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\(\dfrac{1}{2017}\)

The \(a_1, a_2, \dots, a_{2017}\) are \(2017\) distinct odd positive integers. Is it possible that

a) \(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dots+\dfrac{1}{a_{2017}}=\dfrac{2017}{10}\)?

b) \(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dots+\dfrac{1}{a_{2017}}=\dfrac{2017}{100}\)?

Note by Áron Bán-Szabó
1 month, 2 weeks ago

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Neither sum is possible.

The highest sum you can get would be by summing the reciprocals of the 2017 lowest odd positive integers. That gives a bit more than 4.44, so not even close to 2017/100.

I think if you kept going, you could get as large a sum as you desired though.... Steven Perkins · 1 month, 2 weeks ago

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What do you mean by "2,017" and "20,17"? Pi Han Goh · 1 month, 2 weeks ago

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@Pi Han Goh I corrected it, now you know What I mean, I hope. Áron Bán-Szabó · 1 month, 2 weeks ago

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