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
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.
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!
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.
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.
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.
11 is alternating sum
So, first digit minus second plus third minus fourth...
And then check if that is divisible by 11.
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.
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
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
👌
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.
You clearly mean:
14: 1 + 8 = 9 (not db 7)
Someone else in this thread correctly stated:
"Chop of last digit, double it and subtract from what is left"
14: 1 - 8 = -7. (dB 7!)
Math is awesome, I didn't know this trick!
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.
My experience of maths in high school was being taught a trick or method to solve a really specific type of problem every week. Sometimes the method would build off something we'd learnt the previous week.
The whole thing was bottom-up learning where you get given piecemeal nuggets of information but never see the big picture. They completely lost me at around the age of 15. I eventually came back to maths later in life after studying formal logic in my philosophy undergrad degree.
I first saw the nine times finger trick in the movie 'Stand And Deliver.' I remember seeing it written out on a blackboard at some point, but never the finger trick.
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I love how every reply has like the opposite energy to the meme. I also find math to be generally awesome.
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.
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.
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
It "looks" or "feels" prime. And being divisible by a prime like 17 feels even stranger.
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.
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
When you start playing modded minecraft you get really good at multiplying and dividing by 144
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.
Upon closer inspection, yeah. 51 = 17 × 3
= (10 + 7) × 3
= (10 × 3) + (7 × 3)
= 30 + 21 = 51
EDIT: Brackets added.
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?
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.
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.
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....
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By definition to be divisible by a number you have to a have a whole number with no remainder as an answer