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seen Feb 5 at 22:40

Feb
2
comment Proof for exponentiation in modular arithemtic
Thank you very much, now it all makes sense.
Feb
2
accepted Proof for exponentiation in modular arithemtic
Feb
2
comment Proof for exponentiation in modular arithemtic
How do you know it is true for $e = k-1$?
Feb
2
asked Proof for exponentiation in modular arithemtic
Feb
2
comment Avoiding overflow when encrypting with RSA
I really don't understand this answer, what is $\mathbb{Z}/n\mathbb{Z}$, and what is meant by machine words?
Feb
1
asked Encryption of a number with RSA bigger than $n$
Jan
30
comment How do institutions like banks do RSA with big primes?
@CodesInChaos Well, lets say the bank's computer has to calculate (a^b) mod n. Where a is some number, could be 065066067, which is ASCII for ABC and b is their 255 digit long decryption key, lets just say it's 10^254, that means, they have to do 065066067*065066067*065066067... 10^254 times, if their computer's processor can do 5 billion multiplications per second, then it'd take (10^254)/(10^9) seconds, which is 3.2*10^237 years
Jan
30
awarded  Editor
Jan
30
comment How do institutions like banks do RSA with big primes?
Yes, thank you.
Jan
30
revised How do institutions like banks do RSA with big primes?
deleted 1 characters in body
Jan
30
comment How do institutions like banks do RSA with big primes?
Are you sure that's it? I mean, decrypting some data with a decryption key with 255 ciphers, seems rather extreme.
Jan
30
asked How do institutions like banks do RSA with big primes?
Jan
30
accepted Usage of Fermat primes in RSA
Jan
29
asked Usage of Fermat primes in RSA
Jan
28
accepted Avoiding overflow when encrypting with RSA
Jan
27
comment Avoiding overflow when encrypting with RSA
How does that work?
Jan
27
comment Avoiding overflow when encrypting with RSA
But m^e will be a huge number though
Jan
27
asked Avoiding overflow when encrypting with RSA
Jan
27
awarded  Scholar
Jan
27
accepted Method to calculating e in RSA