Let $f$ be a function on the set on non-negative integers and taking values in the same set. Suppose we are given that
$$x-f(x)=19\left\lfloor \frac{x}{19}\right\rfloor-90\left\lfloor\frac{f(x)}{90}\right\rfloor$$ for all non-negative integers $x$.
$1900<f(1900)<2000$
Find all possible values of $f(1900)$.
In solution given, they have found $f(1990)$ and I just want to check if my solution for $f(1900)$ is correct or not.
My solution:
Let $f(1900)=a$. Substitute $x=1900$ in the first relation. We have: $$1900-a=19\left\lfloor\frac{1900}{19}\right\rfloor-90\left\lfloor\frac{a}{90}\right\rfloor$$. Simplifying, $$\frac{a}{90}=\left\lfloor\frac{a}{90}\right\rfloor$$ Now, $x=\lfloor x\rfloor$ only when $x$ is an integer. Hence, $a=90k$ where $k$ is an integer.
Now $90k$ must lie between $1900$ and $2000$ giving $k=22$ and $f(1900)=1980$.
Am I correct?
Both the simplification and the answer are correct.