Question

Find the general solution of the congruence 98x – 1 \[ \equiv \] 0 ( mod 139) .

Answer 1

Hint: We know that if a \[ \equiv \] b ( mod m) i.e. if a is congruent to b , mod m the statement says that a, when divided by m gives a remainder of b or you can say (a-b is divisible by m). Apply this condition to the given statement, then we will get an equation with a single variable.

Complete step by step answer:
The congruence ax\[ \equiv \]b ( mod c ) has a solution for any b when gcd ( a , c ) = 1 .
This is obvious as Euclid tells us that whenever gcd ( a , c ) = 1 . we have \[a{x_0}\]+ \[c{y_0}\]= 1 for some integers \[{x_0}\] and \[{y_0}\].
So we can say that \[ab{x_0}\]+ \[bc{y_0}\]= b so by placing x = b\[{x_0}\]we have the solution .
From this we can say that
The congruence of 98x – 1 \[ \equiv \] 0 ( mod.139) . that means 98x – 1 is divisible by 139 .
Therefore we can write dividend = ( divisor x quotient ) + remainder .
Therefore the above one we can write it as
98x – 1 = 139q where q is the quotient (let us assume) .
We can write it as 98x – 139q = 1 .
Convert \[\dfrac{{139}}{{48}}\] into a continued fraction we get the fraction is just preceding the fraction \[\dfrac{{61}}{{43}}\].
From Euclid’s algorithm, we can say that
      1 = 61 x 98 – 43 x 139
Hence the solution are in form 61 + 139t where t is an integer.
X = 61 + 139t , y = 43 + 98t .

Note:
We can also do the same question in another method. As mentioned in hint We can use that formula to solve the problem. And by taking various values of t in the answer we got, we can find many numbers that can satisfy the given condition.