Question

(a) Solve: 7X ? 1 (mod 19).

(b) Determine [7]^{–1} in
**Z**_{19}

Answer #1

a) Given 7x ? 1 (mod 19)

Here gcd(7,19) = 1. Hence the congruence has a unique solution.

Since gcd(7,19) = 1, there exist integers u,v such that 7u+19v = 1.

Here u = -8, v = 3. Therefore 7*(-8)+19*3 = 1 and this implies 7*(-8) ? 1 (mod 19).

Here x = -8 is a solution.

All solutions are x ? -8 (mod 19) ,i.e., x ? 11 (mod 19).

All the solutions are congruent to x ? 11 (mod 19) and therefore the given congruence has a unique solution.

b) [7]-1 = 7-1 = 6.

Therefore, [7]-1 in z_{19} is 6.

find the following square roots.
A. √8( mod 41)
B. √7(mod 41)
C. √7 (mod 19)
D. √7 (mod 29)
E. √5 (mod 29)

Solve (a) x^3=7 (mod 16), (b) x^3=12 (mod 27).

Solve the linear congruence
x = 2 mod (7)
x = 1 mod (3)

Determine the set of solutions of: X ? 1 (mod 81) and X ? 7 (mod
12), as the solutions to a single congruence.

1) Solve the system by substitution.
{7x−7y=−42
{y=2x+7
a) One solution:
b) No solution
c) Infinite number of solutions
2) Solve the system by substitution.
{5x+4y=27
{y=2x-16
a) One solution:
b) No solution
c) Infinite number of solutions

(a) Solve 14x == 271 (mod 11111) (b) Solve x2 == 91 (mod 77)
Note: "==" mean congruence.

Let a be an inverse of a (mod m)
a) Explain what does it mean for a to be an inverse of a (mod m)
?
b) Find the inverse of 7(mod19)
c) Solve the linear congruence 7x=3(mod19)

Solve the system of congruences
3x+4= 2 (mod 7)
x-13= 24 (mod 29)

(a) Solve
x ≡ 11 (mod 12), x ≡ 4 (mod 5), x ≡ 0 (mod 7)
(b) Find all the solutions of the following system:
x ≡ 5 (mod 6), x ≡ 4 (mod 11), x ≡ 3 (mod 17).

Find all solutions to the system:
2x ≡ 4 (mod 5)
3x ≡ 5 (mod 7)
7x ≡ 2 (mod 13)
need help with discrete math HW, please write solutions clearly,
and please don't just answer wrong solution, cus then i will need
to post the same question twice. i appreciate every help i can get
but please let someone else help me solve the question if you're
not sure about any part to avoid reposting. thanks, will rate best...

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