Talk:Divisibility rule

Divisibility rule was a Mathematics good articles nominee, but did not meet the good article criteria at the time. There may be suggestions below for improving the article. Once these issues have been addressed, the article can be renominated. Editors may also seek a reassessment of the decision if they believe there was a mistake.

Article milestones
Date	Process	Result
June 19, 2006	Good article nominee	Not listed

major revision[edit]

I've noticed there was more or less the only one person still editing on that article within the last 9 months and before that a larger discussion of how to improve it (without being implemented).

I'm considering a major rewrite of the article and actually splitting it up in probably 3 different ones

basic ideas, few rules, generalizations of the ideas/concepts
a somewhat comprehensive list of divisibilitiy rules
proofs & techniques
Proofs & techniques —Preceding unsigned comment added by 198.189.251.26 (talk) 17:35, 8 April 2009 (UTC)[reply]

The latter 2 are basically new articles and don't concern the original articles, however i'd like to use the occasion to completely rewrite (in particular shorten) the original article as well. That means most of its content would be deleted or moved to the other articles. However I don't want to do such extensive modifications without the original contributors being ok with it. So let me know if there any objection or if you have any suggestions for the rewrite.--Kmhkmh 16:53, 17 April 2007 (UTC)[reply]

I would definitely support such a movement and help where I could. As long as you cite what you replace, I have no problem with my work being redone... I'll ask Cryptic C62 to make a statement also, as he did quite a bit here a while back. -- Rmrfstar 22:24, 17 April 2007 (UTC)[reply]

Bah, a statement? Do what you will. All I ever did was crank out original researchy home-made rules. --Cryptic C62 · Talk 23:06, 17 April 2007 (UTC)[reply]

Split from Divisor topic[edit]

OK, I moved the material, although I'd still like to talk about a name change for the "new" topic. --Jay (Histrion) 20:23, 6 August 2005 (UTC)[reply]

how about "Divisibility test"? -- Yannick Gingras 15:30, August 7, 2005 (UTC)

I changed the visible part of the pipe over at Divisor to "properties". If that sticks there then maybe that is another potential name. Walt 00:06, 29 May 2006 (UTC)[reply]

Proofs[edit]

I agree that the proofs should be moved to another page but it is not clear yet if this should be a regular page or a subpage. User:Linas made a request for the subpages, maybe we should wait for the decision. -- Yannick Gingras 15:26, August 7, 2005 (UTC)

I don't think it's appropriate for the proofs to be here. The proofs are not too hard to work out and most mathematics pages on Wikipedia do without. — ciphergoth 22:24, August 7, 2005 (UTC)

I don't think that those are that easy to work out for someone who first learn about divisibility test. I find that knowing why it works helps a lot to remember them. Maybe some explanation with prose would be better but I fear that it might become too long. Most articles don't include the proofs but on W:WikiProject_Mathematics/Proofs it is said that including proofs can be helpfull if they are put aside from the main article. Act how you think will yield the greatest good for everyone, I won't start a revert war. -- Yannick Gingras 02:44, August 9, 2005 (UTC)

I've cut the proofs down to a representative sample and tidied them up a bit. Some were deleted because they would need to be fixed in order to stay: eg where it says "6 = 2 x 3" that's not nearly sufficient justification; as a counterexample to the explicit rule, 4|12 and 6|12 but (4 x 6) does not divide 12. — ciphergoth 07:57, August 9, 2005 (UTC)

I agree with shortening the proofs. About making proofs subpages (se e the very top comment in this section), I think the use of subpages is very discouraged on Wikipedia. Basically, proofs should be in the same text as the theorem, but proofs should be avoided or made sketches unless they add significant value in understanding the article. That's my 2c. Oleg Alexandrov 15:40, 9 August 2005 (UTC)[reply]

So do you think there's merit to my contention that we would probably be wisest to remove them altogether? — ciphergoth 22:09, August 9, 2005 (UTC)

Fine to me either way. Now the proofs are in a separate section, and that section is at the very bottom of the article, so there is not much harm done to keep then in. So I am not sure if I would strongly support removing them.

The situation was of course very different before the article was split off. Then the proofs took a lot of real estate in the middle of the article obscuring more important stuff.

