[Thgx0312555]

Day II ข้อ 15
พิจารณาทีละ $\left\lfloor\,\sqrt{n}\right\rfloor $

Number of integer n possible
= $\left\lfloor\,\frac{3}{1}\right\rfloor -\left\lfloor\,\frac{3}{2}\right\rfloor +\left\lfloor\,\frac{8}{2}\right\rfloor - \left\lfloor\,\frac{8}{3}\right\rfloor + \left\lfloor\,\frac{15}{4}\right\rfloor - ... - \left\lfloor\,\frac{1935}{44}\right\rfloor + \left\lfloor\,\frac{2012}{44}\right\rfloor $

= $(\left\lfloor\,\frac{3}{1}\right\rfloor +\left\lfloor\,\frac{8}{2}\right\rfloor + \left\lfloor\,\frac{15}{4}\right\rfloor + ... + \left\lfloor\,\frac{1935}{43}\right\rfloor + \left\lfloor\,\frac{2012}{44}\right\rfloor) - (\left\lfloor\,\frac{3}{2}\right\rfloor + \left\lfloor\,\frac{8}{3}\right\rfloor + ... + \left\lfloor\,\frac{1935}{44}\right\rfloor )$

= $\sum_{i = 1}^{43}\left\lfloor\,\frac{i\times i+2i}{i}\right\rfloor + 45 - \sum_{i = 2}^{44}\left\lfloor\,\frac{i \times i-1}{i}\right\rfloor$

= $\sum_{i = 1}^{43}\left\lfloor\,i+2\right\rfloor + 45 - \sum_{i = 2}^{44}\left\lfloor\,i-\frac{1}{i}\right\rfloor$

= $\sum_{i = 1}^{43}(i+2) + 45 - \sum_{i = 2}^{44}(i-1)$

= $3+4+5+...+45+45-1-2-...-43$

= $131$