Finding a special prime The prime number $$p=82\ 954\ 517$$ has the property that the numbers $$2!+p,3!+p,\cdots , 11!+p$$ are all prime, but $12!+p$ is composite.
Upto $10^{10}$, the only other prime with this property is $105\ 204\ 557$

Does a prime $p$ exist such that $$2!+p,3!+p,\cdots , 12!+p$$ are all prime ? If yes, which is the smallest such prime ? Such a prime must exceed $10^{10}$

Update : The prime $$p=79\ 017\ 245\ 897$$ is even better than what I wanted. $j!+p$ is prime for $j=2,3,4,\cdots,13$. Now it remains to find the minimum primes for the limit $12$ and the limit $13$

Just a few restrictions:


*

*first leads to p is 5 mod 6

*second eliminates 29 mod 30

*third eliminates 11 mod 30

*fourth eliminates 6 mod 7 ( aka 167,83 mod 210)

*fifth eliminates  1 mod 7 ( aka 197, 113 mod 210)

*sixth eliminates 9 mod 11

*seventh eliminates 6 mod 11

*eighth eliminates 10 mod 11

*ninth eliminates  1 mod 11

*tenth eliminates  12 mod 13

*eleventh eliminates 1 mod 13

*twelvth eliminates  14 mod 17

*thirteenth eliminates 9 mod 17

*fourteenth eliminates 16 mod 17

*fifteenth eliminates 1 mod 17

*sixteenth eliminates 18 mod 19

*seventeenth eliminates 1 mod 19

*eighteenth eliminates 19 mod 23

*nineteenth eliminates 12 mod 23

*twentieth eliminates  22 mod 23

*twenty first eliminates 1 mod 23

*twenty second eliminates 7 mod 29

*twenty third eliminates 23 mod 29

*twenty fourth eliminates 24 mod 29

*twenty fifth eliminates 15 mod 29

*twenty sixth eliminates 28 mod 29

*twenty seventh eliminates 1 mod 29

*twenty eighth eliminates 30 mod 31

*twenty ninth eliminates 1 mod 31

*thirtieth eliminates 4 mod 37


okay I have messed up ( prior) we need -n! mod q# eliminated. table updated (fixed) and extended. 
