Let $A_n$ be the number of arranged n-tuples of natural numbers $(a_1,a_2…a_n)$, such that
$\frac{1}{a_1} +\frac{1}{a_2} +...+\frac{1}{a_n} =1$.
Find the parity of $A_{68}$.
Let $\mathcal{A}$ be the family of all multisets $A$ of exactly $n$ natural numbers for which $\displaystyle \sum_{a\in A}\frac{1}{a}=1$. For each $A\in \mathcal{A}$ let $p_A$ be the number of all distinct permutations (ordered $n$-tuples) of elements of $A$.
Clearly,
$$A_n=\sum_{A\in \mathcal{A}} p_A$$Let $n$ be even. Then each $p_A$ is even except in only one case: when $A$ consists of $n$ equal elements. In that case $p_A=1$. There is only one such multiset $A$ in $\mathcal{A}$, namely when all $n$ elements of $A$ are equal to $\frac{1}{n}$.
Hence $A_n$ is odd, for any even $n$.