So, I think keeping the proofs they way they are now is not too bad. But I would not object to them being removed or even more shortened either. Oleg Alexandrov 22:14, 9 August 2005 (UTC)[reply]

53[edit]

I think 50 is enough. After that, it becomes useless information and Wikipedia is not the place for that. -- Rmrfstar 00:59, 25 March 2006 (UTC)[reply]

I agree, 50 is a good place to stop. There might even be a case for stopping much earlier. — ciphergoth 22:23, 26 March 2006 (UTC)

I somewhat strongly disagree to that, wikipedia can (and mayb even should) a large list of divisibility rules.Keep in mind the math portal is used as reference by mathematicians and for specific math facts as well.From my point of view it is more a question of properly organizing the material in either long article with sensible chapters (which could be skipped by a reader not interested in them) or perhaps even better

split it in serveral separate articles - for instance:

basics including a few rules
comprehensive list of rules
proofs

--Kmhkmh 12:11, 17 April 2007 (UTC)[reply]

I'm fine with that: if there's a dedicated list, I won't mind any extra rules because they won't interfere with the content. -- Rmrfstar 22:26, 17 April 2007 (UTC)[reply]

—The preceding unsigned comment was added by Kmhkmh (talk • contribs) 12:10, 17 April 2007 (UTC).[reply]

There might... -- Rmrfstar 00:05, 27 March 2006 (UTC)[reply]

The general form of a divisibility test is this. If x is not a prime power, then it's a product of prime powers a_1^{p_1} a_2^{p_2} ... and y is divisible by x if it is divisible by a_1^{p_1} and by a_2^{p_2} and so on. If x = 2^n or x = 5^n, then y is divisible by x if the last n digits of y are divisible by x. Otherwise, 10a + b is divisible by x if a + kb is divisible by x, where k is a number such that 10k -1 is divisible by x (ie the multiplicative inverse). k can be negative, which is often more convenient for keeping k small, and this works not just for prime powers but for any x not divisible by 2 or 5. Perhaps we should stop at 20 and then explain the general form in more detail? — ciphergoth 08:23, 27 March 2006 (UTC)

I support stopping at 20, except for notable or easy ones, like 25, 27, 32, and 37 (add 36, 49, 33). The rest can be an exercise for the reader :-) By the way, I'm hoping to make an easier to understand explanation soon. You can do it with algebra without using modular arithmetic. Walt 00:10, 29 May 2006 (UTC)[reply]

Well 50 is a good area for now, but for biggners it should be 20 — Preceding unsigned comment added by 137.186.196.22 (talk) 23:12, 29 July 2015 (UTC)[reply]

Text from Divisor[edit]

I'm storing the following here for integration with this article. It seems that if Divisibility Rule is a separate article, then this should not be there.

If an integer n is written in base b, and d is an integer with b ≡ 1 (mod d), then n is divisible by d if and only if the sum of its digits is divisible by d. The rules for d=3 and d=9 given above are special cases of this result (b=10).

We can generalize this method even further to find how to check divisibility of any integer in any base by any other (smaller integer). Let us say that we want to determine if d | a in base b. Then we first find a pair of integers (n, k) that solves the congruence bⁿ ≡ k (mod d). Now rather than summing the digits, we take a (which has m digits) and multiply the first m-n digits by k and add the product to the last (or more precisely, smallest) k digits and repeat if necessary. If the result is a multiple of d then the original number is divisible by d.

A few examples will help demonstrate this. Since 10³ ≡ 1 (mod 37) then the number 1523836638 gives 1+523+836+638 = 1998 which gives 1×1 + 998 = 999. We know that 999 is divisible by 37 because of the above congruence. Again, 10² ≡ 2 (mod 7) so 43106 gives 431×2 + 06 = 868 which gives 8×2+68 = 84 which is easily noted as being a multiple of 7.

Note that there is no unique triple (n, k, d) since for example 10 ≡ 3 (mod 7) so we could also have done 4310×3 + 6 = 12936 and 1293×3 + 6 = 3885 and 388×3 + 5 = 1169 and 116×3 + 9 = 357 and 35×3 + 7 = 112 and 11×3 + 2 = 35 and 3×3 + 5 = 14 and 1×3 + 4 = 7. Clearly this is not always efficient but note that each number in this series, 43106, 12936, 3885, 1169, 357, 112, 35, 14, 7 is a multiple of 7 and many series could contain trivially identifiable multiples. This method is not necessarily useful for some numbers (for example 10⁴ ≡ 4 (mod 17) is the first n where k < 10) but lends itself to fast calculations in other cases where n and k are relatively small.

