First, you’re right in thinking that it’s essentially the same idea. It’s also basically correct, but it could stand a bit of reorganization. Specifically, it would be better to begin by applying the division algorithm to $\lfloor x\rfloor$ and $d$. And if I were giving a proof more detailed than the one in the book, I’d probably also go into a bit more detail about the manipulation of the floor function. Making those changes but otherwise trying to stay as close as possible to your basic approach, I might end up with something like this:
The positive integers less than or equal to $x$ are precisely the integers $1,2,\ldots,\lfloor x\rfloor$. By the division algorithm there are unique integers $q$ and $r$ such that $\lfloor x\rfloor=dq+r$, and $0\le r<d$, so we can split these $\lfloor x\rfloor$ positive integers into $q$ blocks of $d$ integers each, possibly with a short block of $r$ integers left over: $$\begin{align*}&1,2,\ldots,d,\\&d+1,d+2,\ldots,2d,\\
&2d+1,2d+2,\ldots,3d,\\&\qquad\qquad\vdots\\&(q-1)d+1,(q-1)d+2,\ldots,qd,\\&qd+1,\ldots,qd+r\end{align*}$$ The multiples of $d$ here are the integers at the ends of the first $q$ lines, namely, $d,2d,3d,\ldots,qd$. Clearly there are $q$ of them, so we need to show that $q=\left\lfloor\frac{x}d\right\rfloor$.
Let $\alpha=x-\lfloor x\rfloor$, the fractional part of $x$; of course $0\le\alpha<1$. Then $0\le r+\alpha<d$, so $$\left\lfloor\frac{x}d\right\rfloor=\left\lfloor\frac{dq+r+\alpha}d\right\rfloor=\left\lfloor q+\frac{r+\alpha}d\right\rfloor=q+\left\lfloor\frac{r+\alpha}d\right\rfloor=q\;,$$ as desired.
Best Answer
In general, if an integer is divisible by $a_1,a_2,\ldots,a_k$, then the integer is divisible by $\text{lcm}(a_1,a_2,\ldots,a_k)$. The proof of this claim immediately follows from the fact that if $a \vert bc$ and $\gcd(a,b) = 1$, then $a \vert c$, which follows immediately from the definition of $\gcd$.
Hence, in your case, the integer is divisible by $120$.