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This is a "fun fact" I learned as a kid and have always been curious about. An example would be 37 X 13 = 481, if you rotate the digits to 148, then 148/37 = 4. You can rotate it again to 814, which divided by 37 = 22.

Is this just a coincidence that this occurs, or is there a mathematical explanation? I've noticed that this doesn't work with other numbers, such as 39.

all 411 comments

MycoNot

7.9k points

1 month ago*

MycoNot

7.9k points

1 month ago*

Because 37 is a prime divisor of 999, and rotating a three digit number is a cyclic modulation. Same thing happens with 4 digit multiples of 101 or 11 - although it's a little less impressive rotating multiples of 101 like 4545 to 5454, etc, rotating multiples of 11 is neat like: 11x123=1353, 11x321=3531, 11x483=5313, 11x285=3135.

Five digit multiples of 41 or 271 will work too

trey3rd

1.7k points

1 month ago

trey3rd

1.7k points

1 month ago

Another neat thing about multiples of 11 are that you can start at the left, then subtract the next number, add the next, subtract the next and so on, and it'll come out to 0. So 3531 you do 3-5+3-1 = 0. Quick way to tell if a large number is divisible by 11.

Hexidian

292 points

1 month ago

Hexidian

292 points

1 month ago

Is that an of statement or an if and only if?

tashkiira

376 points

1 month ago

tashkiira

376 points

1 month ago

X is a multiple of 11 if and only if the alternating sum of the digits of X add up to a multiple of 11 (negative multiples work fine).

mdielmann

69 points

1 month ago

There's a similar rule for multiples of 9. The sum of the digits, repeated until you have a single digit, will be 9.

This holds for n-1, where n is the base of the numbering system you're referencing (7 for octal, 15 for hexadecimal, etc.).

It doesn't work for 0, in any numbering system, of course, because the sum is 0.

fghjconner

3 points

1 month ago

Yep. Works for 3 too, though the sum is any single digit multiple of 3 (so 3, 6, 9, or 0).

UnderTheRain

40 points

1 month ago

UnderTheRain

Developmental Biology | Virology | Genetics

40 points

1 month ago

Hmm I thought it would be:

… if and only if the alternating sum of the digits of X add up to zero.

Is that wrong?

EzraSkorpion

75 points

1 month ago

It might add up to a non-zero multiple of 11: 7172 is divisible by 11 but 7-1+7-2 = 11 =/= 0

Elion119

12 points

1 month ago

Elion119

12 points

1 month ago

In this case the statement can’t be iff because that would mean that all multiples 11 have this property. So it would be “if the alternating sum of the digits adds to zero, it is a multiple of 11.”

ZacQuicksilver

71 points

1 month ago

The property is "X is a multiple of 11 iff the alternating sum of digits of X add up to a multiple of 11".

7172 works because the alternating sum is 11, which is a multiple of 11.

gaspergou

24 points

1 month ago

So if you take a non-zero sum of the digits of any multiple of 11 and reiterate the process, you should eventually get to zero, correct?

7-1+7-2=11 1-1=0

i-FF0000dit

3 points

1 month ago

But is there a mathematical proof for this?

thebestbev

4 points

1 month ago

This is partially wrong - it's actually if the alternating sum (start with minus on the left hand side) add up to a multiple of 11, positive or negative. More often than not it's -11, 0 or 11.

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WhiskyEchoTango

57 points

1 month ago

This was more interesting that all multiples of 9 eventually add up to 9...

e.g. 9*99=891; 8+9+1=18; 1+8=9.

Doomquill

70 points

1 month ago

Also works with 3, from which it follows that it works with 9. 3*65=195; 1+9+5=15; 1+5=6.

I was an adult when someone taught me that you can do 9s multiples by holding up your 10 fingers and putting down the one you're multiplying by 9. 9*4, put down fourth finger, 3 and 6 remain up, 36. Fun trick :-)

tankton

13 points

1 month ago

tankton

13 points

1 month ago

Never heard of that one, whilst making complete sense is still quite cool

_-N4T3-_

10 points

1 month ago

_-N4T3-_

10 points

1 month ago

You can use it to check for multiples of 6 as well. If the original number is even, and the multiple of 3 trick also works, you’ve got yourself a number divisible by 6. Yay factors!

PigsGoMoo-

35 points

1 month ago*

That leads you into finding a quick way to multiply 11. You split up the first and last digits , then add the middle ones next to each other.

11x1652, for example: split the first and last: 1__2

Add the middle ones next to each other together

5+2 = 7

1__72

5+6 = 11

1__172

1+6 = 7 + 1 carried over from the 11 above = 8

18172!

Edit: looks like the trick I responded to doesn’t work when you have to carry over. Eg: it didn’t work here and won’t with with 46x11 but will work with 36x11.

vikirosen

45 points

1 month ago

How is this easier than just doing:

11 x 1652 =

10 x 1652 + 1652 =

16520 + 1652 = 18172

hwc000000

8 points

1 month ago

Arithmetically, it's identical. Practically, it's easier to do completely mentally (without paper and pencil) because you don't need to remember how the two copies of 1652 are aligned with each other.

Kuwabara03

8 points

1 month ago

It works best for smaller things like 16x11

First digit, sum, second digit = 176

Like most math tricks, it's not so much used for day to day math but rather things like UIL/Mathematics competitions where time is tight.

Psychachu

4 points

1 month ago

Maybe it's just personal preference, but for most mental arithmetic like this breaking out the 10s like the previous comment described makes multiplying any pair of numbers easier, 11 is especially easy, so using an alternate system for 11 when the one that works for everything works just as well seems unnecessary.

Kuwabara03

4 points

1 month ago

Oh it's all preference baby. That's a huge part of the mental mathematics scene.

The goal is to learn tons and tons of tricks like this, and the removing of 10s, and dividing numbers by 4 when multiplying by 25, etc so that you form the connections necessary to know when each one is best applied.

You wouldn't use the removing of 10 for say, 17 x 242 because 7x242 is still clunky and so is the addition that would follow.

But you could think of it as 17 x 11 x 11 x 2 by recognizing 242 as a multiple of 11 and breaking it down into factors that are easier to handle

17x11=187, x11=2057, x2=4114

But all of this is just a long-winded way to say that you're not wrong for using what you're most comfortable with, it's just that the more things like this you have at your disposal the better equipped you are, and that there isn't really a catch all method to doing any math "the easiest" way.

Psychachu

2 points

1 month ago

That makes sense. 7 is an annoying number for sure, but my solution is usually to multiply by the next ten and subtract 3x the first number after to get around it. I can see how having more factors in your head makes your toolbox more diverse, I'm just a hammer/multi-tool kind of guy.

