6. Two periodic functions $y_{1}$ and $y_{2}$ have the smallest positive periods $a$ and $b$ respectively, and $b=n a(n \geqslant 2, n$ is an integer $)$. If the function $y_{3}=y_{1}+y_{2}$ has the smallest positive period $t$, then among the following 5 scenarios:
(1) $tb$

the number of scenarios that cannot occur is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
6. B
$b$ is a period of $y_{3}$; hence $t \leqslant b$. If $t=a$, then from $y_{2}=y_{3}-y_{1}$ we get $b \leqslant a$, which is a contradiction. Therefore, (2) and (5) are impossible.
The following examples show that the other three cases can occur:
Take $y_{2}=\sin x+\sin \frac{2 x}{3}$, then $b=6 \pi$.
(1) Let $y_{1}=-\sin \frac{2 \pi}{3}$, in this case,
$$
a=3 \pi, y_{3}=\sin x, t=2 \pi, t<a \text {; }
$$
(2) Let $y_{1}=-\sin x$, in this case,
$$
a=2 \pi, y_{3}=\sin \frac{2 \pi}{3}, t=3 \pi, a<t<b \text {; }
$$
(3) Let $y_{1}=\sin x$, in this case,
$$
a=2 \pi, y_{3}=2 \sin x+\sin \frac{2 \pi}{3}, t=6 \pi, t=b .
$$