11 is alternating sum
So, first digit minus second plus third minus fourth…
And then check if that is divisible by 11.
Comment on Panik
logicbomb@lemmy.world 8 months agoI 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.
octoperson@sh.itjust.works 8 months ago
Frozengyro@lemmy.world 8 months ago
I’m sure every digit has rules to figure it out if you get technical enough.
logicbomb@lemmy.world 8 months ago
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.
beckerist@lemmy.world 8 months ago
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…
logicbomb@lemmy.world 8 months ago
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.
tigeruppercut@lemmy.zip 8 months ago
But at least I seems like you could do that trick in your head
logicbomb@lemmy.world 8 months ago
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.