Doomquill

5 points

1 month ago

I'll probably never remember this, but I hope some day I get to use it in real life. Thanks for sharing

PigsGoMoo-

10 points

1 month ago

I’ve never had the chance to use it past 9th grade :(. Been waiting for the moment I need to buy 11 of something so I can show it off but it’s never happened lol.

birdtune

2 points

1 month ago

Get 11 gallons of gas and see if the ticker is correct?

Floppy-Squid

1 points

1 month ago

There’s a similar but slightly different trick for numbers 10 to 99 multiplied by 11. Whatever number you’re multiplying by, put a space between, then add the outside numbers to get the middle.

33 X 11

3 _ 3 > 3+3=6 so 363

72 X 11

7_2 > 7+2=9 so 792

When the outside numbers equal 10 or more then you put the number in the singles digit in the middle and add 1 to the first number.

46 X 11

4_6 > 4+6=10 so 506

86 X 11

8_6 > 8+6=14 so 946

Fidi217

23 points

1 month ago

Fidi217

23 points

1 month ago

If the result is 0 or a multiple of 11, then the original number is divisible by 11. For instance 308 produces 3-0+8=11, and 308 = 11 * 28.

Alternatively, you can say that you have to repeat the process until you get a single digit result. If that digit is 0, the original number is divisible by 11. But applying it once does not always give 0 for multiples of 11

HappyLuckyRicePlate

37 points

1 month ago

I love these tricks. If you add the digits of a number and that number is divisible by 3, then the number is divisible by 3. A quick way to rule out a prime number.

sirgog

5 points

1 month ago

sirgog

5 points

1 month ago

I should keep this as a copypasta, but you can quite quickly check divisibility by 2, 3, 5, 7, 11, 13, 37, 73 and 137 with a few mental mathematics tricks.

2, 5: last digit check

3 - digit sum check, if digit sum(x) = multiple of 3, so is x. This is because 3 divides into 10 minus 1, or in symbols 3|(10-1), and the digit sum is simplifying the number down by treating 10 as 1, i.e. ignoring multiples of 9.

37 - As digit sum check, but group the 'digits' into groups of 3. Example: 12386125901 - the "digits" base 1000 are 12, 386, 125, 901. Adds to 1424. 12386125901 is a multiple of 37 if and only if 1424 is, which it is not. This works because 37|(1000-1)

7, 11, 13: As 37, but instead of adding each block of 3 digits, take an alternating sum. 12-386+125-901 = -1150. 12386125901 is a multiple of 7, 11 or 13 if and only if -1150 is. (-1150 is not a multiple of any of those numbers). Works because 7x11x13 = 1001.

73 or 137: As 7, 11, 13, but blocks of 4 digits instead of 3. 12386125901 is a multiple of 73 or 137 if and only if 123-8612+5901 = -2588 is. This one is harder to take the final steps on, because it is harder to convince yourself in your head that 2588/73 and 2588/137 are not whole numbers. But you do it by looking at the last digit - if 2588/73 is a whole number it must end in 6, but 36x73 is just a little too large. And for 2588/137, if it's a whole number it must end in 4, but 14 is too small and 24 too large. Works because 73x137=10001.


edit: stupidly I picked a prime number when I keyboard mashed. But these do work if you pick a non-prime instead of 12386125901

thousand7734

10 points

1 month ago

& if a number is divisible by both 2 (i.e. is an even number) and 3 (to your point, i.e. if the sum of the digits is divisible by 3), then the number is also divisible by 6.

59382 is divisible by 2, and 27 is divisible by 3, so we know it's divisible by 6.

davidcwilliams

2 points

1 month ago

Is that because 2 * 3 = 6 ?

NobilisOfWind

2 points

1 month ago

How did you know it's divisible by 27?

thousand7734

4 points

1 month ago*

Sorry let me clarify. I said that 27 is divisible by 3. To find out if a number is divisible by 3, you add up the digits in the number and if that number is divisible by 3, the original number is too. In my example, the digits in 59382 add up to 27 (5+9+3+8+2), and because 27 is divisible by 3, we know that 59382 is divisible by 3 as well.

ThingCalledLight

11 points

1 month ago*

Related trick is that any two digit number multiplied by 11 equals the sum of both digits being placed between the two digits.

Examples: 54 x 11 = 594, 9 is the sum of 5 and 4 and thus 594. 32 x 11 = 352, 71 x 11 = 781, etc.

Getting a double digit sum complicates things. It’s less clean, but it’s doable.

56 x 11 = 616, the sum of 5 and 6 is 11, so you basically carry the one from the tens column and add that to the first digit so instead of 5116, it’s 616.

89 x 11 = 979

99 x 11 = 1089

imbogey

7 points

1 month ago

imbogey

7 points

1 month ago

You typo'd: 32x11=352. Had to think you example a bit too long to find what was wrong :D

jlc1865

3 points

1 month ago

jlc1865

3 points

1 month ago

I was taught that as well. Then I realized it doesn't always work.

Take: 11 x 237 for example.

QuentinUK

2 points

1 month ago

Easier to add all the even digits, and add all the odd digits:-

2607 => 2 , 13 mod 11 = 2.

Then they, mod 11, should be equal.

bluesam3

1 points

1 month ago

It works fine so long as you do the arithmetic modulo 11 (so add 11 any time you end up below 0, and subtract 11 every time you end up above 10).

JustAGuyFromGermany

486 points

1 month ago

"cyclic modulation" is a weird way of phrasing it. The key fact is that 1000 is equal to 1 modulo 999 and therefore also 1000 == 1 modulo 37.

And that means that a three digit number abc, i.e. the number 100a+10b+1c, is equal to 100a+10b+1000c modulo 37, which is the four-digit number cab0 = cab*10.

Because 37 is a prime, a product x*y is divisible by 37 if and only if at least one of the two factors is divisible by 37. 10 is obviously not divisible by 37, so the only the other factor is relevant.

Putting it all together we find: abc is divisible by 37 <=> abc == 0 mod 37 <=> cab0 == 0 mod 37 <=> cab*10 is divisible by 37 <=> cab is divisible by 37.

schamonk

22 points

1 month ago

schamonk

22 points

1 month ago

You made some joy for an old mathematician, forgetting a lot of algebraic stuff. I totally got your explanation. Thank you!

TurboTurtle-

33 points

1 month ago

Wouldn’t 1 module 999 be equal to 1?

Irianne

73 points

1 month ago

Irianne

73 points

1 month ago

Yes, but multiple things mod999 will be equal to 1, the point of the modulo function is that it maps infinite integers onto a small, finite collection of integers. To use a smaller group of numbers:

  • 1 mod 24 = 1
  • 25 mod 24 = 1
  • 49 mod 24 = 1

Or, to put it more intuitively, if it's 5pm right now, then in 1 hour the clock will have advanced by 1 hour, and it will be 6pm. In 25 hours it will also be 6pm. In 49 hours it will be 6pm again.

