1. Find an inverse for 41 modulo 660.
2. Find the least positive solution for the following congruence: 41x=125(mod 660).

(a)

To find:

An inverse for 41 modulo 660.

Given information:

41mod660

Concept used:

a mod b = Remainder, when a is divided by b.

Calculation:

The objective is to find an inverse for 41 modulo 660.

First. Find the greatest common divisor using the quotient remainder theorem as,

41=0⋅660+41660=16⋅41+441=10⋅4+14=4⋅1+0

The greatest common divisor when we divide 41 by 660 is the last non zero remainder.

Thus, the greatest common divisor is 1.

Now, express the greatest common divisor as a linear combination of 41 and 660.

1=41−10⋅4=1⋅41−10⋅4=1⋅41−10⋅<

(b)

To find:

The least positive solution for the following congruence: 41x=125(mod660)