Walt 00:18, 29 May 2006 (UTC)[reply]

The above was changed by another editor. Per WP:TPO, we need to leave each other's comments alone and instead respond to them, so I undid the change and instead am describing the change here. In the "We can generalize" paragraph, next-to-last sentence, the following alteration was made (~~removed~~/added):

Now rather than summing the digits, we take a (which has m digits) and multiply the first m-n digits by k and add the product to the last (or more precisely, smallest) k n digits and repeat if necessary.

— Preceding unsigned comment added by 121.244.182.76 (talk) 07:45, 11 November 2014‎ (UTC)[reply]

DMacks (talk) 08:58, 11 November 2014 (UTC)[reply]

Revamping the page[edit]

I'm trying to make the page easier for someone who doesn't know a lot of math, and is looking for the rules to actually use them. The rest of us I'm sure can cope by looking at the proofs. I've removed a few rules and added some I think are easier, while also cleaning up (IMHO) the entries.

I hope anyone who doesn't like these changes will discuss it here first so we can satisfy everyone.

Walt 23:57, 29 May 2006 (UTC)[reply]

I love 'em. -- Rmrfstar 23:51, 14 June 2006 (UTC)[reply]

Thanks. Always nice to get some feedback (esp. positive) -- Walt 12:58, 15 June 2006 (UTC)[reply]

Cut text - trial divisors[edit]

I cut the following: the point of this page is that you can determine divisibility without determining the quotient. This may belong on a page of how to do mental math instead.

Proof using trial divisors For 21:

Choose a number n such that n>0. (2n)*10+n=21n, 21n/21=n

For example, 53: 53*2=106*10=1060+53=1113. 1113/21=53

You can also do this logically, noting that 20n+n=21n. That is to say, if 8 is multiplied by 20, the result is 160. If 8 is added to that product, the result is 168, which is divisible by 21, obviously. This stems from the known fact that multiplication is a simplified form of addition.

This rule for 21 will also apply for all numbers 10n+1 where n is an integer. That is to say, for instance, in order to test to 31, multiply the number by 3, then by 10, and add the number to itself. Thus, if a number can be broken up into two numbers and 1/3 of the first number is equal to the second number, then it is divisible by 31. E.g., 713 = 690 + 23. 69/3 = 23, thus 713 is divisible by 31.

Walt 01:46, 2 June 2006 (UTC)[reply]

overall form[edit]

I feel as if this article can not be made to fit the format of a traditional article... perhaps it should be made into an official list? -- Rmrfstar 03:51, 4 January 2007 (UTC)[reply]

Confusing Divisibility Rule for 17[edit]

216.167.135.24 20:46, 20 February 2007 (UTC) From: Dann 216.167.135.24 20:46, 20 February 2007 (UTC)[reply]

In the section entitled "2 through 20"... a chart lists various divisibility rules (for these 19 whole-numbers-- from 2 through 20)... I have been able to understand (& gratefully utilize) all of these rules... all, EXCEPT FOR ONE-- that one being: the second rule listed for the number 17.

The description of the "Divisibility Condition"... does NOT seem to match the given "Example". The named "Divisibility Condition" is described (with ambiguous wording) as follows: "Alternatively subtract and add blocks of two digits from the end, doubling the last block and halving the result of the operation, rounding any decimal end result as necessary."

Does this mean "doubling the last block" of those 2-digit blocks-of-digits which a person (in reverse order) does "[a]lternatively subtract and add"... or does this mean "doubling the last block" of the original number? Oddly, in the given "Example", though, every "block of two digits" EXCEPT FOR "the last block" of the original number... are ones which the article's readers can witness the said "Example" to be "doubling". Perhaps, the contributor had intended to say "doubling each 2-digit block-of-digits OTHER THAN the last block". The reader is left to guess.

Another wording leaves uncertainty, as well. Specifically, when does "halving the result of the operation" occur? To do so, immediately after doubling any 2-digit block-of-digits... would leave the same 2-digit block-of-digits which had existed prior to the original doubling (making that doubling-- be a pointless waste-of-time).

If, instead, the doubling should occur AFTER a doubled 2-digit block-of-digits has been subtracted from, or added to, another 2-digit block-of-digits... then, should that "halving" occur after EACH subtraction, or addition, of a (possibly, doubled) 2-digit block-of-digits... ONLY after the FIRST subtraction of a (possibly, doubled) 2-digit block-of-digits... or ONLY after ALL of the (possibly, doubled) 2-digit blocks-of-digits have been subtracted, or added?

Looking (for insight into these matters) to the provided "Example"... provides no clear answers... and, instead, only creates more questions.