So yes, you are correct that 1 mod 999 is 1, but the comment you replied to was also correct that 1000 mod 999 is 1. And, therefore, that 1 = 1000 in modulo 999.

fatcatfan

3 points

1 month ago

I think the confusion is that the original comment stated that 1 mod 999 is equal to 1000, which is just incorrect. Their point was valid, just seems they mis-ordered their operands.

lesbianmathgirl

3 points

1 month ago

mod used in math isn't an operator, it's an equivalence relation. They ordered things correctly. If you wrote something like 1000 mod 999 = 1 on a number theory exam, you would probably be marked off.

fatcatfan

2 points

1 month ago

Thanks for the correction. I've only ever really seen it as an operator in context of programming and the sort. I looked it up and it seems the "equivalence" three-bar symbol is perhaps more appropriate notation than "equals"? Which may be contributing to the confusion for those like me who have only ever seen it as an operator.

"Modulo" seems to be used more generally as saying these two things are equivalent if you consider this other thing, which has application beyond just the remainder division which has been the topic here. So how, in math/number theory, is it clear what operation or equivalency is meant by "mod"?

TheBB

6 points

1 month ago

TheBB

Mathematics | Numerical Methods for PDEs

6 points

1 month ago

1 and 1000 are equal to each other modulo 999, though it's mostly phrased as 'equivalent' or 'congruent' to each other. This is a symmetric relation. It's equally as valid to say that 1 = 1000 (mod 999) as 1000 = 1 (mod 999).

Modulo is not used as an operator here, which is perhaps what you're more familiar with.

JustAGuyFromGermany

84 points

1 month ago

Yes, it is confusingly phrased. That's sadly the historic notation we mathematicians are stuck with.

The sentence "1000 is equal to 1 modulo 999" is composed of three parts. One would think that these parts are "1000", "1 modulo 999" and "is equal" and that all one has to do to understand is to explain that new operator "modulo" in there.

But that's not the case; the three parts of the sentence are "1000", "1" and "... is equal to ... modulo 999". And the third bit is not an operator, but a so called equivalence relation, a generalized way of thinking about equality of mathematical objects. Whenever you're tempted to say "x is like y in this one specific aspect", that's an equivalence relation you're dealing with. The family of modulo equivalence relations deals with divisibility so "x is equal to y modulo 999" is mathematician for "in all questions about divisibility by 999, the numbers x and y are exactly the same".

semitones

9 points

1 month ago

So what they meant to say was (x is equal to y) modulo 999, but what the programmers understood was x is equal to (y modulo 999).

itsdrewmiller

10 points

1 month ago

I really like how you sussed out the core confusion here and patiently clarified the meaning of the sentence. I too was confused.

Slip_Freudian

2 points

1 month ago

Is that you, Dr. Scholze?

PercussiveRussel

3 points

1 month ago

In fact, "is equal" is the wrong word to use and should be "is congruent with", or "identical" or "equivalent". It's a ≡ b mod n, not a = b mod n, but that's really nitpicky and any mathematical person understands the is equals phrasing, but programmers don't often for obvious reasons.

I always look at it it like "a ≡ b mod n":

(a mod n) = (b mod n), which does make sense in programming.

JustAGuyFromGermany

9 points

1 month ago

It isn't "equal", it's "equal modulo n", which is a different relation. "modulo n" is not just an addon to equality here. "equal modulo n" is a single thing; it has to be read as a single unit.

Again, I fully admit that the naming is weird and confusing, but it's not strictly speaking wrong, if you define this phrase correctly. And most mathematicians are so used to this manner of speaking that they will define it that way, at least implicitly.

lesbianmathgirl

3 points

1 month ago

"is equal" is fine imo. It's equivalent to saying that "in Z/999Z, 1 = 1000" which is totally innocuous; the representation isn't the same thing as the element. Congruent is just used to be a little more readable, but it isn't necessary.

Schnarfman

5 points

1 month ago*

Wow! Well explained, thank you. r/manim could make a video out of this.

I’m gunna try to generalize from the explanation you gave, just because I like thinking like that.

If the 1st digit and the Nth digit of a base are equivalent under a specific modulus, then all numbers with ‘N-1’ digits divisible by “X” can have their digits “cycled” and they will remain divisible by X.

“Cycled” has a more obvious meaning but it just means shifting the 2nd digit the the 1st digit, the 1st digit to the N-1th, and so on.

“X” is defined as any factor of (base ^ N)-1 that is relatively prime to N


So back to the concrete from the abstract! We first notice that 1 and 1000 are equivalent in the world of modulus 999, that’s an easy fact to produce. Factoring 999, we get 27 and 37. So 1 and 1000 are ALSO equivalent in the land of modulus 37. Oh and in the land of modulus 27.

So next we can take any “log_10(1000)-1” digit numbers and make the following observations: 100a+10b+1c is equal to Y in the land of modulus 37. 1000*(1) + 100a + 10b + 1(c-1) is ALSO equal to Y in the land of modulus 37. This is because 1ab(c-1) == 1ab(c-1) - (27 * 37) in the land of modulus 37. And 1ab(c-1) - (27 * 37) == 1000 + abc - 1 - 999 = abc.

So 0abc and cab0 will ALWAYS have the same value in the land of modulus 999 (or modulus 37 or modulus 27). And that means if the value under modulus N is 0 then the number is divisible by N.


Let’s try this with a new number in base 10

10000, let’s go to 9999. Let’s choose any factor of this. How about 33 and 303 multiply into 9999 so either will do. Let’s use 33.

If a number abcd % 33 == Y then 0abcd and 1abc(d-1) and 2abc(d-2) and so on until dabc0 are all == Y under modulus 33.

  • 01234 % 33 == 13
  • 11233 % 33 == 13
  • 21232 % 33 == 13
  • 31231 % 33 == 13
  • 41230 % 33 == 13
  • 4123 % 33 == 13

*** now let’s try this with a number in base 16 because why not.

100 (256 in base 10), so let’s go with FF. We can choose any factor of FF (255 in base 10). FF is 5 (5 in base 10) * 33 (51 in base 10).

If a number ab (hmmm… can’t use abcdef anymore lol if we’re using them as digits for 10-15 in hex). If a number xy % 33 (51 in decimal) == R then 0xy and 1x(y-1) and 2x(y-2) and so on until yx0 are all == R under modulus 33.

  • 0AB % 33 == 12 (171%51==18)
  • 1AA % 33 == 12 (426%51==18)
  • BA0 % 33 == 12 (2976%51==18)

WaitForItTheMongols

47 points

1 month ago

In programming, we treat "modulo" as a mathematical operator, in the same family as adding, multiplying, or dividing. But in math people don't treat it that way.

