Daniaal Sanchez

2021-02-25

Find the least positive solution for the following congruence: $41x=125(mod\text{}660)$ .

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$41x=125(mod\text{}660)$ .

From the properties of modulo.

$a\equiv (mod\text{}c)$ is equivalent to a mod $c=b$ mod c.

Here$a=41x,\text{}b=125\text{}and\text{}c=660$ .

Then,$41x\text{}mod\text{}660=125\text{}mod\text{}660$ .

Now, the inverse of 41 on both sides of the above equation.

$x\text{}mod\text{}660=161\ast 125\text{}mod\text{}660$ ( From part a. inverse.of 41mod,660 is 161)

$x\text{}mod\text{}660=20125\text{}mod\text{}660$

$x\text{}mod\text{}660=325$ (Since $20125=325\text{}mod\text{}660$ )

Therefore, the least positive solution for the congruence$41x=125(mod\text{}660)$ is 325.

Conclusion:

The least positive solution for the congruence$41x=125(mod\text{}660)$ is 325.

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.

Here

Then,

Now, the inverse of 41 on both sides of the above equation.

Therefore, the least positive solution for the congruence

Conclusion:

The least positive solution for the congruence

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