$52$ cards reciprocal sum probability Imagine a deck of $52$ cards but instead of having suits and ranks, they only have sequential (unique) integer ranks from $1$ to $52$. You could also imagine a standard deck of $52$ cards but convert the ranks and suits to an integer number from $1$ to $52$ so that all the cards have different numbers assigned to them.
So the question is, what is the probability, if you randomly choose exactly $10$ cards from that deck without replacement, that the sum of the reciprocals of the cards ranks (from $1$ to $52$) totals exactly $1$?  For example, if you chose cards $5,10,15,20,25,30,35,40,45$ and $50$, the sum of the reciprocals would only be $0.58579$... so that is too small.
As a hint, I believe if you sort the cards in ascending order, the first card (the lowest rank card) MUST be between a $2$ and $6$ (inclusive) to be a candidate solution.  This is because $1/1$ is already a sum of $1$ so any additional cards will make it too large a sum.  Also $7$ as a lowest card will not work cuz $1/7$ + $1/8$... + $1/16$ = $0.93$...  so the lowest card MUST be a $2,3,4,5$, or $6$.  I am re-running my modified simulation now with that added information to prune the state space it must check.
Also I am seeing multiple solutions so there is not just $1$ solution but the probability is likely very low, only a small fraction of $1$% I would guesstimate.
Also note than many solutions are very close to $1$ but not exactly $1$, thus making computer simulation of this type of problem more difficult.  An example of a "close solution" is $2,4,13,25,35,41,47,50,51,52$ which evaluates to $0.99999995750965$.  The closest sum not equal to $1$ happens with $3,4,8,14,17,22,26,29,46,47$ which evaluates to $1.00000000288991$.  That is $8$ zeros after the $1$.
An interesting note...  That number very close to $1$ is gotten by summing only $10$ terms.  This is impressive since even summing negative powers of $2$ which converges to $1$, ($1/2$ + $1/4$ + $1/8$...), it takes $28$ terms to almost match that closeness to $1$ and $29$ terms to beat it.
 A: The a=6 case can be eliminated by pencil and paper.
Consider that 1/1 = 1, 1/2 = 7, 1/3 = 9, and 1/4 = 10, all mod 13. The only multiple of 13 we can get out of sums of these numbers is 7+9+10=26, which shows that 1/26+1/39+1/52=1/12 is feasible, but 1/13 must not appear in the sum.
Also 1/1 = 1, 1/2 = 6, 1/3 = 4, 1/4 = 3, all mod 11. The only multiple of 11 in this case is 1+6+4=11, so 1/44 is excluded, and if 1/11 is in the sum, then 1/22 and 1/33 must also be.
These two considerations alone mean that a=6 is impossible:
sum(1.0/[6,7,8,9,10,11,12,13,14,15]) =    1.034896
sum(1.0/[6,7,8,9,10,11,12,14,22,33]) =   0.9670634
sum(1.0/[6,7,8,9,10,12,14,15,16,17]) =   0.9883869

Similar considerations exclude a total of 20 cards. With that, along with the exclusion of card 1 by its magnitude, the deck is down to 31 cards and there are only 4 possibilities for the lowest card. Throw in an early-out test at about the seventh card and compiled code gets down to about 30 ms. The interpreted code should go through in tolerable time. Obviously not an improvement on brute force because that is quick enough and less error-prone.
