▲ 1265 ▼ Panik (i.postimg.cc) submitted 2 years ago by ickplant@lemmy.world to c/lemmyshitpost@lemmy.world 146 comments fedilink hide all child comments
[–] beckerist@lemmy.world 17 points 2 years ago (1 child) Same with 9. There are rules for every number at least through 13 that I once knew... permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 23 points 2 years ago (3 children) 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. permalink fedilink source parent hideshow 6 child comments replies: [–] beckerist@lemmy.world 16 points 2 years ago (1 child) 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... permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 18 points 2 years ago (3 children) 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. permalink fedilink source parent hideshow 6 child comments replies: [–] tigeruppercut@lemmy.zip 5 points 2 years ago (1 child) But at least I seems like you could do that trick in your head permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 4 points 2 years 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. permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago* [deleted] permalink fedilink source parent [–] Frozengyro@lemmy.world 5 points 2 years ago (1 child) I'm sure every digit has rules to figure it out if you get technical enough. permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 8 points 2 years ago (1 child) 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. permalink fedilink source parent hideshow 2 child comments replies: [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent [–] octoperson@sh.itjust.works 4 points 2 years ago 11 is alternating sum So, first digit minus second plus third minus fourth... And then check if that is divisible by 11. permalink fedilink source parent
[–] logicbomb@lemmy.world 23 points 2 years ago (3 children) 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. permalink fedilink source parent hideshow 6 child comments replies: [–] beckerist@lemmy.world 16 points 2 years ago (1 child) 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... permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 18 points 2 years ago (3 children) 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. permalink fedilink source parent hideshow 6 child comments replies: [–] tigeruppercut@lemmy.zip 5 points 2 years ago (1 child) But at least I seems like you could do that trick in your head permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 4 points 2 years 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. permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago* [deleted] permalink fedilink source parent [–] Frozengyro@lemmy.world 5 points 2 years ago (1 child) I'm sure every digit has rules to figure it out if you get technical enough. permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 8 points 2 years ago (1 child) 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. permalink fedilink source parent hideshow 2 child comments replies: [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent [–] octoperson@sh.itjust.works 4 points 2 years ago 11 is alternating sum So, first digit minus second plus third minus fourth... And then check if that is divisible by 11. permalink fedilink source parent
[–] beckerist@lemmy.world 16 points 2 years ago (1 child) 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... permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 18 points 2 years ago (3 children) 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. permalink fedilink source parent hideshow 6 child comments replies: [–] tigeruppercut@lemmy.zip 5 points 2 years ago (1 child) But at least I seems like you could do that trick in your head permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 4 points 2 years 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. permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago* [deleted] permalink fedilink source parent
[–] logicbomb@lemmy.world 18 points 2 years ago (3 children) 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. permalink fedilink source parent hideshow 6 child comments replies: [–] tigeruppercut@lemmy.zip 5 points 2 years ago (1 child) But at least I seems like you could do that trick in your head permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 4 points 2 years 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. permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent [+] beckerist@lemmy.world 1 point 2 years ago* [deleted] permalink fedilink source parent
[–] tigeruppercut@lemmy.zip 5 points 2 years ago (1 child) But at least I seems like you could do that trick in your head permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 4 points 2 years 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. permalink fedilink source parent
[–] logicbomb@lemmy.world 4 points 2 years 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. permalink fedilink source parent
[–] Frozengyro@lemmy.world 5 points 2 years ago (1 child) I'm sure every digit has rules to figure it out if you get technical enough. permalink fedilink source parent hideshow 2 child comments replies: [–] logicbomb@lemmy.world 8 points 2 years ago (1 child) 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. permalink fedilink source parent hideshow 2 child comments replies: [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent
[–] logicbomb@lemmy.world 8 points 2 years ago (1 child) 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. permalink fedilink source parent hideshow 2 child comments replies: [+] beckerist@lemmy.world 1 point 2 years ago [deleted] permalink fedilink source parent
[–] octoperson@sh.itjust.works 4 points 2 years ago 11 is alternating sum So, first digit minus second plus third minus fourth... And then check if that is divisible by 11. permalink fedilink source parent
