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176 points1 year agoAlso, 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.

44 points1 year ago* (last edited 1 year ago)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

52 points1 year agothrew up and died while reading this

16 points1 year agoI wish I could read 😞

7 points1 year agoJust squint and wing it.

2 points1 year agoThat is way too accurate. Lol

2 points1 year ago^ This. The thing about Arsenal is they always try and walk it in.

1 points1 year agoAlso works with 9s!

26 points1 year agoThe 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.

8 points1 year agoBut how do I prove it for 6

11 points1 year agoGet 6 apples. Duh.

2 points1 year agoThere is a mathematical proof that 1 + 1 = 2 so surely you could make a proof for 6 ÷ 3 = 2

2 points1 year agoProve it for 2, then un-distribute.

27 points1 year agoWitchcraft! Burn them!

9 points1 year agoShe turned me into a newt!

8 points1 year ago...I got better

21 points1 year agoFuck 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 fistI'll get you NEXT time logicbomb!

3 points1 year agoPosted the same info. Silly me

17 points1 year agoSame with 9. There are rules for every number at least through 13 that I once knew...

23 points1 year 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.

16 points1 year ago7 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...

18 points1 year agoAnother 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.

5 points1 year agoBut at least I seems like you could do that trick in your head

4 points1 year agoIf 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.

1 points1 year ago1 points1 year ago* (last edited 1 year ago)5 points1 year agoI'm sure every digit has rules to figure it out if you get technical enough.

8 points1 year agoI 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.

1 points1 year ago4 points1 year ago11 is alternating sum

So, first digit minus second plus third minus fourth...

And then check if that is divisible by 11.

13 points1 year agoWhat does the proof for this look like?

17 points1 year agohttps://math.stackexchange.com/questions/341202/how-to-prove-the-divisibility-rule-for-3-casting-out-threes

3 points1 year ago90°

10 points1 year ago* (last edited 1 year ago)And since both 3 and 17 are prime numbers, that makes 51 a semiprime number

10 points1 year agoWhich is not really rare under 100.

3 points1 year agoWhich is why it feels

kind ofprime, 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.

1 points1 year agoDo do do, do do do do.

9 points1 year agoDoes 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?

8 points1 year agoIndeed, 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

iffanswers your question then; all multiples of 3 have to, by definition, have digits whose sum is a multiple of 32 points1 year agoThank you for this detailed response 🙏

4 points1 year agoUsername checks out

3 points1 year agoShow off

3 points1 year agoTil thanks

2 points1 year agoOh, neat trick!

2 points1 year agoDamn, logicbomb indeed!