Putnam 1997, Problem A-5

Let $N_n$ denote the number of ordered $n$-tuples of positive integers $(a_1,a_2,\ldots,a_n)$ such that $1/a_1 + 1/a_2 + \cdots + 1/a_n = 1$. Determine whether $N_{10}$ is even or odd.

Solution:
We may discard any solutions for which \( a_1 \neq a_2 \), since those come in pairs; so assume \( a_1 = a_2 \). Similarly, we may assume that \( a_3 = a_4 \), \( a_5 = a_6 \), \( a_7 = a_8 \), \( a_9 = a_{10} \). Thus we get the equation  
\[
2/a_1 + 2/a_3 + 2/a_5 + 2/a_7 + 2/a_9 = 1.
\]  
Again, we may assume \( a_1 = a_3 \) and \( a_5 = a_7 \), so we get \( 4/a_1 + 4/a_5 + 2/a_9 = 1 \); and \( a_1 = a_5 \), so \( 8/a_1 + 2/a_9 = 1 \). This implies that \( (a_1 - 8)(a_9 - 2) = 16 \), which by counting has 5 solutions. Thus \( N_{10} \) is odd.