Let $[x$ denote the integer part of the real number $x$. Suppose $a, b, c$ are positive real numbers, then the equation $a x+b[x]-c=0 \quad(\quad)$
(A) has no solution
(B) has at most one solution
(C) has only a finite number of solutions
(D) has infinitely many solutions
3. (B).

Let $x=[x]+y, 0 \leqslant y<1$, then from the original equation we get $\left[x=\frac{c-a y}{a+b}\right.$. Since $a, b$ are positive, we have $\frac{c-a}{a-b}<[x] \leqslant \frac{c}{a+b}$. Because the length of the interval $\left(\frac{c-a}{a-b} , \frac{c}{a+b}\right)$ is less than 1, if there is an integer in this interval, there is at most one, and this number is $[x]$, in which case the equation has one solution; if there is no integer in this interval, and $\frac{c}{a+b}$ is not an integer, then the equation has no solution.