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?

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

1 Answer 1


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 $\ldots$)

  • $\begingroup$ (+1) This was the method by which I solved it... anyways thanks ;-)!! $\endgroup$
    – Soham
    Oct 20, 2018 at 18:57
  • $\begingroup$ Please go ahead and write the general scheme. It many help future readers! $\endgroup$
    – Soham
    Oct 22, 2018 at 11:06

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