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Let $k$ be a positive integer. Let $n_1,n_2,\ldots,n_k$ and $n$ be integers, each greater than one . Suppose they satisfy

$$\sum_{i=1}^k\left(1-\frac 1 {n_i}\right)=2-\frac 2 n $$

Then the only possible values of $k$ are $\underline{\hspace{1in}}$

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  • $\begingroup$ You can prove that $k<4$ because every element $1-\frac{1}{n_i}\geq 1/2$ hence you can deduce that $k\in \{2,3\}$, for $k=2$ take $n_1=n_2=n$ and for $k=3$ take $n_1=n_2=2$ and $n_3=n$. $\endgroup$ – Elaqqad Sep 22 '15 at 12:19
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HINT

We have $\sum_{i=1}^k (1 - \frac{1}{n_i}) \ge k - \frac{k}{2}$.
On the other hand, the RHS has an obvious upper bound.

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