1971 AHSME Problems/Problem 12

Problem

For each integer $N>1$, there is a mathematical system in which two or more positive integers are defined to be congruent if they leave the same non-negative remainder when divided by N. If $69,90$, and $125$ are congruent in one such system, then in that same system, 8$1$ is congruent to

$\textbf{(A) }3\qquad \textbf{(B) }4\qquad \textbf{(C) }5\qquad \textbf{(D) }7\qquad  \textbf{(E) }8$