Can we just say it isn’t? Like that’s an exception, and then the rest of math can just go on like normal
And 299999999 is divisible by 13
Submitted 1 year ago by m4m4m4m4@lemmy.world to science_memes@mander.xyz
https://lemmy.world/pictrs/image/1434de78-da71-47f6-9542-66fdf62c6061.jpeg
Comments
grrgyle@slrpnk.net 1 year ago
Etterra@lemmy.world 1 year ago
So what? Being a prime number doesn’t mean it can’t be a divisor. Or is it the string of 9s that’s supposed to be upsetting? Why? What difference does it make?
Revan343@lemmy.ca 1 year ago
Thanks, Satan
Klear@lemmy.world 1 year ago
Wait until you leard of 51/17.
Revan343@lemmy.ca 1 year ago
Thanks, I hate it
prime_number_314159@lemmy.world 1 year ago
With 17, I understand that you’re referring to how 299,999 is also divisible by 17. What is the 51 reference, though? I know there’s 3,999,999,999,999 but that starts with a 3. Not the same at all.
m_f@midwest.social 1 year ago
⅐ = 0.1̅4̅2̅8̅5̅7̅
The above is 42857 * 7, but you also get interesting numbers for other subsets:
7 * 7 = 49 57 * 7 = 399 857 * 7 = 5999 2857 * 7 = 19999 42857 * 7 = 299999 142857 * 7 = 999999
chemicalwonka@discuss.tchncs.de 1 year ago
I know you opened your calculator app to check it.
Grandwolf319@sh.itjust.works 1 year ago
49 is divisible by 7, so why not?
henfredemars@infosec.pub 1 year ago
…9999 is exactly equal to -1.
bstix@feddit.dk 1 year ago
But only in base 10.
henfredemars@infosec.pub 1 year ago
Well yes of course. If it was a different base, writing it that way if the symbol was even available would be a different number.
OfficerBribe@lemm.ee 1 year ago
Never realized there are so many rules for divisibility. This post fits in this category:
Forming an alternating sum of blocks of three from right to left gives a multiple of 7
299,999 would be 999 - 299 = 700 which is divisible by 7. And if we simply swap grouped digits to 999,299, it is also divisible by 7 since 299 - 999 = -700.
grubberfly@mander.xyz 1 year ago
That is indeed an absurd amount of rules (specially for 7) !
It should be fun to develop each proof. Specially the 1,3,2,-1,-3,-2 rule
TwilightKiddy@programming.dev 1 year ago
The divisability rule for 7 is that the difference of doubled last digit of a number and the remaining part of that number is divisible by 7.
E.g. 299’999 → 29’999 - 18 = 29’981 → 2’998 - 2 = 2’996 → 299 - 12 = 287 → 28 - 14 = 14 → 14 mod 7 = 0.
It’s a very nasty divisibility rule. The one for 13 works in the same way, but instead of multiplying by 2, you multiply by 4. There are actually a couple of well-known rules for that, but these are the easiest to remember IMO.
candybrie@lemmy.world 1 year ago
I think it might be easier just to do the division.
urda@lebowski.social 1 year ago
[deleted]veroxii@aussie.zone 1 year ago
This math will not stand man!
darkpanda@lemmy.ca 1 year ago
If all of the digits summed recursively reduce to a 9, then the number is divisible by 9 and also by 3.
If the difference between the sums of alternating sets digits in a number is divisible by 11, then the number itself is divisible by 11.
That’s all I can remember, but yay for math right?
levzzz@lemmy.world 1 year ago
The 9 rule works for 3 too The 6 rule is if (divisible by 3 and divisible by 2)
TwilightKiddy@programming.dev 1 year ago
Well, on the side of easy ones there is “if the last digit is divisible by 2, whole number is divisible by 2”. Also works for 5. And if you take last 2 digits, it works for 4. And the legendary “if it ends with 0, it’s divisible by 10”.
SerpentPeaked@lemmynsfw.com 1 year ago
Isn’t every number divisible by 7?
ilovededyoupiggy@sh.itjust.works 1 year ago
Yup, this is just a coordinated smear campaign from Big Integer.
Usernamealreadyinuse@lemmy.world 1 year ago
42857 for those who wonder
And for ops title: 23076923
sem@lemmy.blahaj.zone 1 year ago
Phew, for a moment I worried that 2.9999… was divisible by 7 and I woke up in some kind of alternate universe