Difference between revisions of "2001 Pan African MO Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | First, note that for positive integers <math>n</math>, the list of integers <math>i</math> where <math>\sqrt{i} = n</math> is <math>n^2, n^2 + 1, n^2 + 2, \cdots n^2 + 2n</math>. Thus, there are <math>(n^2 + 2n) - n^2 + 1 = 2n + 1</math> values of <math>i</math> where <math>\sqrt{i} = n</math>. | + | First, note that for positive integers <math>n</math>, the list of integers <math>i</math> where <math>[\sqrt{i}] = n</math> is <math>n^2, n^2 + 1, n^2 + 2, \cdots n^2 + 2n</math>. Thus, there are <math>(n^2 + 2n) - n^2 + 1 = 2n + 1</math> values of <math>i</math> where <math>[\sqrt{i}] = n</math>. |
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Revision as of 15:23, 17 December 2019
Problem
Find the value of the sum:
![\[\sum_{i=1}^{2001} [\sqrt{i}]\]](http://latex.artofproblemsolving.com/1/3/b/13b97ad5b59212b797cd833ceef0381d0e2b5898.png)
where ![$[ {x} ]$](http://latex.artofproblemsolving.com/9/7/9/97988f3618289185c731de2299e17e1b84bef42d.png)
denotes the greatest integer which does not exceed 
.
Solution
First, note that for positive integers 
, the list of integers 
where ![$[\sqrt{i}] = n$](http://latex.artofproblemsolving.com/a/8/d/a8d3f5a43154a55252a621ea2a3d947665ac1963.png)
is 
. Thus, there are 
values of where .
Additionally, note that 
and 
. There are 
values of where 
. Therefore, the value ![$\sum_{i=1}^{2001} [\sqrt{i}] = 1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \cdots + 43 \cdot 87 + 44 \cdot 66$](http://latex.artofproblemsolving.com/d/c/7/dc7876498f5abd466319aebbd09bdb099f70b7bb.png)
.
By rewriting the above value as a summation, we can compute the value to be equal to
See Also
| 2001 Pan African MO (Problems) | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All Pan African MO Problems and Solutions | ||