The "Example" (for 209,865-- which commas divide... into each, said block of 2 digits-- as "20,98,65"), however, does not double the last block (i.e., of 65); but rather, that said "Example" doubles both the middle block (of 98... which becomes "(98x2)")... as well as the first block (i.e., of 20... becoming "40"). Plus, the "Example" does NOT remove the resulting decimal... by "rounding" (but rather, through multiplication by 10).

The said "Example" (which appears as: "20,98,65: (65 - (98x2)) : 2 + 40 = - 25.5 = 255 = 15x17")... CAN work, however, when expressed as follows: "209,865: ( [65 - (98•2) ] / 2) + (2•20) = [ (65 - 196) / 2] + 40 = (-131 / 2) + 40 = -65.5 + 40 = -25.5"... and "-25.5•10 = -255 = -15•17".

Despite my best efforts, though, NONE of my attempts to apply this divisibility rule to larger multiples of 17 (ones with more digits than the given "Example") have ever been successful. Please, either help me to understand this Divisibility Rule... or send this note to the contributor (of the said Divisibility Rule)... so that I'll learn how to apply the Divisibility Condition to this sizable multiple of 17: 9,349,990,820,016,829,983 (a whole-number which is the product of the following prime factors: 3 • 3 • 3 • 3 • 7 • 11 • 13 • 13 • 13 • 17 • 37 • 43 • 557 • 45,293).

Thanking you, in advance, for your prompt reply,

Danny

Hello Danny. I read a book, Vedic Mathematics that gives a general method for divisibility called osculation. It follows the vedic ideal of one-line notation. It requires no division, just multiplication and addition. Larry R. Holmgren March 4, 2007.

Example one: Is 209,865 divisible by 17? First convert the divisor 17 to the nines family. 17x7=119. Add one, drop the zero. The 12 is the Ekādhika, the multiplier. Start on the right, one digit at-a-time, moving to the left. If you finish with 17 (or a multiple of 17) or zero, then the number is divisible by 17, otherwise it is not divisible by 17. Ten multiples of 17: 17, 34, 51, 68, 85, 102, 119, 136, 153, 170.

The process: Multiply the first digit on the right by 12. Add the product to the next digit to the left. Write the result below that next digit. Multiply its units digit of the result by 12 and add that product to the other digits in the result and add to the next digit to the left. Set down this result below that digit. Repeat the process to the end on the left. Mental math: 5x12=60. 60+6=66. 6x12=72. 72+6+8=86. 6x12=72. 72+8+9=89. 9x12=108. 108+8+0=116. 6x12=72. 72+11+2=85.

2   0    9   8   6   5  
85  116  89  86  66 
YES.

Example two: Is 9,349,990,820,016,829,983 divisible by 17? As above for a divisor of 17 the Ekādhika (multiplier) is 12. 18 steps of mental math: 3x12+8=44. 4x12+4+9=61. 1x12+6+9=27. 7x12+2+2=88. 8x12+8+8=112. 2x12+11+6=41. 1x12+4+1=17. 7x12+1+0=85. 5x12+8+0=68. 8x12+6+2=104. 4x12+10+8=66. 6x12+6+0=78. 8x12+7+9=112. 2x12+11+9=44. 4x12+4+9=61. 1x12+6+4=24. 4x12+2+3=53. 3x12+5+9=50.

9   3  4  9  9  9   0  8  2   0  0  1  6  8   2  9  9  8  3
119 29 22 61 44 112 78 66 104 68 85 17 41 112 88 27 61 44  
YES.

Larry R. Holmgren 11:23, 3 March 2007 (UTC)[reply]

Danny Please check the prime factorization of that 19-digit number again. 9,349,990,820,016,829,983 =? 3 • 3 • 3 • 3 • 7 • 11 • 13 • 13 • 13 • 17 • 37 • 43 • 557 • 45,293 ? I checked 9 by casting out nines. OK

Larry R. Holmgren 18:36, 3 March 2007 (UTC)[reply]

**********************************************************************************************

Hello, Larry.

Thank you for posting your reply.