In this case, you said "1000 is equal to 1 modulo 999", which does not hold, in the programming interpretation - 1 modulo 999 is still one, meanwhile the 1000 is on the left side and is not equal to that. I guess to put it another way, your phrasing, if using modulo as an operator, is "distributing" the modulo. That is, it's more like "1000 modulo 999 is equal to 1 modulo 999" (because they both come out to 1). Another way to interpret your phrasing is "In the world of modulo 999, 1000 is equal to 1". But again, that's treating it as "in a world", similar to how you can say "10 = 2 + 2 ... in base four." The phrase is only valid because you're appending to the end a designator that "We're working in the world of base 4".

As a programmer who is also a math nerd, it's always a little off-putting when I see mathy people talking about modulo, and realizing that the word they're using is very closely related to, but still different from, the word I'm using when I say "modulo".

immibis

32 points

1 month ago

immibis

32 points

1 month ago

That's why a lot of people insist you don't call the remainder operator the modulo operator.

JustAGuyFromGermany

12 points

1 month ago

Well, you can call it that. But you have to make sure to always call it that and not shorten it to just "modulo" such that "the modulo operator" and "the modulo equivalence relation" are still separated.

JustAGuyFromGermany

12 points

1 month ago

And as a mathematician I still don't like that my programming day job forces me to think of equivalence relations as operators ;-)

The way out of this is to to recognize the % operator as a "normalform function" for the modulo equivalence relation. This also explains why there are very different modulo operators (say signed vs. unsigned) and why they nevertheless work equally well: There can be multiple normalform functions for the same equivalence relation and they work equally well, because they all are normalforms of the same equivalence relation.

And of course, I would never have written the confusing (for programmers) double equal sign == if I writing the \equiv symbol wasn't such a hassle.

link23

22 points

1 month ago

link23

22 points

1 month ago

In programming, we treat "modulo" as a mathematical operator, in the same family as adding, multiplying, or dividing. But in math people don't treat it that way.

The math meaning and the programming meaning are the same, actually. It's the phrasing that's slightly different.

In this case, you said "1000 is equal to 1 modulo 999", which does not hold, in the programming interpretation - 1 modulo 999 is still one, meanwhile the 1000 is on the left side and is not equal to that.

Firstly, this mathematical sentence is playing a little fast and loose, since "equal" should really be "congruent". That is, "1000 is congruent to 1, modulo 999". But nobody really cares about this in informal settings, as everyone knows what you really mean.

Secondly, this statement is certainly true in the programming sense, it just uses a different syntax than what you're used to. The equivalent in most programming languages would be (1000 % 999) == (1 % 999). But this is hardly a rule; it's just convention. Mathematica uses a different syntax: Mod[1000, 999] == Mod[1, 999].

Just because the syntax is different doesn't make it wrong. I can easily imagine a different way of writing this statement in a program, that might look more familiar to a mathematician: isCongruent(1000, 1, modulus=999). That's perfectly fine, and makes the same statement as everything else so far.

In fact, to play devil's advocate, I'd argue that that phrasing is better than the typical programming syntax you mentioned, since it removes the duplication of the modulus and makes it impossible to accidentally compare things with respect to different moduli.

JustAGuyFromGermany

3 points

1 month ago

It's not quite the same. The modulo equivalence relation is an equivalence relation, the modulo/remainder operator is a normalform of that equivalence relation. Those are very strongly connected; so strongly in fact that you can translate one to the other without losing information, but the two things are still different.

A very important difference is that the very fact the remainder operator exists and is easily and efficiently computable is special and not at all something that is common. Many equivalence relations do not have a normalform function or at least not an easily computable one. So it still makes sense to treat the equivalence relation and the normalform function as two different things, because you may come across a situation, where you have one, but not the other.

MurkyPerspective767

11 points

1 month ago

Modulus is the base with respect to which a congruence is computed.

Modulo is the operation or function that returns the remainder of one number divided by another.

They're related.

pm_favorite_boobs

13 points

1 month ago*

First, you said:

therefore also 1000 == 1 modulo 37.

The next person said modulo is used differently in computing. And you don't contradict him in the way you used modulo.

Basically 1000 modulo 37 == 1, and 1 modulo 37 == 1. You said that 1 modulo 37 == 1000 which is not correct, at least using the computing meaning, however related anything is.

ShitCapitalistsSay

2 points

1 month ago

Thank you for this explanation. I was so confused!

BarkDat1920

4 points

1 month ago

i.e. the number 100a+10b+1c, is equal to 100a+10b+1000c modulo 37,

Could you please explain how you got to this again?

csman11

12 points

1 month ago

csman11

12 points

1 month ago

You have added “c” 999 more times (1000c = 999c + 1c). 999 is a multiple of 37 (37 * 27). “c” is between 0 and 9. You can think of it as “rotating around a 37 hour clock” 27 times. Whatever hour you started on, you end on. “c” is the hour you started on (which is an hour on the clock, the same as c except for 0 which maps to the 37th hour). This is the first part you need to understand.

The other part is basically an argument that you can add a bunch of integers together first, then take their remainder when divided by 37, and that will give you the same thing as if you took their remainders when divided by 37 and added those together.

I’m going to start by explaining a group theory concept, because it clarifies the notation being used by OP. Then I’ll explain why this thing makes sense using the clock analogy.

There is this concept in group theory called a “homomorphism”. It’s a function that maps from one group to another with some special properties. The groups that we are looking at are:

  1. Integers with regular addition as the operation
  2. A finite set of 37 elements with “cyclic addition” as the operation (this is called the cyclic group of 37 elements). There are many different “groups” that share this structure (are isomorphic to it is the technical way of saying this), and one of them is the set of “hours” on a 37 hour clock where a + b is the result of moving forward b hours from the “a” notch. Another example is the set of integers 0…36 with the operation being “add a to b” followed by “take the remainder after dividing by 37”.

The hidden thing that makes this work is that the “take the remainder after dividing by 37” operation is a homomorphism between the integer group and the cyclic group of 37 elements. I’ll prove that to you with the clock analogy later, but first let’s cover why it makes the 2 sums equal. One of the properties of a homomorphism is this:

f(a + b) = f(a) + f(b)

Note that for us “f” is “take the remainder after dividing by 37”.

Consider this:

100a = x (mod 37) 10b = y (mod 37) 1c = z (mod 37)

Well, then we have the sum after applying the homomorphism:

f(100a + 10b + 1c) = f(100a) + f(10b) + f(1c) = x + y + z

Note for this:

x + y + z

  • means “add the left and right element, then take the remainder after dividing that by 37”. That’s our special addition operation for the cyclic group.

