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.
122,300,223รท3 = 40,766, 741
1+2+2+3+2+2+3 = 15
threw up and died while reading this
I wish I could read ๐
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!
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!
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.
What does the proof for this look like?
And since both 3 and 17 are prime numbers, that makes 51 a semiprime number
Which is not really rare under 100.
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
๐
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
First non fibonacci prime
Nobody told her that 100,000,001 is also divisible by 17
Holy crapballs
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.
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
I love how every reply has like the opposite energy to the meme. I also find math to be generally awesome.
Math is hard, so I'm just going to assume that's true and move on with my day.
Why did she share this? Does she hate us? I don't even know her.
So is 100,000,001.
When you start playing modded minecraft you get really good at multiplying and dividing by 144
Big, if true
Large if factual
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.
Upon closer inspection, yeah. 51 = 17 ร 3
= (10 + 7) ร 3
= (10 ร 3) + (7 ร 3)
= 30 + 21 = 51
EDIT: Brackets added.
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