# Find $n$ such that $\int_1^n \lfloor{x}\rfloor\lfloor{\sqrt x}\rfloor dx>60$

Find the smallest positive integer $$n$$ such that $$\int_1^n \lfloor{x}\rfloor\lfloor{\sqrt x}\rfloor dx>60$$

where $$\lfloor.\rfloor$$ is the GIF. I couldn't find any decent method rather than explicitly evaluating it to a few terms by brute force.

Is there a better method out?

• It's not very relevant for the question at hand, but with some work the integral can be evaluated in closed form $$\frac{1}{20}\lfloor \sqrt{n}\rfloor\left(-2\lfloor\sqrt{n}\rfloor^4-5\lfloor\sqrt{n}\rfloor^3+5\lfloor\sqrt{n}\rfloor+10n^2-10n+2\right)$$ Oct 20, 2018 at 18:16

The function $$f(x):=\lfloor x\rfloor\cdot\lfloor\sqrt{x}\rfloor$$ satisfies f(x)=\left\{\eqalign{\lfloor x\rfloor\qquad&(1\leq x<4)\cr 2\lfloor x\rfloor\qquad&(4\leq x<9)\ .\cr}\right. It follows that $$\int_1^8 f(x)\>dx=1+2+3+8+10+12+14=50\>,$$ and $$\int_1^9f(x)\>dx=50+\int_8^9f(x)\>dx=66$$. The answer to your question therefore is $$9$$.
(Remark: If you had written $$6382$$ instead of $$60$$ I would have set up a general scheme $$\ldots$$)