Question

Use extended Euclid algorithm to find the multiplicative inverse of 27 modulo n, if it exists, for n = 1033 and 1035. Show the details of computations.

Answer #1

a. Show that if a has a multiplicative inverse modulo N,then
this inverse is unique (modulo N).
b. How many integers modulo 113 have inverses? (Note: 113 =
1331.)
c. Show that if a ≡ b (mod N) and if M divides N then a ≡b (mod
M).

Using the extended Euclidean algorithm, find the multiplicative
inverse of a. 135 mod 61 b. 7465 mod 2464 c. 42828 mod 6407

Q1. Using Euclideanalgorithm find GCD(21, 1500). Show you
work
.Q2. Using Extended Euclidean algorithm find the multiplicative
inverse of 8 in mod 45 domain .Show your work including the
table.
Q3. Determine φ(2200). (Note that 1,2,3,5, 7, ... etc.are the
primes). Show your work.
Q4. Find the multiplicative inverse of 14 in GF(31) domain using
Fermat’s little theorem. Show your work
Q5. Using Euler’s theorem to find the following exponential:
4200mod 27. Show how you have employed Euler’s theorem here

We say that x is the inverse of a, modulo n, if ax is congruent
to 1 (mod n). Use this definition to find the inverse, modulo 13,
of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.
Show by example that when the modulus
is composite, that not every number has an inverse.

Use extended Euclidean algorithm to find x and y such that 20x +
50y = 510.

a) Use the Euclidean Algorithm to find gcd(503, 302301).
(b) Write gcd(503, 302301) as a linear combination of 38 and
49.
(c) What is an inverse of 503 modulo 302301?
(d) Solve 503x ≡ 2 (mod 302301)

Use the Euclidean Algorithm to find the smallest nonnegative
integer x, s.t. there exists integers k_1, k_2, such that x = 11k_1
+ 3 and x = 49k_2 + 5.

For solving the problems, you are required to use the following
formalization of the RSA public-key cryptosystem.
In the RSA public-key cryptosystem,
each participants creates his public key and secret key according
to the following steps:
· Select two very large
prime number p and q. The number of bits needed to represent p and
q might be 1024.
· Compute
n = pq
(n) = (p – 1) (q – 1).
The formula for (n) is owing to...

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