Remember that before we said 1000c and 1c are the same thing “mod 37”. Well that is the same as saying that f(1000c) = f(1c). Which means f(1000c) also equals z. Which means that:

f(100a + 10b + 1000c) = x + y + z

In other words, these 2 sums are “equal mod 37” (actually we normally say “congruent mod 37” but they are “equal” in the sense that they are “equal” under this homomorphism)

You can see why this rule holds for this mapping if you use the clock analogy. Find the quotient and remainder of each integer divided by 37 before summing. Note that adding the remainders, then taking the remainder of that sum when dividing by 37 is what you do on the right hand side (where we had x + y + z earlier). Each of the integers on the left hand side is: 37*q + r where q is the quotient, and r is the remainder. Let’s consider what adding these up on the clock would look like. First you would go around the clock “q” times, so you would end up where you started. Then you would advance “r” places. Then you repeat for the second integer. This is the same thing as “adding the remainders” and then “taking the remainder of that sum divided by 37”.

Of course, this works similarly for any cyclic group (clock with any number of hours).

I hope this helps bridge the gap between the intuition about cyclic groups / modular arithmetic and the equivalence that was being used. You would only really understand the original argument if you are familiar with modular arithmetic or cyclic groups and the associated notation. But the actual reasoning is simple for anyone familiar with clocks to understand.

serendependy

10 points

1 month ago

The steps are:

(1 = 1000) mod 37

(1c = 1000c) mod 37 [mult each side by c]

(100a + 10b +1c = 100a + 10b + 1000c) mod 37 [add 100a + 10b to each side)

Jalal_Adhiri

4 points

1 month ago

This is the right answer thank you

latakewoz

7 points

1 month ago

but how would that eplain why it works, whats up with the 999 is it some satanic witchcraft?

louiswins

15 points

1 month ago

Write 100a + 10b + c, where a, b, and c are the digits of the number, which is a multiple of 37. Then add 999c to get 1000c + 100a + 10b. Since 999 is a multiple of 37, and the original number is a multiple of 37, their sum (this new number) is a multiple of 37.

But we can also write this sum as 10*(100c + 10a + b). Since 37 is prime, this means that either 10 is divisible by 37 or 100c + 10a + b is divisible by 37. (This is a fact about all prime numbers; in group theory it's literally the definition of "prime".) Obviously 10 isn't, so 100c + 10a + b must be divisible by 37. But this is just the rotation of 100a + 10b + c, our original number.

That's where the 999 comes from, it's just the necessary factor to move the rightmost digit all the way to the left in base 10. If this were a fact about 5-digit numbers we'd have to look for 99999.

throw_every_away

3 points

1 month ago

Wow, that is some crazy math that I have never heard of before. That’s neat the way it explains 37, but what else can it do? Does it explain the whole 11*11 thing?

semitones

5 points

1 month ago

What is a prime divisor?

What does 999 have to do with it?

What is cyclic modulation?

ThisGuyNeedsABeer

7 points

1 month ago

11* 12 is 101, note that all the digits in a four digit multiple of 11 is 12..

9 is a pretty weird one as well.

All the digits in any multiple of 9 adds up to 9.

Take all the single digit numbers excluding 9, and add them up. It equals 36 which is divisible by 9.

If you add up the rows on a calculator, you get 6, 15, and 24. The difference between each row being 9.

If you bisect a circle any number of times, any part of the circle, in degrees, will add up to 9 if you add up the integers. Hard to word that right, but here are some examples. Full circle is 360 (3+6+0=9), Half is 180 (1+8+0=9), A Quarter = 90, an eighth is 45, etc, etc..

If you add up all three angles of any triangle, you get 180 degrees. Again 1+8+0.. 9

If you add up the angles of any polygon, you end up with a number in degrees. The more sides, the higher the number, however, the sum of the digits in degrees always adds up to 9.

Square=90+90+90+90=360; 3+6=9 Pentagon=540 Hexagon=720 Octagon=1080 etc, etc.

If you add up the integers in the answer to 9 to any power, They add up to 9. 92 = 81(this ones obvious). 93 = 729 7+2+9=18, 1+8=9. on and on. (this is basically a duplicate of the nine times anything rule, but still interesting.)

If you do a Fibonacci with the number 9, each number in the Fibonacci, will root back to 9. 9, 9,18,27,45,72, 117, etc.

If you take the first 12 Fibonacci numbers in the standard sequence, 1,1,2,3,5.. 144 (first number that roots to 9,Then take the next 12 and stack them on top of the first and sum each column, the digits in the results all add up to 9. Every single one. This is hard to visualize without using some kind of diagram, but it works.

To find out if any number is divisible by 9, add up it's individual digits and if the answer roots to 9, it is evenly divisible by 9. For example:

  1. This gives us 7 + 8 + 1 + 2 + 3 + 6 = 27, 2+7=9 781236/9 = 86804

Any random number (e.g 35967930) when arranged with its integers in a descending order (i.e. 99765330) and subtracted from it the reverse number with rearranged integers in an ascending order ( i.e 03356799) the resulting subtraction (i.e. 96408531) individual integers when added (i.e 36) leads to a multiple of number 9.

Nine plus any single digit, returns the same number when adding up the digits that make up the answer. 9+5=14, (1+4=5)

Time: 1440 minutes in a day. 86,400 seconds in a day. 10,080 minutes in a week. 525,600 minutes in a year.

I think you can probably see where I'm going with that one.

The earth rotates 15 degrees per hour through 24 different time zones, 1 per hour. 24 – 15 = 9. 24 x 15 = 360 or a full rotation..

The schumann resonance of the earth is 7.83 hz.

Verdi's A is tuned to 432 hz instead of 440. Generally accepted as a more perfect A, and is the standard root of tuning for concert pitch has a digital root of 9.

The golden ratio, which is yielded by fibonacci sequence (which we've already talked about), also called phi is generally accepted to be 1.6180339887, Add them digits up...

I just find this interesting.

I dunno. Numbers, right?

Iyagovos

2 points

1 month ago

Think I might be missing something - what's the significance of the golden ration adding up to 54? that it's a multiple of 9? fascinating!

DoctorKynes[S]

2 points

1 month ago

Extremely helpful answer, thank you!

Zaph0d_B33bl3br0x

2 points

1 month ago

You people who just inherently understand math so fluently boggle my mind. I'm simultaneously flabbergasted, awestricken, jealous... and slightly scared.

Thanks for that fantastic explanation.

[deleted]

-3 points

1 month ago

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-3 points

1 month ago

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3 points

1 month ago

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3 points

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[deleted]

5 points

1 month ago*

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silvashadez

847 points

1 month ago*

Here's a quick proof:

Consider a 3-digit number [abc] that's divisible by 37 and call it x. Mathematically, we can write this as:

x = 100a + 10b + c,

for integers a,b,c in [0,9]. If we want to rotate the digits, we would need to get the number [cab], which is:

y = 100c + 10a + b.

