a)
Since, 4x≡3(mod7)
i.e., 4x−3=7k for k∈I.
⇒4x=7k+3
The values 6 and 13 satisfy this equation (when k=3 and k=7 ).
Answer {6,13}
b)
Since, 9x≡11(mod26)
i.e., 9x−11=26k for k∈I.
⇒9x=26k+11
The values 7 and 33 satisfy this equation (when k=2 and k=11 ).
Answer {7,33}
c)
Since, 8x≡6(mod14)
i.e., 8x−6=14k for k∈I.
⇒8x=14k+6
⇒4x=7k+3
The values 6 and 13 satisfy this equation (when k=3 and k=7 ),
while 8,14 and 16 do not.
Answer {6,13}
d)
Since, 8x≡6(mod422)
i.e., 8x−6=422k for k∈I.
⇒8x=422k+6
⇒4x=211k+3
The values 159 and 370 satisfy this equation (when k=3 and k=7 ),
Answer {159,370}
Comments