Find the least positive solution for the following congruence: 41x=125(mod 660).

Concept used:
a mod b = Remainder, when a is divided by b.
Calculation:
The objective is to find the least positive solution for the congruence $41x=125(mod660)$.
From the properties of modulo.
$a\equiv (modc)$ is equivalent to a mod $c=b$ mod c.
Here $a=41x,b=125andc=660$.
Then, $41xmod660=125mod660$.
Now, the inverse of 41 on both sides of the above equation.
$xmod660=161\ast 125mod660$ ( From part a. inverse.of 41mod,660 is 161)
$xmod660=20125mod660$
$xmod660=325$ (Since $20125=325mod660$)
Therefore, the least positive solution for the congruence $41x=125(mod660)$ is 325.
Conclusion:
The least positive solution for the congruence $41x=125(mod660)$ is 325.