Although I do appreciate the fact that you did provide a response to my inquiry... I am afraid that your analysis has something amiss about it. Using long-division, I have been able to verify my factoring. To start with, 9,349,990,820,016,829,983 (in fact) IS evenly divisible by 17. The quotient that results is 549,999,460,000,989,999. When 549,999,460,000,989,999 becomes the dividend... with a divisor of 45,293... the next quotient (again, using long-division) is 12,143,144,856,843. Then, next, when 12,143,144,856,843 is divided by 204,709,197 (i.e., the product of... 3 • 7 • 11 • 37 • 43 • 557)... the next quotient (again, using long-division) is 59,319. Finally, when a person divides 59,319 by 4,563 (i.e., the product of... 3 • 3 • 3 • 13 • 13)... the next quotient (again, via long-division) is 13.

Although, initially, I did have some difficulty in grasping the mechanics of these fascinating principles of Vedic Mathematics... eventually, I did comprehend them. You have taught me an invaluable lesson... for which I am very grateful; however, I do believe that I have identified the reason for you reaching the mistaken conclusion of the number not being evenly divisible by 17 (when, in fact, it is). Near the end of the process, you stated (perhaps, from a typographical error)... that (1 • 12) + 6 + 4 = 24. In fact, however, (1 • 12) + 6 + 4 = 22 (instead of 24). This tiny little glitch... led to two subsequent mistakes. Had this minor slip not happened... no doubt, you would have followed the result of 22... with the next two statements: (2 • 12) + 2 + 3 = 29 and (9 • 12) + 2 + 9 = 119... thereby, verifying the divisibility.

Even though I am glad to have learned this new technique (at least, that is, new to me)... the divisibility test (or rule)-- which I had referenced in my original inquiry... still, remains a mystery to me. Anyone, at all... please, post information about how to apply (to 9,349,990,820,016,829,983)... the divisibility rule (for 17) which says to: "Alternatively subtract and add blocks of two digits from the end, doubling the last block and halving the result of the operation, rounding any decimal end result as necessary."

Thanking you, in advance, for your prompt reply,

Danny

More on General Divisibility Rules[edit]

Interesting work there... though I warn you that yellow on white is barely legible. I certainly can't read it.

--Changed the fonts and presentation to make it readable. 2007.06.08

I think another component to general divisibility rules deserves some discussion. In particular, alternate bases than '10'. E.g. if a number is represented in binary or hexadecimal. For young programmers working on big-number representations and optimizations of associated functions (like 'isprime?') that would be quite a boon. Not that it hasn't been solved in the open-source world already... but a decent explanation would be nice.

Thank a lot for your contribution, really I don't know nothing about isprime, if you know about this we can collaborate in this, I am murad aldamen, who made this rule...

no problem for the font Donquimico 01:55, 9 June 2007 (UTC)[reply]

Danny, perhaps the algorithm for divisibility by 17 is the negative osculation algorithm, Vedic Mathematics. I can look it up later. Divisor, D=17. 17x7=119. Hence, the positive osculator, P = 12. One nine means use one digit at-a-time. Next, generate a double nine ending. 17x47=799. Hence, P=8, taking 2-digits at-a-time. And the negative osculator (multiplier) is 4? but alternating + and - signs, osculating right to left. If we let the negative osculator = Q, then P+Q=D. There is both positive and negative osculation as well as digit-wise, whole number, complex, and multiplex osculation.Larry R. Holmgren 17:36, 22 June 2007 (UTC)[reply]

I am the man 2859 03:22, 24 October 2007 (UTC) i still don't get how you do divisibility rule 11![reply]

it is easily, see general divisibility test,... for examples: 121=120+1--> 12-1=11; another example, 143,...140+3=>14-3=11;...etc Donquimico —Preceding unsigned comment added by 81.203.145.59 (talk) 00:44, 2 March 2008 (UTC)[reply]

General rule[edit]

Thanks for working on this topic! I'd like to add a suggestion about adding some wording regarding a general rule, especially if you plan to cap the divisors at 50. I see that you've noted on the table for 7 that 'Double the number with the last two digits removed and add the last two digits' works. Let's extend that a little, assuming 'b' equals the last two digits and 'a' everything else.

If x is a coefficient equal to 100 - (the greatest multiple of the divisor less than 100), then divisor d is evenly divisible if xa+b is. For 7 that is obviously the 2a+b that you mention. For 19 it would be 5a+b, etc. For some numbers like 17, it's more trouble than it's worth (see below), but there are clearly some numbers where this is handy. If b is extended to three digits, then x = 1000 - (greatest multiple <1000), etc and the same formula works.

Or, you can subtract using the test xa-b, where x is a coefficient equal to (the smallest multiple of the divisor greater than 100) - 100. So for 17 the formula 2a-b is preferable to 15a+b.

