2002 IMO Problems/Problem 4

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 $d_xd_{r-x+1} = d_{x+1}d_{r-x} = n$ for all x < r, $d_xd_{x+1} = \frac{n^2}{d_{r-x}d_{r-x+1}}$ . Then $d_1d_2 + d_2d_3 \cdots

2002 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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2002 IMO Problems/Problem 4

Solution

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