2y ago

Panik

146 comments

Also, any number whose digits sum to a multiple of 3 is divisible by 3. For 51, 5+1=6, and 6 is a multiple of 3, so 51 can be cleanly divided by 3.
- I'd forgotten this trick. It works for large numbers too.
  122,300,223÷3 = 40,766, 741
  1+2+2+3+2+2+3 = 15
  
  threw up and died while reading this
  
  The neat part is that if you add the numbers together and they're still too large to tell, you can do it again. In your example, you get 15. If you do it again, you get 6, which isn't the best example because 15 is pretty obvious, but it works.
- Witchcraft! Burn them!
  
  She turned me into a newt!
- Fuck you and take an upvote for coming here to state what I was going to when I immediately summed 5+1 to 6 and felt clever thinking "well I do know it's not prime and divisible by 3" Shakes fist
  I'll get you NEXT time logicbomb!
  
  Posted the same info. Silly me
- Same with 9. There are rules for every number at least through 13 that I once knew...
  
  I only know rules for 2 (even number), 3 (digits sum to 3), 4 (last two digits are divisible by 4), 5 (ends in 5 or 0), 6 (if it satisfies the rules for both 3 and 2), 9 (digits sum to 9), and 10 (ends in 0).
  I don't know of one for 7, 8 or 13. 11 has a limited goofy one that involves seeing if the outer digits sum to the inner digits. 12 is divisible by both 3 and 4, so like 6, it has to satisfy both of those rules.
- What does the proof for this look like?
  
  https://math.stackexchange.com/questions/341202/how-to-prove-the-divisibility-rule-for-3-casting-out-threes
  
  90°
- And since both 3 and 17 are prime numbers, that makes 51 a semiprime number
  
  Which is not really rare under 100.
  
  Which is why it feels kind of prime, imho. I don't know if other people get this, but I get a sense of which two-digit numbers are prime probably because of how often they show up in times tables and other maths operations.
  3*17 isn't a common operation though and doesn't show up in tables like that, so people probably aren't generally familiar with it.
  
  Do do do, do do do do.
- Does this also work the other way round, i.e. do all multiples of three have digits that sum to a multiple of 3? All the ones I've checked so far do, but is it proven?
  
  Indeed, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
  For proof, take the polynomial representation of an integer n = a_0 * 10^{k + a_1 * 10}{k-1} + ... + a_k * 1. Note that 10 mod 3 = 1, which means that 10^{i mod 3 = (10 mod 3)}i = 1. This makes all powers of 10 = 1 and you're left with n = a_0 + a_1 + ... + a_k. Thus, n is divisible by 3 iff a_0 + a_1 + ... + a_k is. Also note that iff answers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 3
- Username checks out
- Show off
- Til thanks
- Damn, logicbomb indeed!
- Oh, neat trick!

51 = 3*17
3*17 = 17 + 17 + 17
17 + 17 + 17 = (10+7) + (10+7) + (10+7)
(10+7) + (10+7) + (10+7) = 30 + 21
30 + 21 = 51
yup, math checks out
- I think you skipped a step:
  1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
  Edit: Ohhhh, math by tens, I totally missed it. In that case, my mind wants to break it down to (10 * 5) + 1, and I'd still totally miss 17 as a possible factor.
  
  You miss a couple os steps too.
  First, lets define the axioms, we're using Peano's for this exercise.
  Axiom 1: 0 is a natural number.
  Jump to axiom 6, define the succession function s(n) where s(n) = 0 is false, and for brevity s(0) = 1, s(s(0)) = 2 and so on...
- 51 = 3*17
  3*17 = 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3
  3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1)
  (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) + (2+1) = 34 + 17
  34 + 17 = 51
  👌
- Math is mathing

This is why I love the number 7. It's the first real prime number. All the others are "first"...1?2?3?5? No, those aren't prime numbers, they're "first" in a long line of not-prime numbers.
Then you get to 7. Is 27943 divisible by 7? If you take away 3 is it? If you add 4 is?
I have no clue, give me 10 minutes or a calculator is the only answer
That's what a real prime number is.
- Take the last digit of the number, double it and subtract it from the rest. If that new number is divisible by 7, the original one is as well. For your example:
  2794 - 6 = 2788
  I know 2800 is divisible by seven, so 2788 is not. Thus 27943 is not divisible by 7.
  Quick maff shows that neither subtracting 3 or adding 4 will make the original number divisible by 7. Adding 1 or subtracting 6 will tho.
  
  Our plan to find the witch has worked, boys! Get her!
  
  Quick check for divisibility: subtract 7 from it. If the new number is divisible by 7, then the original number is too
  
  ~~But what about 14, 21 and 28?~~
  ~~14 - 4*2 = 6, not divisible by 7~~
  ~~21 - 1*2 = 19, not divisible by 7~~
  ~~28 - 8*2 = 12, not divisible by 7~~
  ~~Or did I misunderstand the algorithm?~~
  EDIT: I didn't realize that you remove the last digit when subtracting, got corrected in the replies.
  
  okay I understand that this works, but is there a mathematical proof for this?
- First non fibonacci prime
- Was this comment made by the timecube guy?
- 27943 - 7*1000 = 20943
  20943 -731000 = 20943 - 21000 = -57
  -57 is not divisible by 7 therefore 27943 is not divisible by 7.
  
