In an RSA cryptosystem, a particular A uses two prime numbers p = 13 and q =17 to generate her public and private keys. If the e part of the public key of A is 35. Then the private key of A is?

Answers

Answer 1

The correct answer is the private key of A is (11, 221).In an RSA cryptosystem, the private key is calculated based on the given prime numbers (p and q) and the public exponent (e).

To find the private key of A, we can follow these steps:

Calculate the modulus (n):

n = p * q = 13 * 17 = 221

Calculate Euler's totient function (φ(n)):

φ(n) = (p - 1) * (q - 1) = 12 * 16 = 192

Find the modular multiplicative inverse of e modulo φ(n).

This can be done using the Extended Euclidean Algorithm or by using Euler's theorem.

In this case, e = 35.

Using the Extended Euclidean Algorithm:

35 * d ≡ 1 (mod 192)

By solving the equation, we find that d = 11.

The private key of A is (d, n):

The private key of A is (11, 221).

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Related Questions

Consider the 90Sr source and its decay chain from problem #6. You want to build a shield for this source and know that it and its daughter produce some high energy beta particles and moderate energy gamma rays. a. Use the NIST Estar database to find the CSDA range [in cm) and radiation yield for the primary beta particles in this problem assuming a copper and a lead shield. b. Based on your results in part a, explain which material is better for shielding these beta particles.

Answers

a. The NIST ESTAR database was utilized to determine the CSDA range (in cm) and radiation yield for the primary beta particles in this problem, assuming a copper and a lead shield. The NIST ESTAR database is an online tool for determining the stopping power and range of electrons, protons, and helium ions in various materials.

For copper, the CSDA range is 0.60 cm, and the radiation yield is 0.59. For lead, the CSDA range is 1.39 cm, and the radiation yield is 0.29.

b. Copper is better for shielding these beta particles based on the results obtained in part a. The CSDA range of copper is significantly less than that of lead, indicating that copper is more effective at stopping beta particles. Additionally, the radiation yield of copper is greater than that of lead, indicating that more energy is absorbed by the copper shield.

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