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.
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.
7 is double the last number and subtract from the rest
749 (easily divisible by 7 but for example sake)
9*2=18
74-18=56
6*2=12
5-12= -7, or if you recognize 56 is 7*8…
I’ll do another, random 6 digit number appear!
59271
1*2=2
5927-2=5925
5*2=10
592-10=582
2*2=4
58-4=54, or not divisible
I guess for this to work you should at least know the first 10 times tables…
Another way to tell if 59271 is divisible by 7 is to divide it by 7. It will take about the same amount of time as the trick you’re presenting, and then you’ll already have the result.
But at least I seems like you could do that trick in your head
If you have no interest in the result of the division, then you can also do the division in your head, without retaining the result, with about the same effort.
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I’m sure every digit has rules to figure it out if you get technical enough.
I looked up a rule for 7, and it seems like it would take about the same amount of time as actually dividing the number by 7.
Meanwhile, it looks like the rule for 8 is to see if the last 3 digits are divisible by 8, which seems like a real time save for big numbers.
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11 is alternating sum
So, first digit minus second plus third minus fourth…
And then check if that is divisible by 11.
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
I wish I could read 😞
Just squint and wing it.
That is way too accurate. Lol
^ This. The thing about Arsenal is they always try and walk it in.
Also works with 9s!
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.
But how do I prove it for 6
Prove it for 2, then un-distribute.
There is a mathematical proof that 1 + 1 = 2 so surely you could make a proof for 6 ÷ 3 = 2
Get 6 apples. Duh.
Oh, neat trick!
Witchcraft! Burn them!
She turned me into a newt!
…I got better
Show off
Damn, logicbomb indeed!
What does the proof for this look like?
90°
Username checks out
Til thanks
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
Thank you for this detailed response 🙏
And since both 3 and 17 are prime numbers, that makes 51 a semiprime number
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.
Which is not really rare under 100.
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?
19*3=57. is that one gross too?
It’s the Grothendieck prime!
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.
Here is an alternative Piped link(s):
Piped is a privacy-respecting open-source alternative frontend to YouTube.
I’m open-source; check me out at GitHub.
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!
How the hell does this have over a mil views?
Get off my lawn you stupid kids
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!
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
Obtouse if precise
Humongous if honest
Fat if fact
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.
You can also prove it a different way if you allow the use of the formula for finding the limit of the sum of a geometric series on a non-convergent series.
Sum(ar^n, n=0, inf) = a/(1-r)
So,
…999999
= 9 + 90 + 900 + 9000…
= 9x10^0 + 9x10^1 + 9x10^2 + 9x10^3…
= Sum(9x10^n, n=0, inf)
= 9/(1-10)
= -1
But why would you allow it?
Because you could argue that the series converges to …999999 in some sense
Yes, you’re right this doesn’t work for real numbers.
It does however work for 10-adic numbers which are not real numbers. They’re part of a different number system where this is allowed.
I don’t get it, why is this a big deal?
Fuck if I know, but look at the discussion this generated!
https://lemmy.ca/pictrs/image/58b19948-2d75-4d18-a2bb-fbf9362dc85b.jpeg
This I assume
Though the title of that meme should have been when millennials see zoomer memes
This is a shitpost
I know…I don’t get what the joke is supposed to be though.
It “looks” or “feels” prime. And being divisible by a prime like 17 feels even stranger.
It’s pretty common for numbers to be divisible by primes tho
You don’t say haha :D But there is no reason why people who don’t work or study in a math related field would have an intuition for that.
Nobody told her that 100,000,001 is also divisible by 17
This one does more than the one OP showed
Holy crapballs
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?
Restore a saved game to restore the weave of fate, or persist in the doomed world you have created
By definition to be divisible by a number you have to a have a whole number with no remainder as an answer
17 * 3 baby!
Math is hard, so I’m just going to assume that’s true and move on with my day.
my palms are sweaty
Mom’s spaghetti
17
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
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
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 + 1Edit: 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…
Curse you OP! Why did you post this?
