And 299999999 is divisible by 13

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.

Thanks I hate it
- This math will not stand man!
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?
- 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".
- The 9 rule works for 3 too The 6 rule is if (divisible by 3 and divisible by 2)
I think it might be easier just to do the division.

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

42857 for those who wonder

And for ops title: 23076923

Actually disgusting

Why

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.

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.

Thanks, Satan

Wait until you learn of 51/17.
- Thanks, I hate it
- 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.

I know you opened your calculator app to check it.

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

49 is divisible by 7, so why not?

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?

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

…9999 is exactly equal to -1.

But only in base 10.
- 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.

Isn’t every number divisible by 7?

Yup, this is just a coordinated smear campaign from Big Integer.
Yes technically almost every number is divisible by another in some way and you're left with a remainder that spans plenty of decimal places.

But common parlance when something is said to be divisible is that the end results is a round number...