Solve for x, verify by division.

Question

a. 3 𝑥 ≡ 1 (𝑚𝑜𝑑 11)

𝑏. 162𝑥 ≡ 1 (𝑚𝑜𝑑 5)

Answer 1

a) 3𝑥 ≡ 1 (𝑚𝑜𝑑 11) => 4*3 𝑥 ≡ 4*1 (𝑚𝑜𝑑 11) => 12𝑥 ≡ 4 (𝑚𝑜𝑑 11) => 11𝑥+x ≡ 4 (𝑚𝑜𝑑 11) => x ≡ 4 (𝑚𝑜𝑑 11), i.e.

x=4+11*k with k being any integer. Verification is straightforward: 3*(4+11*k)-1=11*(k+1) and 11 divides 11*(k+1) .

b) 162𝑥 ≡ 1 (𝑚𝑜𝑑 5) => 160x+2x ≡ 1 (𝑚𝑜𝑑 5) => 5*2*16x+2x ≡ 1 (𝑚𝑜𝑑 5) => 2x ≡ 1 (𝑚𝑜𝑑 5) =>

=> 3*2x ≡ 3*1 (𝑚𝑜𝑑 5) => 5x+x ≡ 3 (𝑚𝑜𝑑 5) => x ≡ 3 (𝑚𝑜𝑑 5), i.e. x=3+5*m with m being any integer(verification works in the same manner as in a) ).

Remark: it's not quite clear whether you are considering the two equations separately, or they represent a system of equations. If it is a system of equations then the solution would be the following:

from a) we got x=4+11*k, k-any integer, and from b) we got x ≡ 3 (𝑚𝑜𝑑 5) => 4+11*k ≡ 3 (𝑚𝑜𝑑 5) =>

11*k ≡ 4 (𝑚𝑜𝑑 5) => k ≡ 4 (𝑚𝑜𝑑 5) => k=4+5*m, m-any integer, and substituting this k to the solution for a) we obtain the solution of the system: x=4+11*(4+5*m) => x=48+55m (equivalently x=48 (mod 55) ).