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Apr 13th, 2003, 01:00 PM
#1
Thread Starter
Hyperactive Member
Modulus Question
Sorry if I sound stupid asking this, but what is meant by the expression "1 (mod p)", where p is greater than one? I'm sure it must make more sense than I can figure out.
If the context is important, I'm reading up on RSA, quote's here:
Let p and q be two different large primes. A large number p can be tested for primeness by applying Fermat's theorem, which states that if p is prime, then
a^p-1 == 1 (mod p)
for every positive integer a not divisible by p. (A double equal sign (==) is used instead of the usual symbol for congruence, which is not supported by all Web browsers.) A number p which passes this test for several hundred randomly selected values of a in the range (1, p-1) is almost surely prime.
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Apr 13th, 2003, 04:46 PM
#2
transcendental analytic
the rest after division with p, for instance 13=3 (mod 5)
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Apr 16th, 2003, 10:50 AM
#3
Fanatic Member
x mod y = z
3 mod 2 = 1
4 mod 2 = 0
X= its the amout between X and the nearest multiple of Y
Don't pay attention to this signature, it's contradictory.
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Apr 18th, 2003, 05:15 PM
#4
Isn't 1 mod P where P>1 always going to be 1?
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Apr 18th, 2003, 07:29 PM
#5
Fanatic Member
for integers a,p,q, a>=0, p>1
q= a (mod p)
means q can be represented as k*p+a, where k is some integer, 0<=a<p
Massey RuleZ! ^-^__  Cheers!  __^-^ Massey RuleZ!
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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