21 and 27 are divisible by three, not primes. Same goes for 201 and 207, 2001 and 2007.
33 and 39 are divisible by three, not primes. Same for 303 and 309, 3003 and 3009.
Now, forties: the ones divisible by three are 42 (already out since even), 45 (already out since a multiple of five), 48 (already out since even). Theoretically one might imagine that 41, 43, 47 and 49 - and so on by adding one or two zeroes between - might all be primes. However, no. 49 is 7*7. But 41, 43 and 47 are all primes. What if you "add"* a zero between or two zeroes between?
401?
| 4 | 0 | 1 |
| 2 | 1 | |
| 3 | 8 |
No, 38 is not divisible by seven. 403, then?
| 4 | 0 | 3 |
| 6 | 3 | |
| 3 | 4 |
Neither is 34. So what about 407? 407 - 7 = 400, not divisible by seven. Same as with 4007. If it were divisible by seven, by taking away seven, you would get a number that is also divisible by seven, and 4000 is not. So, 409?
| 4 | 0 | 9 |
| 4 | 9 | |
| 3 | 6 |
Nope, 409 was not divisible with seven since 36 (actually 360) is not. Lets get higher up. 4001?
| 4 | 0 | 0 | 1 |
| 2 | 1 | ||
| 3 | 9 | 8 | |
| 2 | 8 | ||
| 3 | 7 |
As little as 37. So, 4003?
| 4 | 0 | 0 | 3 |
| 6 | 3 | ||
| 3 | 9 | 4 | |
| 1 | 4 | ||
| 3 | 8 |
No, 38 is not divisible by seven (it is 28 which is so), and neither is 4003. 4007 has already been dealt with, so let us take 4009.
| 4 | 0 | 0 | 9 |
| 4 | 9 | ||
| 3 | 9 | 6 | |
| 5 | 6 | ||
| 3 | 4 |
So, no. 7 is out of the way when it comes to blowing prime numbers in the areas 401 - 409 or 4001 - 4009. I wondered what might be the case with 11, 13, 17, 19 ... and yes, I got results.
| 4 | 0 | 3 |
| 1 | 3 | |
| 3 | 9 |
403 = 13*31
| 4 | 0 | 0 | 9 |
| 1 | 9 | ||
| 3 | 9 | 9 | |
| 1 | 9 | ||
| 3 | 8 |
4009 = 19*211
So, I suppose that between 40.001 and 40.009 one prime position will not be so, since a multiple of ... 23? 29? I do not know. Your turn ...**
Hans-Georg Lundahl
Bpi, Georges Pompidou
Epiphany 2014
* Add to numerals in writing, not to numbers as in making additions. Sorry for pun. Hope you can cope anyway ...
** I actually did another check meanwhile. Like the tens, hundreds and thousands with two numerals possibly adding zeros between on 1 and on 4, also those on 7 lack interference from the prime number 3, since they are either divisible by 5 (75 ...) or by 2 (72, 78) anyway. However those that also end in 7 go away as divisible by 7, of course. That leaves three positions - "start 7 and end 1", or "and end 3", "or and end 9". Of the thousands I have so far found no number except 7007 which goes away, but for hundreds there is 703. I reproduce the check for 19:
| 7 | 0 | 3 | |
| 1 | 3 | 3 | (since 9*7=63 and 10*7=70) |
| 5 | 7 | ||
| 5 | 7 | (since 3*19=57) |
This means 703 is 19*37. In the hundreds, there is no need to go on beyond 29 for checks before you get to 961 (square of 31). That leaves 701 and 709 as prime numbers, unless I overlooked something. In the thousands I would have to check on prime numbers up to 83 (which is the highest whole number root below the "pseudo-root" - or geometric side - of the nonsquare number 7010).
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