Difference between revisions of "2001 Pan African MO Problems/Problem 5"
Rockmanex3 (talk | contribs) (Solution to Problem 5 -- summation and floors) |
Rockmanex3 (talk | contribs) m |
||
Line 7: | Line 7: | ||
==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>. |
<br> | <br> |
Revision as of 16:23, 17 December 2019
Problem
Find the value of the sum: where denotes the greatest integer which does not exceed .
Solution
First, note that for positive integers , the list of integers where is . Thus, there are values of where .
Additionally, note that and . There are values of where . Therefore, the value .
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 |