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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.
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?
It goes like this
create 2 distinct numbers by isolating the last digit from the other. For example, 154 becomes 15 and 4.
double the number derived from the last digit. So, the four becomes 8.
subtract from the number derived from the preceeding digits. 15 - 8.
the resulting number is 7. Seven is divisible by 7, so we know 154 is divisible by 7.
Yeah you got it wrong, it's
1 - 4*2 = -7
2 - 1*2 = 0
2 - 8*2 = -14
27943 - 7*1000 = 20943
20943 -7*3*1000 = 20943 - 21000 = -57
-57 is not divisible by 7 therefore 27943 is not divisible by 7.
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.
Girl.
double it and ~~add it to~~ subtract it from what’s left
14: 1 + 8 = 7. Not db7
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.
I guess I was one of the lucky few who learned this in elementary school. And later again.
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.
I knew that worked with 9. Hmm, does it work with 6?
Doesn’t look like it.
my palms are sweaty
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!
So is 100,000,001.
I don't get it, why is this a big deal?
This is a shitpost
I know...I don't get what the joke is supposed to be though.
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?
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.
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
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.
Curse you OP! Why did you post this?
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.
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):
Piped is a privacy-respecting open-source alternative frontend to YouTube.
I'm open-source; check me out at GitHub.
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
I wish I could read 😞
Just squint and wing 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