# Difference between revisions of "2001 Pan African MO Problems/Problem 5"

## Problem

Find the value of the sum: $$\sum_{i=1}^{2001} [\sqrt{i}]$$ where $[ {x} ]$ denotes the greatest integer which does not exceed $x$.

## Solution

First, note that for positive integers $n$, the list of integers $i$ where $[\sqrt{i}] = n$ is $n^2, n^2 + 1, n^2 + 2, \cdots n^2 + 2n$. Thus, there are $(n^2 + 2n) - n^2 + 1 = 2n + 1$ values of $i$ where $[\sqrt{i}] = n$.

Additionally, note that $44^2 = 1936$ and $45^2 = 2025$. There are $2001-1936+1 = 66$ values of $i$ where $\sqrt{i} = 44$. Therefore, the value $\sum_{i=1}^{2001} [\sqrt{i}] = 1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \cdots + 43 \cdot 87 + 44 \cdot 66$.

By rewriting the above value as a summation, we can compute the value to be equal to \begin{align*} \sum_{i=1}^{43} (i(2i+1)) + 44 \cdot 66 &= \sum_{i=1}^{43} (2i^2 + i) + 44 \cdot 66 \\ &= 2 \sum_{i=1}^{43} (i^2) + \sum_{i=1}^{43} (i) + 44 \cdot 66 \\ &= \frac{2 \cdot 43 \cdot 44 \cdot 87}{6} + \frac{44 \cdot 43}{2} + 44 \cdot 66 \\ &= (86 \cdot 22 \cdot 29) + (22 \cdot 43) + (22 \cdot 132) \\ &= (22 \cdot 2494) + (22 \cdot 175) \\ &= 22 \cdot 2669 \\ &= \boxed{58718}. \end{align*}