Answer to Question #232259 in Combinatorics | Number Theory for paa

Question

Find solution of:

a)$4x$ is identical to 3 mod 7

b)$9x$ is identical to 11 mod 26

c)$8x$ is identical to 6 mod 14

d) $8x$ is identical to 6 mod 422

Answer 1

a)

Since, $4x≡3(mod 7)$

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, $9x≡11(mod 26)$

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, $8x≡6(mod 14)$

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, $8x≡6(mod 422)$

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 $\{159, 370\}$