if I had this content you’re putting out 3-4 years ago I can’t imagine how much more math I would’ve been able to explore in a more organized way. Keep going with this stuff, it’s a beautiful asset to everyone with any relationship to math!
Perhaps extends this to all bases? All of these rules are based upon 0 (mod 10^n) [2, 5, 4, 25] and ± 1 (mod 10^n) [3, 7, 9, 11, 13, 37]. You can pretty easily switch to base b and have the same set of divisibility rules for numbers which is 0 (mod b^n) and ± 1 (mod b^n) using the same logic. EDIT: You are pretty much guaranteed to have at least one rule for every prime from Fermat's little theorem. This is because if the prime does not belong to the base's prime factors, b^p-1 will be congruent to 1 (mod p).
11:45 well, actually there is a way better way to know if some number is a multiple of 7: 7|N iff (if and only if) 7|a-2b, where N=10a+b (note that every number can be written this way) proof: note that 10a+b=7(a+b)+3(a-2b) 7|10a+b iff 7|10a+b-7(a+b)=3(a-2b) iff 7|a-2b example: 7|343 iff 7|34-6=28 but since 7|28 so 7|343
That reminds me of a little bit I noticed the other day when thinking about hexadecimal numbers: If a number is divisible by three, it has a decimal numerical expression which, when reinterpreted in base 16, is also a number that's divisible by three. 21 base 10 is divisible by 3, and so is 21 base 16 (= 33 base 10). And so on. A corollary to that is that the divisibility-by-three rule works the same in base 10 as it does in base 16. And the same rule also works for divisibility-by-five in base 16. But ... not for divisibility-by-nine.
3 is -1 (mod 10) and also -1 (mod 16) so it should be the same divisibility rule. 9 is -1 (mod 10) but not -1 (mod 16) so in hexadecimal would need a different rule. Also since 5 is -1 (mod 16) and 15 is also -1 (mod 16) so you can use a similar divisibility rule for 5 and 15 in hexadecimal (digit sum) as divisibility rule for 3.
25:10 If a and b are between 0 and 9 then (2a-b) has a maximum of 18 and cannot be 22.
if I had this content you’re putting out 3-4 years ago I can’t imagine how much more math I would’ve been able to explore in a more organized way. Keep going with this stuff, it’s a beautiful asset to everyone with any relationship to math!
Sorry this wasn't out 2,000 years ago Euclid.
10:04 He meant (mod 37) in this line and the next two lines.
5:30, one 3 being changed to 9 got missed!
Perhaps extends this to all bases? All of these rules are based upon 0 (mod 10^n) [2, 5, 4, 25] and ± 1 (mod 10^n) [3, 7, 9, 11, 13, 37]. You can pretty easily switch to base b and have the same set of divisibility rules for numbers which is 0 (mod b^n) and ± 1 (mod b^n) using the same logic.
EDIT: You are pretty much guaranteed to have at least one rule for every prime from Fermat's little theorem. This is because if the prime does not belong to the base's prime factors, b^p-1 will be congruent to 1 (mod p).
11:45 well, actually there is a way better way to know if some number is a multiple of 7:
7|N iff (if and only if) 7|a-2b, where N=10a+b (note that every number can be written this way)
proof:
note that 10a+b=7(a+b)+3(a-2b)
7|10a+b iff 7|10a+b-7(a+b)=3(a-2b) iff
7|a-2b
example: 7|343 iff 7|34-6=28 but since 7|28 so 7|343
That reminds me of a little bit I noticed the other day when thinking about hexadecimal numbers: If a number is divisible by three, it has a decimal numerical expression which, when reinterpreted in base 16, is also a number that's divisible by three. 21 base 10 is divisible by 3, and so is 21 base 16 (= 33 base 10). And so on.
A corollary to that is that the divisibility-by-three rule works the same in base 10 as it does in base 16. And the same rule also works for divisibility-by-five in base 16. But ... not for divisibility-by-nine.
3 is -1 (mod 10) and also -1 (mod 16) so it should be the same divisibility rule. 9 is -1 (mod 10) but not -1 (mod 16) so in hexadecimal would need a different rule. Also since 5 is -1 (mod 16) and 15 is also -1 (mod 16) so you can use a similar divisibility rule for 5 and 15 in hexadecimal (digit sum) as divisibility rule for 3.
divisibility does rule!!!
2a-b can’t be 22 since at most 2a-b could be 2(9)-0=18
In the first example, most folks would start by noticing 25 is a factor of N.