Difference between revisions of "2002 IMO Problems/Problem 4"

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==Solution==
 
==Solution==
 
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Since 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}}
 
==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 15:28, 17 June 2024

Problem: Let $n>1$ be an integer and let $1=d_{1}<d_{2}<d_{3} \cdots <d_{r}=n$ be all of its positive divisors in increasing order. Show that \[d=d_1d_2+d_2d_3+ \cdots +d_{r-1}d_r <n^2\]

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it. Since 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}}

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