this post was submitted on 04 Nov 2024
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top 32 comments
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[–] sem@lemmy.blahaj.zone 4 points 10 hours ago

Phew, for a moment I worried that 2.9999... was divisible by 7 and I woke up in some kind of alternate universe

[–] grrgyle@slrpnk.net 5 points 12 hours ago

Can we just say it isn't? Like that's an exception, and then the rest of math can just go on like normal

[–] m_f@midwest.social 56 points 19 hours ago* (last edited 19 hours 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

Related to cyclic numbers:

142857 * 1 = 142857
142857 * 2 = 285714
142857 * 3 = 428571
142857 * 4 = 571428
142857 * 5 = 714285
142857 * 6 = 857142
142857 * 7 = 999999
[–] Etterra@lemmy.world 3 points 12 hours 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 15 points 18 hours ago (1 children)
[–] Klear@lemmy.world 9 points 14 hours ago* (last edited 14 hours ago) (1 children)

Wait until you learn of 51/17.

[–] prime_number_314159@lemmy.world 1 points 8 hours ago (1 children)

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.

[–] Klear@lemmy.world 2 points 8 hours ago (1 children)

57 / 17 = 3. That messes up with my brain.

[–] scutiger@lemmy.world 3 points 4 hours ago (1 children)
[–] Klear@lemmy.world 1 points 2 hours ago

That it is. And it's not just my math that is wrong.

[–] TwilightKiddy@programming.dev 82 points 1 day ago (3 children)

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 5 points 10 hours ago

I think it might be easier just to do the division.

[–] urda@lebowski.social 59 points 1 day ago (1 children)
[–] veroxii@aussie.zone 9 points 21 hours ago

This math will not stand man!

[–] darkpanda@lemmy.ca 13 points 1 day ago (2 children)

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 points 13 hours ago

The 9 rule works for 3 too The 6 rule is if (divisible by 3 and divisible by 2)

[–] TwilightKiddy@programming.dev 8 points 23 hours ago (2 children)

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".

[–] darkpanda@lemmy.ca 6 points 23 hours ago* (last edited 20 hours ago) (1 children)

There’s also the classic “no three positive integers a, b, and c to satisfy a**n + b**n = c**n for values of n greater than 2“ trick but my proof is too large to fit in this comment.

[–] JamesStallion@sh.itjust.works 2 points 20 hours ago

Fucking lol

[–] Scubus@sh.itjust.works 3 points 22 hours ago (1 children)

Its never divisible by zero, and its always divisible by one

[–] wicked@programming.dev 3 points 15 hours ago (1 children)
[–] LordTrychon@startrek.website 1 points 11 hours ago

Interesting read. Thank you.

[–] chemicalwonka@discuss.tchncs.de 13 points 20 hours ago

I know you opened your calculator app to check it.

42857 for those who wonder

And for ops title: 23076923

[–] AI_toothbrush@lemmy.zip 33 points 1 day ago (1 children)
[–] OfficerBribe@lemm.ee 20 points 1 day ago* (last edited 23 hours ago) (1 children)

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.

And as for 13:

Form the alternating sum of blocks of three from right to left. The result must be divisible by 13

So we have 999 - 999 + 299 = 299.

You can continue with other rules so we can then take this

Add 4 times the last digit to the rest. The result must be divisible by 13.

So for 299 it's 29 + 9 * 4 = 65 which divides by 13. Pretty cool.

[–] grubberfly@mander.xyz 1 points 10 hours ago* (last edited 10 hours ago)

That is indeed an absurd amount of rules (specially for 7) !

It should be fun to develop each proof. Particularly the 1,3,2,-1,-3,-2 rule, which at first sight seems could be easily expanded to any other number.

[–] Grandwolf319@sh.itjust.works 8 points 23 hours ago

49 is divisible by 7, so why not?

[–] henfredemars@infosec.pub 3 points 23 hours ago (1 children)

…9999 is exactly equal to -1.

[–] bstix@feddit.dk 2 points 18 hours ago (1 children)
[–] henfredemars@infosec.pub 1 points 10 hours 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.