mod b is an *integer* (iff a and b are
> relatively prime).
> http://www.mtholyoke.edu/~mpeterso/Applets/CalculatorApplet.html . I
> list member can.
> -- Asheesh.
> You're definitely on their list. The question to ask next is what list it is.
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