Thus, do there exist $n$ distinct primes whose summed reciprocals fall short of $1$ by the reciprocal of their product, for some $n\geqslant6$? I can get as far as $n=5$:
$$\dfrac{1}{2}=1-\dfrac{1}{2},$$
$$\dfrac{1}{2}+\dfrac{1}{3}=1-\dfrac{1}{2\cdot3},$$
$$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}=1-\dfrac{1}{2\cdot3\cdot7},$$
$$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{43}=1-\dfrac{1}{2\cdot3\cdot7\cdot43},$$
$$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{11}+\dfrac{1}{23}+\dfrac{1}{31}=1-\dfrac{1}{2\cdot3\cdot11\cdot23\cdot31}.$$
But I can't see the way beyond that. If there is a fraction for some $n\geqslant6$, then one may naturally ask: is there a bound on such $n$?



How long can a primal egyptian fraction be, that optimally approaches unity?
This may be related to [Giuga numbers](http://oeis.org/A007850): composite numbers $n$ such that $p$ divides $(n/p)-1$ for every prime divisor $p$ of $n$. A 10-factor Giuga number is given in the comments on that page: $$\eqalign{&420001794970774706203871150967065663240419575375163\cr&060922876441614
2557211582098432545190323474818\cr&= 
2 \cdot 3 \cdot 11 \cdot 23 \cdot 31 \cdot 47059 \cdot 2217342227 \cdot 1729101023519 \cdot \cr&8491659218261819498490029296021 \cdot 58254480569119734123541298976556403\cr}$$


Also possibly related to [primary pseudoperfect numbers](http://oeis.org/A054377). 


