We will prove a more general result about the solutions of equations of the type $\lfloor x\rfloor+\lfloor 2x\rfloor+...+\lfloor nx\rfloor = n+1$. First, we look at the case $n=3k$. We denote the term on the LHS by $f(k,x)$. We first see that for any fixed $k$, $f(k,x)$ is increasing in $x$. Also, $f(k,\frac{1}{k})=k+k+k+3=3k+3 > 3k+1$ by trivial calculation. Let's look at $f(k, \frac{1}{k}- \epsilon)$. For small $\epsilon$, almost all the terms of $f(k, \frac{1}{k}-\epsilon)$ are the same as the corresponding ones of $f(k,\frac{1}{k}$ except for $\lfloor kx\rfloor, \lfloor 2kx\rfloor$ and $\lfloor 3kx\rfloor$ who are one less than before. Hence, for sufficiently small $\epsilon$ we have $f(k, \frac{1}{k}- \epsilon)=3k<3k+1$. Hence, there can't be any real $x$ such that $f(k,x)=3k+1$ which also solves part b). In a similar way, one can prove that the equation has solutions when $n$ is not divisible by 3. They are given by $\frac{2}{2k+1} \le x < \frac{3}{3k+1}$ in the case $n=3k+1$ and $\frac{2}{2k+2} \le x < \frac{3}{3k+2}$ in the case $n=3k+2$. The special case for $k=670$ (part a)) now follows easily...