smc

Consider each option, one at a time. For option A, let $x=y-1$ and $z=y+1$. That means $3y=46$, so $y=\frac{46}{3}$. That is not an integer, so option A is eliminated. For option B, let $x=y-2$ and $z=y+2$. That also means $3y=46$, so $y=\frac{46}{3}$. That is also not an integer, so option B is eliminated. For option C, let $y=x+1$, $z=x+2$, and $w=x+3$. That means $4x+6=46$, so $x=10$. Option C works. For options D and E, let $y=x+2$, $z=x+4$, and $w=x+6$. That means $4x+12=46$, so $x=\frac{17}{2}$. Since the result is not an integer, options D and E are eliminated. Thus, the answer is $\boxed{\text{II can be solved in consecutive integers}}$.