  The other posters algorithm was better, but I was exaggerating - ultimately my point is you have to math it out

Nobody told her that 100,000,001 is also divisible by 17
- Holy crapballs
- This one does more than the one OP showed
  
  Amirite

Any number where the individual digits add up to a number divisible by '3' is divisible by 3.
51 = 5+1 = 6, which is divisible by three.
Try it, you'll see it always works.
- There are tricks like that for a lot of numbers. For 7, chop off the last digit, double it and add it to what's left. Repeat as required. If the result is divisible by 7 then the original number was. eg: 356 -> 35+12=47 not db7. 357 =>35+14 both db7 so we don't even need to do the add.
  
  Clever.
  
  double it and ~~add it to~~ subtract it from what’s left
  
  14: 1 + 8 = 7. Not db7
- One of the reasons why I love the number 3. There are other neat digit sum tricks, see for example for the numbers 1 to 30 here: https://en.m.wikipedia.org/wiki/Divisibility_rule
  
  They didn't teach stuff like this in school, which is silly. This is the kind of thing that a kid would eat up. It's like they wanted to make sure people hated math.
- I knew that worked with 9. Hmm, does it work with 6?
  Doesn’t look like it.
  
  Technically it does work for 6, more literally, still aiming for 3, not 6. That's half of it, if the starting number is even and divisible by 3 then it is also divisible by 6.

my palms are sweaty
- Mom's spaghetti
  
  17

wait till she finds out that 0.99999... 9's to infinity is the same as 1
- Lmao how about ...99999 = -1?
  
  This one has always bothered me a bit because ....999999 is the same as infinity, so when you're "proving" this, you're doing math using infinity as a real number which we all know it's not.

Math is hard, so I'm just going to assume that's true and move on with my day.

So is 100,000,001.

Why did she share this? Does she hate us? I don't even know her.

I love how every reply has like the opposite energy to the meme. I also find math to be generally awesome.
- That's Lemmy for you!

When you start playing modded minecraft you get really good at multiplying and dividing by 144
- Why 144? You mean a Minecraft stack of 64?
  
  Let me introduce you to the wonders of 144 millibuckets ingots

I used to do this thing where I would figure out if a number was prime or not and it kept me sane. Realizing this isn't, may have just caused my whole world to fall apart.
- If you skipped checking divisibility by 3 you already messed up.

Big, if true
- Large if factual
  
  Humongous if honest
  
  Obtouse if precise

Upon closer inspection, yeah. 51 = 17 × 3
= (10 + 7) × 3
= (10 × 3) + (7 × 3)
= 30 + 21 = 51
EDIT: Brackets added.
- = 10 × 3 + 7 × 3
  Careful, that's how you get a Facebook thread with people claiming 51 = 111... /s
  
  Would it be better to add brackets? I think I'm gonna edit to add brackets.
  EDIT: I edited to add brackets.

weird how ppl are getting all excited over this. weirder all the random math facts on the comments. and everyone checking with long math as if it might not be lol. I guess I'll throw a few math facts in?
17 is a prime number. 3 is a prime number.
all numbers can be factored down to primes.
19 is a prime number.
19*3=57. is that one gross too?
- Oh yeah? Then factor zero into primes, poindexter
  
  shut up, fascist! 😭
- What's weird is that 17 feels like a small enough number where it seems like we should know intuitively what its multiples are. And it feels like by this point in our lives we should at least know all numbers up to 100 or so that are composite vs prime. But yeah it's actually not that weird when you consider that the multiplication table usually stops at 12. And also that we really don't get that much exercise in multiplication in daily life.
- Nah, 17 is underage so its extra creepy.
- 19*3=57. is that one gross too?
  It's the Grothendieck prime!

This is the only one kind of math that professional darts players will dominate.

I actually really like this. 17 is three less than 20, 20x3 is 60, 3x3 is 9, 51 is 60-9. It just feels nice how it all fits together.
- Numberwang!

Curse you OP! Why did you post this?
- Didn’t want to be the only one to suffer.

Oh no! 😰
That made my back hurt.

Technically, isn't everything divisible by any number? You just get remainders and/or fractions in the result?
I mean, I still didn't want to know this, but....
- The definition for "divisible" is being able to be divided without a remainder.
- isn't everything divisible by any number?
  Do 5/0
  
  Okay, now what?
- By definition to be divisible by a number you have to a have a whole number with no remainder as an answer

Obligatory
- WTF, we are making videos from text posts now ? It feels so weird... Instead of reading a post in 10s, I get to wait 35s for the video to unfold the text discussion, and youtube gets to puts ads on top of it, what a time to be alive..
  
  But it's in a British accent, it automatically makes you smarter!
- Here is an alternative Piped link(s):
  Obligatory
  Piped is a privacy-respecting open-source alternative frontend to YouTube.
  I'm open-source; check me out at GitHub.

17 * 3 baby!

I wasn't that surprised to be honest... but than again, I'm divergent 🤷.
- Your limit goes to infinity???
- Same same. I was like "its 3x20 minus 3x3, 60 minus 9...?"
  
  Yeah, I would be like 😲 for 101, but 51... yeah, not that surpised.
- What the fuck does divergent mean? Are you divisible by 17 or not?
  
  lol 🤣🤣🤣 i might be, depends how you look at things 🤣

146 comments