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.
Never realized there are so many rules for divisibility. This post fits in this category:
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:
So we have 999 - 999 + 299 = 299.
You can continue with other rules so we can then take this
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.