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


  • $\begingroup$ 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.) $\endgroup$ – 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 $19$.


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