[sf-lug] Ex 11.7 Exponentiation (understandable) & RSA algorithm (not so!)
jim at well.com
Mon Jan 12 15:08:10 PST 2009
i assume the crazy world of mod, which is
not an operator, is part of integer-land.
extrapolating, i take it that 3 * 8 is 2.
if the above is correct, maybe i get it.
On Mon, 2009-01-12 at 10:31 -0800, Asheesh Laroia wrote:
> On Mon, 12 Jan 2009, jim wrote:
> > i'm sorry too: i still don't get it. for their
> > example:
> > 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
> You place the (mod 11) in the wrong place, I think.
> 12 = 1 (mod 11)
> means, "In the crazy world we call (mod 11), twelve equals one."
> (mod 11) is not an operator, it's just the name of a mathematical world.
> > 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.
> But I read it as:
> "One (in the crazy world we call mod 11)".
> > if my assumption is incorrect, then the mystery is
> > 3 * x = 1(mod 11)
> "In the crazy world we call (mod 11), three times what equals one?"
> "Well, in that world, three times four equals twelve, and twelve equals
> one. Three times four equals one, in that world, so we know we can
> substitute four and x. Therefore, x equals four."
> Does that make more sense?
> -- Asheesh.
> P.S. You're becoming a computer scientist now, not a programmer. (-:
More information about the sf-lug