$\dfrac{1}{2017}$

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

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Comments

What do you mean by "2,017" and "20,17"?

Pi Han Goh - 3 years, 11 months ago

I corrected it, now you know What I mean, I hope.

Áron Bán-Szabó - 3 years, 11 months ago

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 - 3 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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

Comments