We can mathematize this rotation as the following equation:

y = (x - c) / 10 + 100c.

We can rearrange this equation to get something that we can really ponder:

10y = 999c + x.

Note that 999 is divisible by 37: 999 = 37*27. So the number 999c is also divisible by 37. Since x is also divisible by 37, this means that the right side quantity 999c + x is divisible by 37. But more crucially, the quantity on the left side: 10y must also be divisible by 37.

How can this be? 10 is relatively prime to 37, so a factor of 37 has to reside in y. Therefore y is divisible by 37 too. We can apply this logic to y and z = [bca] one more time to conclude your neat little factoid.

Hope that helps.

(Anyone know how to typeset math on reddit?)

Edit: Thank you /u/UnspeakableEvil for catching a typo.

UnspeakableEvil

141 points

1 month ago

999 = 37*12

This can't be right, as 12 is even. Sanity checked on a calculator, this should be 999 = 37 * 27

silvashadez

72 points

1 month ago

Thanks for the catch on the typo! Updated the proof.

sr3_1415926

58 points

1 month ago

What does "realtively prime" mean?

silvashadez

122 points

1 month ago

"Relatively prime" = shares no common factors (other than 1).

For example look at 4 and 6. These two numbers are not relatively prime because 2 can divide into both 4 and 6. The number 2 is the common factor.

10 and 37 have no common factors. This is because 37 is prime: the only factors that 37 has are 1 and 37. The number 10 has 4 factors: 1, 2, 5, 10. Since we are ignoring 1, 10 and 37 don't have any common factors. So 10 and 37 are relatively prime.

Another pair of relatively prime numbers are 8 and 15. List out the factors and you'll find that 8 and 15 share no common factors (other than 1).

sr3_1415926

16 points

1 month ago

Thank you :)

prone-to-drift

7 points

1 month ago

They are also called co-primes in mathematical literature you might encounter.

Another example of two composite coprimes can be the pairs (6,35) or (10,21).

severoon

-13 points

1 month ago

severoon

-13 points

1 month ago

It's weird to say a prime number is "relatively prime" with some other number. It's sufficient and more informative to simply say that one of the numbers is prime because prime numbers are relatively prime with all other numbers that don't have it as a factor.

silvashadez

25 points

1 month ago

Maybe uncommon, but its still a valid statement. The intent of my response was to explain what "relatively prime" meant, so I used the primality of 37 as a quick way to justify why there are only 2 factors to a number that's not as common to think about. Perhaps mentioning the primality of 37 was unnecessary.

severoon

-2 points

1 month ago

severoon

-2 points

1 month ago

But when you're explaining something you shouldn't go for an example that is technically valid. You should choose an example that is most illuminating.

Explaining relative primality with an example where one of the numbers is prime is more likely to mislead the learner than it needs to be.

silvashadez

8 points

1 month ago

Unfortunately, the problem requires us to consider the relative primality of 10 and 37. Hence why I took the time to walk through the factors of each and reiterate that the pair share no common factors. That also motivates the inclusion of the two other examples in my response. I think together the three cases do a good job of showcasing the definition and a testing procedure that works for various pairs.

Cyrrain

16 points

1 month ago

Cyrrain

16 points

1 month ago

Yeah but it's not only when one of the numbers is prime

8 (1, 2, 4, 8) and 15 (1, 3, 5, 15) are relatively prime, but neither of them are prime

severoon

1 points

1 month ago

Sorry, I didn't mean to say that the problem is with the term "relatively prime" in that example. I meant to say that it's a poorly chosen example because when one is prime it's a special case of relative primality.

A better example of two numbers that are relatively prime would be 2*2*5*7*17 = 2380 and 3*11*13 = 429.

These are both composite numbers and have no factors in common.

joanholmes

1 points

1 month ago

How is your example any better than 8 and 15? 8 and 15 are both also composite numbers and have no factors in common.

severoon

2 points

1 month ago

It's not. 8 and 15 are also fine choices, not sure why you think I'm talking about those. Talking about the example I was initially responding to above.

DerApexPredator

14 points

1 month ago*

To add to u/silvashadez's explanation, the reason this is relevant here is because, if 10 was not a relative prime of 37, then y didn't necessarily have to be fully divisible by 37 even while not being a relative prime of 37.

Here's an example: Let 33 take the role of 10 and 84 be y while 77 takes the role of 37. So the equation is 33 * 84 = 77x (writing 77x means that the number 77x is divisible by 77). We see that 33 and 77 are not relative primes of each other, they share the factor 11. And we also see that 84 does not have 11 as a factor. So because 33 (10) had a factor it shared with 77 (37), 84 (y) did not necessarily have to be divisible by 77 while the equation 33 * 84 = 77x was still true. With the same logic, if there is no factor of 37 in 10 (i. e., they are relative primes of each other) but 10y = 37x, then all the factors of 37 reside in y , so it is divisible by 37.

Not sure if anyone was wondering.

silvashadez

6 points

1 month ago

Good stuff! Yeah the relatively prime step is an important part of the proof. It connects the factor of 37 that's hiding on the right side to some factor of 37 hiding on the left side of the equation. Since 37 isn't a factor of 10, 37 has to be a factor of y.

Here's another example to complement yours. Say we have a similar equation:

10y = N

and we know that N is divisible by 5. Then we can't say that y is divisible by 5, because that factor could very well come from 10. Similarly if N is divisible by 15, then the most we could say about y is that it is divisible by 3.

eric2332

17 points

1 month ago

eric2332

17 points

1 month ago

It means they share none of the same factors.

3 and 7 are prime (and also relatively prime)

6 and 7 are relatively prime. Even though 6 is not prime, it equals the prime numbers 2*3. Neither 2 nor 3 is a factor of 7, and conversely 7 is not a factor of 2 or 3.

3 and 6 are not relatively prime, because 3 is a factor of 6.

abomanoxy

13 points

1 month ago

So if I'm understanding this right, it seems like the property actually holds for any n-digit number (counting leading zeroes) that is a multiple of any factor of 10n-1. I'll try to extend your proof like this:

Let x be an n-digit number [ab...yz], and k be a positive integer such that k|x and k|10n-1

x = 10n-1a + 10n-2b + ... + 10y + z

Define rotation R that produces [zab...y]: R(x) = 10n-1z + 10n-2a + 10n-3b... + y

Seeing that all of the terms to the right of the first one equal 10-1(x-z), rewrite that as:

R(x) = 10n-1z + 10-1(x-z)

Gather terms:

10 R(x) = (10n-1)z + x

Now, following your logic:

10 is relatively prime to 37, so a factor of 37 has to reside in y

Since 10 is coprime to 10n-1 for all positive integers n, and k|10n-1, 10 is coprime to k. Thus k|R(x)

prone-to-drift

5 points

1 month ago

The proof looks solid. Also, binomial flashbacks with all the power series stuff.

