### Instruction: For how many values of $c$ in the interval $[0, 1000]$ does the equation \[7 \lfloor x \rfloor + 2 \lceil x \rceil = c\]have a solution for $x$?### Response: I notice that the equation involves the floor and ceiling functions, which round down and up to the nearest integer respectively.
I wonder what happens if I try some values of $x$ and see what $c$ I get.
For example, if $x = 0$, then $c = 0$; if $x = 1$, then $c = 9$; if $x = 2$, then $c = 18$; and so on.
I observe a pattern: every time $x$ increases by $1$, $c$ increases by $9$.
This makes sense, because the floor and ceiling functions increase by $1$ as well, and their coefficients are $7$ and $2$, so $7 + 2 = 9$.
I realize that this means that $c$ is always a multiple of $9$, and the smallest and largest possible values of $c$ are $0$ and $999$ respectively.
So I can find the number of values of $c$ by dividing the interval length by $9$ and adding $1$.
I calculate the number of values of $c$ as $\frac{999 - 0}{9} + 1 = 111 + 1 = 112$.