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

Daniaal Sanchez

Daniaal Sanchez

Answered question

2021-02-25

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

Answer & Explanation

Nathalie Redfern

Nathalie Redfern

Skilled2021-02-26Added 99 answers

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(mod 660).
From the properties of modulo.
a(mod c) is equivalent to a mod c=b mod c.
Here a=41x, b=125 and c=660.
Then, 41x mod 660=125 mod 660.
Now, the inverse of 41 on both sides of the above equation.
x mod 660=161125 mod 660 ( From part a. inverse.of 41mod,660 is 161)
x mod 660=20125 mod 660
x mod 660=325 (Since 20125=325 mod 660)
Therefore, the least positive solution for the congruence 41x=125(mod 660) is 325.
Conclusion:
The least positive solution for the congruence 41x=125(mod 660) is 325.

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