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 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.
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.
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.
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.
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!
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.
deleted by creator
deleted by creator
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.
deleted by creator
11 is alternating sum
So, first digit minus second plus third minus fourth…
And then check if that is divisible by 11.
And since both 3 and 17 are prime numbers, that makes 51 a semiprime number
Which is not really rare under 100.
Do do do, do do do do.
Til thanks
Show off
Username checks out
Damn, logicbomb indeed!
Oh, neat trick!