Track / Stage 5 / 227 of 400 #827 of 2000

Problem 827

AIME late

Number theory Difficulty 5.5 Find the answer

15. For any real number $x$ , let $\lceil x\rceil$ denote the smallest integer that is greater than or equal to $x$ and $\lfloor x\rfloor$ denote the largest integer that is less than or equal to $x$ (for example, $\lceil 1.23\rceil=2$ and $\lfloor 1.23\rfloor=1$ ). Find the value of
$\sum_{k=1}^{2010}\left[\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor\right] \text {. }$

Your answer

Official solution

15. Answer: 1994
Consider $k=1,2, \ldots, 2010$ . If $k \mid 2010$ , then $x:=\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor=0$ , so $\lceil x\rceil=0$ . If
$k \backslash 2010$ , then $0<y:=\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor<1$ , so $\lceil y\rceil=1$ .
Since the prime factorization of 2010 is $2 \times 3 \times 5 \times 67$ , we see that 2010 has 16 distinct divisors. Hence
$\begin{aligned} \sum_{k=1}^{2010}\left\lceil\frac{2010}{k}-\left\lfloor\frac{2010}{k}\right\rfloor\right\rceil & =\sum_{k: 2010}\lceil x\rceil+\sum_{k, 2010}\lceil y\rceil \\ & =\text { number of non-divisor of } 2010 \text { among } k \\ & =2010-16=1994 \end{aligned}$

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.