# sum of digits of $k$ in given expression.

The number of $$n,$$ such that $$1991$$ is the minimum value of $$\displaystyle k^2+\lfloor \frac{n}{k^2} \rfloor\;,$$ where $$k$$ ranges over

all positive integers, is $$l,$$ Then sum of digits of $$l$$ is (where $$\lfloor x \rfloor$$ represent floor of $$x$$.)

I did not understand how can i solve it, Help me

Thanks

• How can there be $k$ such values for $n$ when $k$ ranges over all positive integers in the expression itself? (I think you have two different variables with the same name here, and you should fix it.) – Arthur Nov 11 '16 at 11:59

First, if $n\leq 0$ then there is no minimum and the problem is symmetric in $k$, so we need consider only positive $n$ and $k$.
Second, note that you can put the $k^2$ inside the floor function. So we need to think about the minimum value of $$\left\lfloor k^2+\frac{n}{k^2}\right\rfloor.$$
Use calculus to minimize $f(x) = x^2 + n/x^2$ to find the critical point at $x=\sqrt[4]{n}$ and $f(\sqrt[4]{n}) = 2\sqrt{n}$.
So we need to find out how many values of $\lfloor 2\sqrt{n} \rfloor =1991$.
This reduces to $991021 \leq n < 992016$. That gives $955$ values for $n$, and the sum of the digits is $19$.