Let's start with the equation given:
\[ \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \frac{1}{a_4} + \frac{1}{a_5} + \frac{1}{a_6} = 6. \]
We can rearrange this equation to:
\[ \frac{a_2 a_3 a_4 a_5 a_6 + a_1 a_3 a_4 a_5 a_6 + a_1 a_2 a_4 a_5 a_6 + a_1 a_2 a_3 a_5 a_6 + a_1 a_2 a_3 a_4 a_6 + a_1 a_2 a_3 a_4 a_5}{a_1 a_2 a_3 a_4 a_5 a_6} = 6. \]
To simplify, multiplying both sides by \(a_1 a_2 a_3 a_4 a_5 a_6\) yields:
\[ a_2 a_3 a_4 a_5 a_6 + a_1 a_3 a_4 a_5 a_6 + a_1 a_2 a_4 a_5 a_6 + a_1 a_2 a_3 a_5 a_6 + a_1 a_2 a_3 a_4 a_6 + a_1 a_2 a_3 a_4 a_5 = 6 a_1 a_2 a_3 a_4 a_5 a_6. \]
Now introduce a change of variables: let \(x_i = a_i\) for \(i = 1, 2, \ldots, 6\). Then the equation rewrites itself to finding the ordered 6-tuples such that each \(x_i\) corresponds to a positive integer in the denominators.
To approach this comprehensively, redefine the equation further:
Let \(b_i = \frac{1}{x_i}\) for \(i = 1, 2, \ldots, 6\). Thus, we reframe the goal:
\[ b_1 + b_2 + b_3 + b_4 + b_5 + b_6 = 6, \]
where \(b_i\) can be written as \(b_i = \frac{k_i}{1}\) with \(k_i\) being a positive integer (\(k_i = \frac{1}{x_i}\)).
To generate valid solutions to \(b_1 + b_2 + b_3 + b_4 + b_5 + b_6 = 6\) we can let \(y_i = b_i - 1\), transforming our equation into:
\[ (y_1 + 1) + (y_2 + 1) + (y_3 + 1) + (y_4 + 1) + (y_5 + 1) + (y_6 + 1) = 6 \Rightarrow y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 0, \]
where \(y_i \geq 0\). The only solution to this is \(y_1 = y_2 = y_3 = y_4 = y_5 = y_6 = 0\), giving \(k_1 = k_2 = k_3 = k_4 = k_5 = k_6 = 1\). Therefore, we have one way to select \(x_i\):
\[ (x_1, x_2, x_3, x_4, x_5, x_6) = (1, 1, 1, 1, 1, 1), \]
completing our enumeration of all \(x_i\) becoming 1.
However, we must assert the number of ways to arrange these values while keeping their sum consistent. Using combinatorics, the problem actually becomes particular when considering distributions of '1' across '6', effectively leading to \(N\).
More formally, by applying the stars and bars theorem or integer partitions, recently adapted to counting ordered partitions:
This turns out to be a traditional formation known as partitioning integers:
\[ N = \binom{6 + 6 - 1}{6 - 1} = \binom{11}{5}. \]
Calculating \(N\):
\[ N = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 462. \]
Next, let's find \(462 \mod 5\):
Finding \(462\) itself via division \(462 \div 5 = 92\) with a remainder of \(2\):
Thus, we have:
\[ \boxed{2} \]
as the final remainder under consideration.
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