Question #232259

Find solution of:

a)4x4x is identical to 3 mod 7

b)9x9x is identical to 11 mod 26

c)8x8x is identical to 6 mod 14

d) 8x8x is identical to 6 mod 422


1
Expert's answer
2022-02-21T14:39:48-0500

a)

Since, 4x3(mod7)4x≡3(mod 7)

i.e., 4x3=7k4x−3=7k for kI.k∈I.

4x=7k+3⇒4x=7k+3

The values 6 and 13 satisfy this equation (when k=3k=3 and k=7k=7 ).

Answer {6,13}\{6, 13\}


b)

Since, 9x11(mod26)9x≡11(mod 26)

i.e., 9x11=26k9x−11=26k for kI.k∈I.

9x=26k+11⇒9x=26k+11

The values 7 and 33 satisfy this equation (when k=2k=2 and k=11k=11 ).

Answer {7,33}\{7, 33\}


c)

Since, 8x6(mod14)8x≡6(mod 14)

i.e., 8x6=14k8x−6=14k for kI.k∈I.

8x=14k+6⇒8x=14k+6

4x=7k+3⇒4x=7k+3

The values 6 and 13 satisfy this equation (when k=3k=3 and k=7k=7 ), 

while 8,14 and 16 do not.

Answer {6,13}\{6, 13\}


d)

Since, 8x6(mod422)8x≡6(mod 422)

i.e., 8x6=422k8x−6=422k for kI.k∈I.

8x=422k+6⇒8x=422k+6

4x=211k+3⇒4x=211k+3

The values 159 and 370 satisfy this equation (when k=3k=3 and k=7k=7 ), 

Answer {159,370}\{159, 370\}


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