08-08-2017, 03:53 AM

(08-04-2017, 06:28 AM)Zijkhal Wrote:(07-16-2017, 09:25 AM)belevid Wrote:To introduce the topic, I would like to give the reader a nice trick I've used many times in my life. If you want to tell if a number is divisible by 3, add up the digits and if the result is divisible by 3, then so is the original number (the same property applies to 6...

I dont think so. 18 is divisible by 6, but the sum of its digits, 9 isnt. The division rule for 6 is if its divisible by both 3 and 2, then its divisible by 6, using the rule that if b = c*d, c and d are relative primes, then a mod b = 0 if and only if a mod c = 0 AND a mod d = 0.

Or if we replace 3 in your proof with 6, we get 4^n mod 6, so it does matter which digit is which, so we cant simply sum them to determine divisibility. Based on this (unless I'm missing something), we cant create a rule for divisibility for any number based solely on the sum of its digits. (for 6, its more like a0 + a1*4 + a2*16 + ... + an*4^n, which is a smaller number that what we begun with, but not as convenient as simply summing the digits)

Apart from that, I think its good, but I would add a statement at the start of the modular arithmetic part (for beginners to the topic) that modular arithmetic deals with whole numbers, so all numbers (in the modular arithmetic segment / chapter) from now on are to be assumed whole, unless stated otherwise.

Ah good catch. The fact that I glossed over in that statement is that for a number to be divisible by 6, the sum of its digits must also be divisible by 3 AND it must be even. The former property is what I was referring to. For succinctness, I'll just eliminate 6 from the statement. Thanks for the help!