Is 75986⋯69 ever prime or is there a 6 padded prime for every prime $p > 5$ ?

Question

I came across this question: How can we turn any number into a prime number by simply adding more digits? While trying different approaches to find an algorithm that increases the chances of finding a prime this way I discovered the six padded primes:

Start with an prime $p > 5$ then add a digit 6 before the last digit, repeat the process until the result is prime. This table shows the results for p up to 100:

\begin{aligned} 7 & 67 \\ 11 & 16661 \\ 13 & 163 \\ 17 & 167 \\ 19 & 1669 \\ 23 & 263 \\ 29 & 269 \\ 31 & 3666661 \\ 37 & 367 \\ 41 & 461 \\ 43 & 463 \\ 47 & 467 \\ 53 & 563 \\ 59 & 569 \\ 61 & 661 \\ 67 & 666667 \\ 71 & 761 \\ 73 & 76666663 \\ 79 & 769 \\ 83 & 863 \\ 89 & 8669 \\ 97 & 967 \end{aligned}

There were only 2 numbers below

200000

for which I could not find a result:

15731

and

75989

. My computer finally found a solution for the first padded with 7460 sixes resulting in a 7465 digit prime number.

75989

was checked up to 15000 digits.

Question:

Is $75986 \dots 69$ ever prime or can it be proved there is a six padded prime for every prime > 5

For an odd number of $6$ 's, it is easy to prove that the number is divisible by $11$ . For an even number of added $6$ 's of the form $n = 2 + 6 k$ Mathematica suggests the number to be a multiple of $13$ , while for $n = 4 + 6 k$ the number seems to be a multiple of $37$ . For $n = 6 k$ I couldn't detect any easy pattern. — Intelligenti pauca
– Intelligenti pauca, Commented Aug 5, 2017 at 10:35
75986⋯69 is a strong possible prime for n=32190. The next p with a larger solution is 212627 no result for n<=27877. — pietfermat
– pietfermat, Commented Aug 15, 2017 at 7:05
I just tried that with Mathematica and it tells it's indeed a prime! I don't know if Mathematica's function PrimeQ can be considered a conclusive test, but that's a strong indication that for $n = 32190$ the number is prime. — Intelligenti pauca
– Intelligenti pauca, Commented Aug 15, 2017 at 7:35

Intelligenti pauca · Accepted Answer · 2017-08-07 17:24:41Z

If you insert an odd number of $6$ 's, it is simple to check that the resulting number is a multiple of $11$ (just use the divisibility criterion).

For an even number $n$ of $6$ 's, notice that

759866 \equiv 3 (\mod 13), 1000000 \equiv 1 (\mod 13), 666666 \equiv 0 (\mod 13),

and

75986666 \equiv 25 (\mod 37), 1000000 \equiv 1 (\mod 37), 666666 \equiv 0 (\mod 37) .

Hence when

n = 2 + 6 k

the number is a multiple of

13

, and when

n = 4 + 6 k

the number is a multiple of

37

.

At the moment, I can say nothing for the case $n = 6 k$ .

EDIT.

Playing around with Mathematica I noticed some other regularities (which I didn't bother to prove):

for $n = 6 (1 + 7 k)$ the number is a multiple of $43$ ;
for $n = 6 (4 + 8 k)$ the number is a multiple of $17$ ;
for $n = 6 (4 + 5 k)$ the number is a multiple of $31$ .

On the other hand, there are several cases where the number has only two prime factors, both large. At the moment, however, I could find no trace of a prime number.

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