SOLUTION: Let $x$ and $y$ be integers. If $x$ and $y$ satisfy $41x + 5y = 31$, then find the residue of $x$ modulo $5$.

You are given this equation

    41x + 5y = 31,    (1)

where x and y are integers.  Hence

    41x = 31 - 5y.    (2)


They want you find residue  {x mod 5} from modular equation (2).


It means that you consider equation (2)  as an equivalence by modulo 5.
If so, you may forget about the term -5y, because it is 0 (zero) by the modulo 5.


So, your task is to find the residue {x mod 5} from the modular equation

    41x = 31 mod 5,

which is equivalent to 

    41x = 1  mod 5.    (3)


The set of possible solutions is  x = {0 mod 5}, {1 mod 5}, {2 mod 5}, {3 mod 5} and {4 mod 5}.


    x = {0 mod 5} turns (3)  into  41*0 = 1 mod 5,  which is obviously wrong. 

    x = {1 mod 5} turns (3)  into  41*1 = 1 mod 5,  which is obviously correct modular equation ( since 41 mod 5 = 1 mod 5). 

    x = {2 mod 5} turns (3)  into  41*2 = 1 mod 5,  which is obviously wrong. 

    x = {3 mod 5} turns (3)  into  41*3 = 1 mod 5,  which is obviously wrong. 

    x = {4 mod 5} turns (3)  into  41*4 = 1 mod 5,  which is obviously wrong. 


Thus of five possible solutions only one is valid:  x = {1 mod 5}.    ANSWER


So, in this problem, you can easily find the answer by the brute force method.