Horse 1405 - Multiples Of 6 And Primes

"I don't like to be pedantic (well, actually I do)," writes Geoff Poulton, of Westleigh (all primes adjoin a number divisible by 6, Column 8, since Monday), "but the rule should be 'all primes greater than 3',", and while we did deal with that on Monday, Geoff goes on to explain why the rule works. "Since all such primes are odd, both adjacent numbers are divisible by 2. Also, in any group of three consecutive numbers, one MUST be divisible by 3, so this has to be true for one of the two numbers bordering a prime (since this can't be true for the prime). Since this number must also be divisible by 2, it's divisible by 6. A corollary is that for paired primes (differing by 2 like 41 and 43) the number in the middle is always divisible by 6." We've had a torrent of correspondence on this subject, which we will carefully dole out over the coming days in bite-sized chunks.
- Column 8, Sydney Morning Herald, 21st Nov 2012
http://www.smh.com.au/opinion/column-8/column-8-20121120-29o2w.html

I saw this in Column 8 and thought that if this proposal is true, there has to be a way to prove it; thankfully there is an exceptionally easy way to do this and it isn't all that complicated.

Pick any number: I pick 247 because as I typed this on the bridge overlooking Military Rd in Mosman, the number 247 bus went underneath.
247 in expanded notation is 2 hundreds, 4 tens and 7 units. 2 * 10² plus 4 * 10¹ plus 7 * 10°

There isn't really any good reason why 10 is so magical. You could use any number system and maths still functions perfectly. If that seems hard to wrap your head around, the Mad Hatter's hat cost 10/6, that is 10 shillings and sixpence.
10 * 12¹ plus 6 * 12° = 10 shillings and 6 pence (pennies being the base unit)
Mrs Rollo is 5'7" tall, 5 * 12¹ plus 7 * 12° = 5 feet and 7 inches (inches being the base unit)
A T20 cricket match at the 5.4 over mark is 5 * 6¹ plus 4 * 6° = 5 overs and 4 balls (balls being the base unit)
People have been using all sorts of bases for numbers whenever it suits them for reasons of usefulness. In this case I'm picking base 6, since that's what's being asked.

The first thing I notice is that apart from 2, all even numbers are not prime. 14 = 2 * 5 for instance.
The second thing I notice is that apart from 3, all numbers ending in three are not prime. 53 for example is 15 * 3. The multiples of 3 beyond 10 are 13, 20, 23, 30, 33, 40, 43, etc. Still confused? Start thinking in base 6.
Having established that all multiples beyond 3 are not prime and all even numbers beyond 2 are also not prime, the only possible end digits that primes can exist in are 1 and 5 and since 1 and 5 happen to be either side of 10 in base 6, all primes will occur adjacent to multiples of 6.

Either Ernie or Bert must be prime. They are adjacent to a multiple of 6.

Geoff Poulton's explanation that for paired primes (differing by 2 like 41 and 43) the number in the middle is always divisible by 6 also holds true. 42 in base 6 is 110; 41 and 43 in base 6 are 105 and 111 just as I'd suggested.