I search the smallest positive integer $k$, such that $40!+k$ splits into three primes having $16$ decimal digits. The smallest solution I found is
$$k=553\ 276\ 187$$
You can see the factorization here : http://factordb.com/index.php?id=1100000001197449481
According to my calculations, no $k\le 2\cdot 10^6$ does the job.
Is my $k$ the optimal solution ?
Best Answer
The answer to the question is no, here is a slightly smaller solution: $$ 494\;804\;473\ . $$
It was obtained also in sage, the program was running one day, and this is in this intermediate situation a found solution. (It is still running. It was not written to touch the numbers one by one starting with $40!$ in order, so this may also not be the minimal number doing the job.)
Note: As i also said in the comment, the fact that $40!^{1/3}$ is so close to the number $999\dots 9$ with $16$ digits makes it harder to get such decompositions with three factors of $16,16,16$ digits each. Because the range of the primes is not really between the $16$ digits numbers $1000\dots 0$ and $999\dots 9$. The computer was founding for instance also
with factors having $\ge 16$ digits, and also insisting that the smaller prime factor among all three of the three is bigger than $7000\dots 0$ ($16$ digits)...