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

Find the least positive solution for the following congruence: .
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Nathalie Redfern
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 .
From the properties of modulo.
is equivalent to a mod $c=b$ mod c.
Here .
Then, .
Now, the inverse of 41 on both sides of the above equation.
( From part a. inverse.of 41mod,660 is 161)

(Since )
Therefore, the least positive solution for the congruence is 325.
Conclusion:
The least positive solution for the congruence is 325.