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And 299999999 is divisible by 13 (lemmy.world)

submitted 3 weeks ago by m4m4m4m4 to c/[email protected]

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[–] [email protected] 22 points 3 weeks ago* (last edited 3 weeks 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.

[–] [email protected] 2 points 3 weeks ago* (last edited 3 weeks 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.