#if you alternate adding and subtracting the digits of a multiple of 11 you get 0

[hazellight11]

Last night for some reason i was struggling to fall asleep so (at like 1:30) i decided to text adrian (who had gone to sleep at about midnight) an explanation of how most letters in the english alphabet have some meaning that makes picking variables difficult. So that's approximately the state of my mind nowadays i guess

[rotationalsymmetry]

Happy Out Of Touch Thursday, or whatever your preferred "it's thursday" meme is. Like every other day, I'm gonna walk you through how to factor today's date in your head, in case anyone reading this is as much of a math weirdo as I am and also doesn't already know this. (Or does know but finds it entertaining anyways.) (There will be a lot of gratuitous "we" talk as though idk I'm teaching a class or something and of course you're doing this along with me.)

It's April 28, and 428 is divisible by, hmm, well at least 4, leaving 107. Yeah, 107 is the end of the line, all the 10_ numbers that could be prime (ie, numbers ending in 1, 3, 7, or 9) are prime. (I just know that, but if you don't you can use the approach I'm going to describe below.) 428's prime factors are 2^2*107.

It's 28 April, and 2804 is divisible by 4, leaving 701, which is probably also prime but I’m not sure, so let’s go, we’ll potentially have to check up to something in the 20’s. (Factors come in pairs: any factor higher than the square root of the original number must have another factor lower than the square root of the number. So checking every prime number up to the square root of the number you’re factoring is enough.) 25^2 is 625, 26^2 will be under 701 but 27^2 will be over it, and both of those are composite, so we have to check (potentially) up to 23.

(What is 27^2? That would be 3^6. Do I know what 3^6 is? 3, 9, 27, 81, 243, 729. OK, well that looks familiar anyways.)

701 is not divisible by 2, because it ends in an odd digit. It’s not divisible by 5 for similar reasons. The digits sum to 8, which is not 3, 6 or 9 so it’s not divisible by 3. It’s not divisible by 7, because 700 is, and 701 is only 1 over 700. It’s not divisible by 11, because the sum of alternate digits differ by 8, not by 0 or 11 (or 22 etc.)

What about 13? The general pattern is to do a bunch of basic arithmetic that preserves the divisible by 13/not divisible by 13 quality until we get a number that is unmistakably either divisible or not divisible by 13. We can do this by adding or subtracting multiples of 13, and dividing by numbers that are relatively prime with 13 -- ie any other number. Generally that's going to be 10, because that's the easiest number to divide by. So the first step is adding or subtracting a number that will give us a number ending in zero.

Well, adding or subtracting by 13 won't get us there, but adding 39 (13*3) will. That gets us 740, which we can divide by 10 to get 74. Do I know all the multiples of 13? I do not. But, I can right now multiply 39 by 2 to get 78, there’s no way both 74 and 78 can be divisible by 13. So 701 isn’t divisible by 13.

(Of course if you’re writing it out, you can do long division, but we don’t need to know the quotient right now, just whether there is a whole-number quotient, and this method is easier to do in your head without losing track of something important.)

17? Using similar arcane methods (subtract 51 getting 650, divide by 10, divide by 5, 13 is not 17) we can determine that 701 is not divisible by 17.

With 19 we don’t have to multiply 19 by anything, we can just add it, getting 730. 19*4 is 76, so 19 isn’t a factor of 701 either.

It’ll be incredible if we find out at the last minute that 701 is divisible by 23. Is it? Well, we can multiply 23 by 3 to get 69 and add that to 701 to get 770, dividing by 10 we get 77 which is 7*11, which you will note doesn’t have a 23 anywhere. So no.

701 is prime — its only factors are 1 and itself.

2804's prime factors are 2^2*701.

Today is the 118th day of the year. 118 is (just doing normal division, 2 goes into 11 five times carry the 1 etc, or 118 is 100+18 and 100/2 is 50 and 18/2 is 9, adding up that's 59) 2*59.

[rotationalsymmetry]

All right. It’s April 21. Here goes.

I’m walking you through how to find the prime factors of the date in month-first then day-first notation, because why wouldn’t you want to know that? This is the sort of thing I do when I’m on a bus and for some reason can’t use my phone. Or, in this case, when I’ve woken up but am putting off actually getting out of bed and starting my day.

421 is divisible by…well, the last digit is not even, so it’s not divisible by 2. The last digit is not a 0 or 5, so it’s not divisible by 5. The digits add up to 7, which is not a multiple of 3, so it’s not divisible by 3. (This trick is specific to the number 3.)

For most primes we use the idea that if a number is divisible by n, you can add or subtract any multiple of n and divide anything, and it stays divisible by n, and same if it isn’t. If 421 was divisible by 7, we could subtract 21 and get another number divisible by 7, so 400 would be, and 4 would be, and that just doesn’t bear thinking about. So 421 is not divisible by 7.

It’s not divisible by 11, or 13, or 17, or 19. That’s as high as we have to check. We only have to check up to the square root of 421, because factors come in pairs: any factor greater than the square root of the original number is paired with a factor less than the square root (if the number is a perfect square, the square root is effectively paired with itself.) Since the square root of 421 is between 20 and 21 (20^2 is 400, 21^2 is (20+21) more than that), we only have to check to the last prime number before 21. And we’re only checking primed because if a factor of the original number is not prime, that number also divisible by a smaller number that is prime. For instance, 24 is divisible by 6, but it’s also divisible by the smaller prime numbers 2 ad 3, and if you know a number isn’t divisible by 2 or 3 then it’s not going to be divisible by 6 either.

Therefore 421 is divisible by nothing at all other than one and itself — it’s prime. (And we got that with only checking 8 actual numbers.) (Yes, doing an internet search is faster.)

If we put the day first, 2104 is not prime because it is at least divisible by 4 (last digit is a 4 or 8 and next to last is even = divisible by 4, if you don’t like that you can divide by 2 twice), leaving 526. That’s divisible again by 2, leaving 263. Is 263 prime?

It’s not divisible by 2 or 5 (last digit) or 3 (digits don’t sum to a multiple of 3.) it’s not divisible by 11, because alternate digits don’t add up to the same thing or numbers that are a multiple of 11 apart. It’s not divisible by 7, because if you add 7 to 263 you get 270, divide by 10 you get 27, which is close to 28 but not there. It’s not divisible by 13, because you can do a similar thing with subtracting 13.

Do we have to check 17? Hmm. 16^2 is 256 (…I just know that), so there’s no way 17^2 is less than 263, which is only a little bit higher than 256. That’s the end of the line, 263 is prime. 2104 is 2^3*263. Two to the three times two sixty-three. Ok. Not bad. Especially since, I mean this has nothing to do with math but is kind of neat, the ^ sign is written by hitting shift-6 on physical keyboards. I’m typing this on my phone, but it’s still kind of cool.

it’s the 111th day of the year, which besides looking awesome (one one-hundred, one ten, one one) is also divisible by 3, leaving 37. In base 8, 111 is equivalent to 73 in base 10. In base 2, 111 is 7 in base 10. In hexadecimal (base 16), it’s…yuck. 256+16+1, 273, huh, that’s a neat number because it happens to be 3*7*13. As far as I can tell that’s a coincidence. In base 6 (why base 6? Why not) it’s 43. In base 12, it’s 157. In base 3, it’s 13.