# Is there a maximum number of consecutive sexy prime pair sums that are all divisible by $10$?

I was finding the sums of pairs of sexy primes (prime numbers that differ by 6) and noticed that there are a lot of pairs who's sum is divisible by $$10$$.

Ex. $$(7, 13)$$ as $$7+13=20$$ and $$20$$ is divisible by $$10$$.

I then wondered if there are consecutive sexy prime pairs who's sum is divisible by $$10$$ and ran some PARI/GP code to find these kind of consecutive pairs. I found upto $$17$$ consecutive pairs divisible by $$10$$ on PARI and tested for a search limit of $$10^{12}$$.

Here is the output I got after running my code, the first column displays the smallest prime of the lowest pair of consecutive sexy primes and the second column displays the number of consecutive pairs divisible by $$10$$.

7             1
167           2
2237          3
2267          4
108187        5
1004057       6
3281777       7
32895377      8
65947927      9
569959037    10
602817437    11
5476396897   12
16842019627  13
16842019637  14
17004549137  15
312318208577 16
382560132847 17


This leads to my question:

If there are an infinite number of sexy primes, then is there a maximum number of consecutive sexy prime pair sums that are all divisible by $$10$$?

• Sum $10$ can only occur , if the smaller prime has ending digit $7$ and the larger ending digit $3$ – Peter May 18 at 15:37
• code is flawed. 7 and 13 are not consecutive primes. – Roddy MacPhee May 21 at 16:34
• @RoddyMacPhee yes they are a pair of sexy primes – Mathphile May 21 at 16:35
• okay then so are 17 and 23 ... 2 pairs consecutively. – Roddy MacPhee May 21 at 16:36
• @RoddyMacPhee yes, I have found 17 sexy prime pairs that are consecutive. – Mathphile May 21 at 16:38

A third of sexy primes has the right sum (ending in 7 and 3, but not 1 and 7, nor 3 and 9), so expect a run of $$n$$ with the right sum every $$3^n$$ sexy primes.