Question

Need help with following maths problem

1.Solve for x in 82x ≡ 36 mod 37? so that x is between 0 and
36.
2.Solve for x in 73x ≡ 6 mod 38? so that x is between 0 and
37.

Answer 1

Explanation:

1.This is a basic linear congruence equation solving:

Convert the first equation into: 8x≡36(mod 37) which simplifies into: x≡23(mod 37), so our answer is x=23 for 0<x<36.

2. Another equation: 35x≡6(mod 38) which simplifies to: x≡36(mod 38), so our answer is x=36 for 0<x<37

Answer 2

Thus, the solution to the congruence 82x≡36(mod37), with x between 0 and 36, is x=17.

To solve the congruence 82x≡36(mod37), we'll use the properties of modular arithmetic to find the value of x within the range of 0 to 36. Firstly, we need to find the modular multiplicative inverse of 82 modulo 37. The modular multiplicative inverse of a modulo m is a number b such that ab≡1(modm). In this case, find the modular multiplicative inverse of 82 modulo 37:

Using the extended Euclidean algorithm or by trial and error, the modular multiplicative inverse of 82 modulo 37 is 9, because 82×9≡1(mod37). Now, multiply both sides of the congruence by the modular multiplicative inverse (9) to solve for

82x≡36(mod37)

9×82x≡9×36(mod37)

x≡324(mod37)

To ensure x is between 0 and 36, divide 324 by 37:

x≡324(mod37)

x≡17(mod37)

Thus, the solution to the congruence 82x≡36(mod37), with x between 0 and 36, is x=17.

Complete Question:

Solve for x in 82x ≡ 36 mod 37? so that x is between 0 and 36.