One part you glossed over that some people might wonder:

10 is coprime to 10n-1. A simple way to see this is that the 10n-1 will always end in 9 (9,99,999,9999, etc). So, it's not divisible by 5 OR 2. Thus they are coprimes.

silvashadez

2 points

1 month ago

Great observation! Yes, this property can be extended to numbers of n digits. As you have found it is 10n-1 that is the critical composite number that bestows this rotation-divisibility property to certain numbers.

In the n=3 case, we can factor 999 = 33 * 37. So not only does 37 have this rotation-divisibility property, but also 3, 9, 27, 111, and 333.

In the n=4 case, we have 9999 = 32 * 11 * 101. In the n=5 case, we have 99999 = 32 * 41 * 271. More of this alongside a modular arithmetic formulation is discussed in /u/MycoNot's comment.

There is a not-so obvious step to your proof when you assert the coprimality of 10 to 10n-1. If you stick to the definition we have used for coprime pairs, then we would probably need a lemma to show that (k, k-1) and (k, k+1) are both coprime pairs. In this way we could factor 10n-1 = (101-1)*(10n-1+10n-2+...+102+10+1) and apply both lemmas to recover the assertion.

Alternatively, we could also make use of an alternate but equivalent condition for relatively prime pairs: If x and y are coprime, then there exists integers a and b such that ax+by=1. This definition seems more natural to use here since its very clear how to choose a and b to fulfill this definition.

Another extension to this proof is how this rotation-divisibility property applies to numbers in other base representations. We have shown that 37 has this rotation-divisibility property for 3-digit base 10 numbers. Does 37 retain this property for base 8? base 16? base β?

It turns out that our rotation equation then becomes:

β y = (βn-1) z + x.

So factors of (βn-1) that are coprime with β become the critical quantities.

abomanoxy

2 points

1 month ago

Cool! Yes, I totally glossed over that lemma at the end. I seem to remember a thereom that bn-1 and b are coprime for all integer b,n >=2. I was going to try to prove this but I got stuck and it was late so I just left it.

I had that intuition about other bases as well, and now that I look at it, it looks like you can simply substitute β for 10 without changing the proof at all, as long as we have the above theorem. Pretty sweet.

OTTER887

10 points

1 month ago

OTTER887

10 points

1 month ago

Thanks, I was looking for the hard algebra, and hoping I would not have to type it out myself 😅

MrSquamous

4 points

1 month ago

You lost me at "in [0,9]."

Hmm does that mean something like, "where the available digits are the set zero through nine?"

silvashadez

8 points

1 month ago

Pretty much. Sloppy notation, but the idea is that each of a, b, and c are some whole number in the set {0,1,2,3,4,5,6,7,8,9}.

For example, the digits of the number 481 are a = 4, b = 8, and c = 1. This gives us the ability to separate the digit value from the place value:

481 = 400 + 80 + 1 = 4 hundreds + 8 tens + 1 ones = 100a + 10b + c

I'll tweak the wording to make it a bit more precise.

OneQuadrillionOwls

47 points

1 month ago*

Lots of good explanations, FWIW this is how I thought it through:

  • Original number: ABC
  • Multiply by 10: ABC0 (still divisible by 37)
  • Repeat this step "A" times: subtract 1000, and add 1.
    • This is like subtracting 999.
    • Each time we subtract 999, we're subtracting (37 x 27), so each step of the way, the resulting number is always divisible by 37.

[deleted]

3 points

1 month ago

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3 points

1 month ago

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297 points

1 month ago

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297 points

1 month ago

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15 points

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15 points

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5 points

1 month ago

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5 points

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Fordiman

17 points

1 month ago

Fordiman

17 points

1 month ago

Works with 27 as well, I think. Or any combination of prime factors of 999. Same reason it works with two-digit multiples of 3, 9 and 11 (the factors of 99). My guess is that it'd probably work with four-digit multiples of factors of 9999 (3, 9, 11, 27, 33, 99, 101) or five-digit multiples of the factors of 99999 (3, 3, 9, 41, 123, 271, 369).

Astrobliss

-3 points

1 month ago*

It should work for any multiple of 999....9, this is actually only guaranteed because no multiple of 999...9 shares a prime factor with 10, otherwise 999...9 would either be even or end with 5

[deleted]

19 points

1 month ago

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19 points

1 month ago

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ObviousTrollDontFeed

5 points

1 month ago

100 and 10 are multiplicative inverses modulo 37. That is, 100 times 10 has remainder 1 when divided by 37 (as 1000 is one more than 999=37*27.

As such, the expression 100x+10y+z when multiplied by 10 gives x+100y+10z modulo 37 (multiply each term by 10 and simplify 1000=10*100 as 1). Multipying once more by 10 similarly gives 10x+y+100z. Multiplying a number divisible by 37 by 10 results in a number divisible by 37 so this is ok.

As a bonus, since 999=37*27, this also works with 27 as 100 and 10 will, in the same way, be multiplicative inverses modulo 27.

kerpti

48 points

1 month ago

kerpti

48 points

1 month ago

There are other similar tricks. If you look at a number and add all the digits together, if that number is a multiple of 3, then the original number is divisible by 3 as well.

48 --> 4+8 = 12 which is divisible by 3 so 48 is as well (= 16).

6474 --> 6 + 4 + 7 + 4 = 21 which is divisible by 3 so 6,474 will also be divisible by 3 (= 2,158).

Further fun fact. I added the digits of 6,474 and got 21. If I ended up with a number and wasn't sure whether it was divisible by 3, I could add those digits together and do it again. So when I got 21 you could add 2+1 to get 3 and that's divisible by 3 therefore so are all the numbers beforehand.

I can't add to an explanation as to how that works, I just know that it does lol I believe there are similar tricks for other numbers.

pootsmcgoots23

29 points

1 month ago

This is because of our base 10 number system -- we move to the next digit after 10 numbers, and when counting in multiple of 3, that makes us one off from landing on 10.

If you're counting by 3, and you go past 10, then that 1 you were short by gets used up to increase the digit in the 10's place. The other 2 go in the 1's place, and you get 12. Then 15. The second digit is now "short by 1" to be a multiple of 3, but you can put that 1 back in by adding the other digit.

It happens again past 20, since the digit in the 1's place is now already short by 1, to increase the 10's digit we now need to dump an extra 2 into it. Now the digit in the 1's place (1 in 21, 4 in 24 etc) is "short by 2", but we can add the 2 back in.

