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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.
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
Get 6 apples. Duh.
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
Witchcraft! Burn them!
She turned me into a newt!
...I got better
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.
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.
11 is alternating sum
So, first digit minus second plus third minus fourth...
And then check if that is divisible by 11.
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.
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
Thank you for this detailed response 🙏
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Til thanks
Damn, logicbomb indeed!
Oh, neat trick!