Is $k=553276187$ the smallest solution?

Question

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 ?

Answer 1

The answer to the question is no, here is a slightly smaller solution: $$ 494\;804\;473\ . $$

sage: factor( factorial(40) + 494804473 )
8912658466556113 * 9232286052422441 * 9915818101022081

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.)

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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

k=    33377 40!+k = 1056334373719901 * 22413347482811693 * 34461718591042889
k=    42269 40!+k = 1859373333290887 * 2719883435120179 * 161334845671400153
k=    47189 40!+k = 1822739844251111 * 15851904710438411 * 28238324719193609
k=    55049 40!+k = 1101465899839957 * 3378301824756127 * 219268155834100091

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)...

k=300722963 40!+k = 7073009263847467 * 8126384986908847 * 14195263159044887
k=349245887 40!+k = 7209029739646403 * 8613116168531887 * 13140380656971067
k= 77639477 40!+k = 7252468942583363 * 7992027613367573 * 14076744068584523
k=300731429 40!+k = 7321097013439289 * 7404841114350401 * 15050577417998861
k=339499973 40!+k = 7398940260351053 * 10008305130239719 * 11018310576981439
k= 96999373 40!+k = 7532964349652891 * 9232067350535819 * 11732219831109637
k= 48531323 40!+k = 7564931816003599 * 10225749107742383 * 10547387873068219
k=465611681 40!+k = 7866432514968563 * 7912332084219151 * 13108794049712437

k=465608309 40!+k = 8221566296360779 * 9835813811956651 * 10089745359967421
k=126205139 40!+k = 8278202201940731 * 9790182376796351 * 10067421590328119
k=388112077 40!+k = 8601556182258047 * 8687613919887803 * 10918614312712297

Answer 2

For the record, after some algorithm saving/loading to continue, the smallest example seems to be for $$k=34\;773\;149$$ having $$40!+k=9115826756817059\cdot 9197548818733747\cdot 9731435389001413.$$ However I am afraid you can't do much better than brute force search. Generating primes directly does not work because small difference in one of the primes still leads to large difference in the overall number, eventually moving you far away from these $k \approx10^7$ examples you are after.