Difference between revisions of "2002 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
{{solution}} | {{solution}} | ||
| − | Since <math>d_xd_{r-x+1} = d_{x+1}d_{r-x} = n, d_xd_{x+1} = \frac{n^2}{d_{r-x}d_{r-x+1}}</math> | + | Since <math>d_xd_{r-x+1} = d_{x+1}d_{r-x} = n</math> for all x < r, <math>d_xd_{x+1} = \frac{n^2}{d_{r-x}d_{r-x+1}}</math>. Then $d_1d_2 + d_2d_3 \cdots |
==See Also== | ==See Also== | ||
{{IMO box|year=2002|num-b=3|num-a=5}} | {{IMO box|year=2002|num-b=3|num-a=5}} | ||
Revision as of 14:31, 17 June 2024
Problem:
Let 
be an integer and let 
be all of its positive divisors in increasing order. Show that
![\[d=d_1d_2+d_2d_3+ \cdots +d_{r-1}d_r <n^2\]](http://latex.artofproblemsolving.com/8/e/2/8e201bb9274ec9bd46a38d9681bf1b22e42d24da.png)
Solution
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Since 
for all x < r, 
. Then $d_1d_2 + d_2d_3 \cdots
See Also
| 2002 IMO (Problems) • Resources | ||
| Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
| All IMO Problems and Solutions | ||
