Comment on Panik
beckerist@lemmy.world 1 year agoSame with 9. There are rules for every number at least through 13 that I once knew…
Comment on Panik
beckerist@lemmy.world 1 year agoSame with 9. There are rules for every number at least through 13 that I once knew…
logicbomb@lemmy.world 1 year ago
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.
beckerist@lemmy.world 1 year 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 1 year 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 1 year ago
But at least I seems like you could do that trick in your head
octoperson@sh.itjust.works 1 year ago
11 is alternating sum
So, first digit minus second plus third minus fourth…
And then check if that is divisible by 11.
Frozengyro@lemmy.world 1 year ago
I’m sure every digit has rules to figure it out if you get technical enough.
logicbomb@lemmy.world 1 year 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.