#standupmaths

every number in base 6 that doesn't end with 1 or 5 is divisible by 2 or 3, so to come up with a divisibility rule for some number 'n':

. divide 'n' by 10, 2, and 3 as much as possible (10 while it ends with 0, 2 while it ends with 2/4, 3 while it ends with 3) so it now ends with 1 or 5, call the result 'a', and call the product of all those 6s 2s and 3s 'b'

. (note that dividing by 2 or 3 is easy in base 6, just triple or double respectively and remove a zero - eg, 32/2 = 160/10 = 16, 15/3 = 30/10 = 3)

. split 'a' into two numbers, 'a1' (all but the last digit) and 'a2' (the last digit)

. if 'a2' is 1, the divisibility rule for 'a' is "take off the last digit of the number, multiply it by 'a1', and subtract it from the rest, and repeat until you get to a number small enough to recognise from multiplication tables"

. if 'a2' is 5, the divisibility rule for 'a' is "take off the last digit of the number, multiply it by 'a1', and add it to the rest, etc"

. split 'b' into two factors: all but the trailing zeroes, which is gonna be either 2^x or 3^x for some 'x', and 1 followed by all the trailing zeroes, which is equivalently 10^y for some 'y'

. given 2^x, define 'c' as 3^x, given 3^x, define 'c' as 2^x

. the divisibility rule for 'b' is to first multiply by 'c' then check if it ends with at least x+y zeroes

. the divisibility rule for 'n' is to do the divisibility rules for both 'a' and 'b', and if it's divisible by both 'a' and 'b', it's divisible by 'n'

example: is 11003535 divisible by 1220?

1220 = 41×20, so a=41, a1=4, a2=1, c=3, x=1, y=1

test for divisibility by 'b': 3×11003535 = 33015453, nope we wanted two zeroes and we found none, sad!

unnecessary now, but test for divisibility by 'a': 11003535 -> 1100353 - 4×5 = 1100321 -> 110032 - 4×1 = 110024 -> 11002 - 4×4 = 15534 -> 1553 - 4×4 = 1525 -> 152 - 4×5 = 120 -> 12, nope, 12 clearly isn't a multiple of 41