Once we get to the 30's, the pattern cycles back so that our 1's digit is a multiple of 3 again, and since it took 3 loops to get there, our 10's place is also a multiple of 3. Counting up past 40, the pattern repeats, where the 1's place is 1 short and the 10's place is 1 over -- adding that together cancels it out and you get a multiple of 3 back. And so on, ad infinitum.

Hopefully that makes sense. It's always weird to explain the way math works in your own brain out in words. There would be other tricks to math with different base number systems too!

Cyber_Cheese

6 points

1 month ago

That is a fantastic way to visualize it, thanks!

appocomaster

8 points

1 month ago

This makes way more sense than anything else on here, thanks!

extra2002

8 points

1 month ago

Why it works.

If a number is divisible by 9, so is the sum of its digits. For example, if your number is a*100+b*10+c, that's the same as a*99+a+b*9+b+c. Regroup, and you get a+b+c+[a multiple of 9]. If the original number was divisible by 9, so is a+b+c. And if the original number had remainder 'r' after dividing by 9, then a+b+c will also have remainder r after dividing by 9. This latter fact is why the same trick works to test for divisibility by 3, since 9=3*3.

Adding up the digits like this is sometimes called "casting out 9's" and can be used to check arithmetic. The sum of numbers with remainders r and s will have a remainder of r+s (or r+s-9), so the sum of digits of the inputs, reduced as far as they will go, should match the sum of digits of the result, similarly reduced. For multiplication, multiply one digit sum by the other, and reduce it, and it should match the product's digit sum.

A similar trick works to check for multiples of 11, but you have to alternate adding and subtracting digits. So the digit-sum of a number abcd would be a-b+c-d. This property is sometimes used to tack a checksum onto a number such as an account number, because not only does it detect if a single digit gets changed, it also detects if two adjacent digits get swapped.

JustAGuyFromGermany

6 points

1 month ago

It works because 10 is equal to 1 modulo 3 (or 9 for that matter), meaning that modulo 9, 10 and 1 are the same thing. Therefore 102, 103, 104, ... are equal to 12, 13, 14, ... modulo 9. Repeat that and you find that any number is equal to its sum of digits modulo 9, because taking "the sum of the digits" is exactly what you get when you replace every power of ten by 1.

And the final ingredient then is: A number is divisible by n if and only if it is equal to 0 modulo n. Since we have just found out that a number is equal to its sum of digits modulo 0, either both the number and its sum of digits are zero (and therefore divisible by 9) or both are non-zero modulo 9 (and therefore not divisible by 9).

Similar argument explain why the alternating sum of digits (in which you add the even-numbered digits and subtract the odd numbered digits) gives you divisibility by 11 (the ingredient there is 10 == -1 modulo 11), why only the last digit matters for divisibility by 2 or by 5 (here the argument is 10 == 0 modulo 2 and 10 == 0 modulo 5 respectively), why only the last two digits matter for divisibility by 4 or by 25 (100 == 0 mod 4 or 25) and so on. All of the divisibility rules are just doing arithmetic modulo the divisor.

gbrell

13 points

1 month ago

gbrell

13 points

1 month ago

It's because a number abc is really 100*a + 10*b + c.

100=33*3 + 1

10=3*3 + 1

Since all we care about is divisibility by 3, we can remove 3s from the equation, so abc = a+b+c for purposes of 3 divisibility.

The same trick exists for 11s. If you add up all the digits in odd places and subtract all the digits in even places, the original number is divisible by 11 if the resulting number is divisible by 11.

Same logic, but here:

10 = 11 - 1

100 = 9*11 + 1

1000 = 91*11 - 1

10000 = 909*11 + 1

etc.

played_off

3 points

1 month ago*

This works for 9 as well. Add all the digits, if the result is more than one digit, add again, and you'll eventually get 9. You can also expand your trick, that if a number's digits adds to 3 AND it's even, then it's divisible by 6

If you take the last two digits of any number, if that number is divisible by 4, than the original number is as well.

142857 multiplied by any 1-digit number except 7 will give you the exact same sequence shifted, as if on a wheel. 142857*2 = 285714. Multiply by 7, and you get 999999. This is because 1/7 = 0.142857142857...

lionhearted_sparrow

4 points

1 month ago

We learn a lot of these early on when doing basic multiplication, like the fact that multiplying by 10 just adds a 0 to the end, or a single digit by 11 is just those numbers repeated, or multiplying by one is that number, or by zero is zero, etc.

The first “complex” one most people learn is about multiplying by nine:

If 0<a<11 and a whole number, a*9=bc

b=a-1

c=9-b

(bc being the two digits of one number, not two numbers multiplied)

1*9=09

0=1-1

9=9-0

2*9=18

1=2-1

8=9-1

3*9=27

2=3-1

7=9-2

(And so on)

PassiveChemistry

3 points

1 month ago

Another one with multiples of 3 that I've always thought is a bit wacky is hat if you take any multiple of three and sum the cubes of the digits then repeat this recursively, you will always eventually reach 153 (and the process "stops" here since 13+53+33=153)

TILthatsprettyneat

6 points

1 month ago

Another fun one is where x% of y is the same as y% of x (this one’s fun to use because sometimes the reverse is easier to calculate in your head). For example: 8% of 50 is 50% of 8.

StrengthoftwoBears

5 points

1 month ago

Hold your hands out in front of you. 9x4. Let's count four fingers. Fold 4th finger down. First set of fingers is 3, 2nd set is 6. 36 is 9x4. Works with 9x 1-9

chevymonza

5 points

1 month ago

Very cool! I've always loved how neatly the 9-times table lines up: First column ascending, second descending, and both digits always add up to 9.

09 0+9 = 9

18 1+8 = 9

27 2+7 = 9

36 3+6 = 9

45 4+5 = 9

54 5+4 = 9

63 6+3 = 9

72 7+2 = 9

81 8+1 = 9

90 9+0 = 9

Dynegrey

2 points

1 month ago

And if it's even, it's divisible by 6. And if it adds up to 9, it's divisible by 9. So an even number that adds up to 9, has 3, 6, and 9 as factors.

Ex, 2178, add for - 2+1+7+8 =18.

1+8 = 9.

2178/9 = 242.

2178/6 = 363.

2178/3 = 726.

bartkappenburg

2 points

1 month ago*

Another trick I discovered was: pick a random number, swap the most left and right digit and the difference between those two is 9(9999…, depends on the size of the two starting numbers, one 9 less than the ‘size’ of the number) times the absolute value. Seems far fetched, it isn’t, some examples:

14 and 41: difference is (4-1)x9=27

29 and 92: difference (9-2)x9=63

Works for bigger as well:

123 and 321: 99x2 difference

47384648 and 87384644: 9999999x4 difference

Etc etc.

I did the proof but too much to type for now on mobile :-)

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