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  • Category : Crack Hack
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  • Update : 2015-04-29
  • Size : 9kb
  • Downloaded :0次
  • Author :eng.yass********
  • About : Nobody
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Introduction - If you have any usage issues, please Google them yourself
We begin with choosing two random large distinct primes p and q. We also pick e, a random integer that is relatively prime to (p-1)*(q-1). The random integer e is the encryption exponent. Let n = p*q. Using Euclid s greatest common divisor algorithm, one can compute d, the decryption exponent, such that: e*d = 1 (mod (p-1)*(q-1)) Both plaintext m and ciphertext c should be in the set of nonnegative integers. Furthermore, before encrypting a plaintext message m, we need to make sure that 0 <= m < n. If m is greater than the modulus n, the result c of the encryption will not be a unique one-to-one mapping m to c. From one of the theorems of Euler s, we know that for all integers m, med = m (mod n)
Packet file list
(Preview for download)


RSA.frm
frmView.frm
frmView.frx
Project.vbp
Project.vbw
RSA.bas
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