more then three times correct answers

Question

i have such kind of question,first of all let us consider following problem

The positive integers $m > n$ leave a remainder of $2$ and $3$, respectively, when divided by $6$.

we should compare remainder when $(m+n)$ is divided by $6$ and $m-n$ is divided by $6$,i have tried several values,like $14$ and $9$,$26$ and $21$,$32$ and $3$,in all case i am getting same value,if such kind of question would be given on GRE ,after several numbers plugging,i got same value,could i assume that they are equal or what could i do more?if we do it algebraicaly,we get $m=6*k+2$ and $n=6*s+3$,

$m+n=6*(k+s)+5$

$m-n=6(k-s)-1$,remainder of first is $5$,for second we have negative number and $-1$ modulus $6$ is remainder $5$,but my question basically is can i assume after three or four number plugin that i can trust given result?

Answer 1

We have $m+n\equiv 2+3\equiv 5\pmod{6}$ and $m-n\equiv 2-3\equiv -1\equiv 5\pmod {6}$. So the remainders are indeed the same.

Maybe more nicely, maybe not, $(m+n)-(m-n)=2n\equiv (2)(3)\equiv 0\pmod{6}$. So the key thing was the value of $n$ modulo $6$. We have equality modulo $6$ whatever $m$ may be.

As to whether you should plug in values, that is a reasonable strategy. When you have uncertainty, it can certainly quickly eliminate several of the multiple choice options.

Unfortunately the GRE people know this, and may sneakily make a superficially attractive option fail at values you are unlikely to try.