Find minimum $n$ that satisfies $\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=\frac{12}{13}$

Question

From the test: We have the following equation:

\begin{equation}
\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=\frac{12}{13}
\end{equation}

where $a_i$ are distinct natural numbers not equal to $13$.

What is the minimum $n$ that satisfies the above equation?

I have tried to find a solution for $\frac{1}{a_1}+\frac{1}{a_2}=\frac{12}{13}$ and $\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}=\frac{12}{13}$ but got nowhere. Only thing I found out, is that $13$ should divide at least one of $a_i$'s. (Not sure if tags are used correctly)

Answer 1

We assume $a_1 < a_2 < \dots$.

It is easy to get an answer with $n = 4$: $\frac 1 2 + \frac 1 3 + \frac 1 {12} + \frac 1 {156}$.

To see that $n = 2$ doesn't work, just note that $\frac 1 2 + \frac 1 3 = \frac 5 6 < \frac{12}{13}$.

To see that $n = 3$ doesn't work:

Assume there is a solution. Note that $\frac 1 3 + \frac 1 4 + \frac 1 5 = \frac{47}{60} < \frac{12}{13}$, so we must have $a_1 = 2$. This leads to $\frac 1 {a_2} + \frac 1{a_3} = \frac{11}{26}$.

Therefore $\frac 2 {a_2} > \frac {11}{26}$ and hence $a_2 \leq 4$. Checking $a_2 = 3$ and $a_2 = 4$, we see that $a_3$ is not an integer in each case.