[sf-lug] Ex 11.7 Exponentiation (understandable) & RSA algorithm (not so!)
Asheesh Laroia
asheesh at asheesh.org
Mon Jan 12 10:31:40 PST 2009
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)
Agreed.
> 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. (-:
--
You have a will that can be influenced by all with whom you come in contact.
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