Hope this is helpful. I've been looking around the web to see if anyone has a 'unified theory' that laypeople like myself would understand, and perhaps it's out there. Since I'm not planning to publish my childhood musings, here it is if you can use it.

Best of luck, George Wunder —Preceding unsigned comment added by 69.115.168.129 (talk) 17:55, 28 May 2008 (UTC)[reply]

Quote:

"14

It is divisible by 2 and by 7.

224: it is divisible by 2 and by 7.

Add the last two digits to twice the rest. The answer must be divisible by 7.

364: (3 × 2) + 64 = 70."

Well, I think there is a mistake. Obviously 371 is not divisible by 14 but (3×2)+71=77.

Where it is written "The answer must be divisible by 7" it should be written "The answer must be divisible by 14".

I corrected it, so I think you shouldn't undo the change.

Divisibility by 13[edit]

It is like 7, but instead of subtracting double the last digit, you add 4 times the last digit. Eg. 13286: 1328 + 4(6)=1352 => 135 + 2(4) = 143 = 13 * 11. They prove this rule as part of Ring Theory in university. — Preceding unsigned comment added by 184.105.71.2 (talk) 19:56, 25 May 2016 (UTC)[reply]

A Better Algorithm for "Divisible by 3" Rule[edit]

I would like to put forward another more efficient algorithm for the "divisible by 3" rule, to add to (ie. not replace) the current "large numbers" algorithm.

Now, since all multiples of 3 can be composed of "the sum of multiples of 3" (eg. 9=6+3), it is more efficient to keep track of each digits variation from a multiple of 3, than it is to actually complete the full sum of digits to get a resulting number (which may or may-not be divisible by 3). This is better explained by this series of examples:

Is 3333333333 divisible by 3? Now 3+3+3+3+3+3+3+3+3+3 = 3×10, ∴ the sum of the digits is divisible by 3.

Is 6969696969 divisible by 3? Now 6+9+6+9+6+9+6+9+6+9 = 6×5 + 9×5 = 3×(2×5) + 3×(3×5) = 3×(2×5 + 3×5), ∴ the sum of the digits is divisible by 3.

Is 123456 divisible by 3? Now 1+2+3+4+5+6 = (0+1)+(3-1)+3+(3+1)+(6-1)+6 = (0+3+3+3+6+6) + (1-1+1-1) = (0+3+3+3+6+6) + (0) = 3×(0+1+1+1+2+2+0), ∴ the sum of the digits is divisible by 3. This example introduces the "variations from a multiple of 3", which was previously mentioned.

And now for the random-number example; is 235418983207893 divisible by 3?
Now 2+3+5+4+1+8+9+8+3+2+0+7+8+9+3 = (3-1)+3+(6-1)+(3+1)+(0+1)+(9-1)+9+(9-1)+3+(3-1)+0+(6+1)+(9-1)+9+3 = (3+3+6+3+0+9+9+9+3+3+0+6+9+9+3) + (-1-1+1+1-1-1-1+1-1) = (3+3+6+3+0+9+9+9+3+3+0+6+9+9+3) + (-3) = 3×(1+1+2+1+0+3+3+3+1+1+0+2+3+3+1-1), ∴ the sum of the digits is divisible by 3. It is important to note from this example that as long as these "ones" sum to a multiple of 3 (or a negative multiple), the sum of the digits will be a multiple of 3.

The Algorithm[edit]

To summarise what I am basically doing in the last example: if you count the quantity of the digits 2, 5 and 8 in the whole number; subtract this from the quantity of the digits 1, 4 and 7 in the whole number; the result will be a (positive or negative) multiple of 3, if and only if, the original number is divisible by 3.

- Ricketts 123.243.217.67 (talk) 02:34, 30 April 2010 (UTC)[reply]

Massive cleanup needed on Aisle 7459634...[edit]

Okay, seriously, this article is a mess...

For one, what is the table of divisibility rules for 2-20 doing ABOVE the table of contents? It should be below it...
And the table of rules for numbers beyond 20 is missing a ton of numbers, specifically prime numbers: 53, 67, 73, 89, 97...and so many more between 91 and 989 that I can't even pin down...
There are text boxes all over the place that look more like website hacks than information boxes...such things belong on an encyclopedia site such as this like bologna belongs in a birthday cake...

I've fixed the first problem (and cleaned up a lot of smaller issues), but the last two are beyond me... -Black Yoshi (talk) 21:27, 25 May 2010 (UTC)[reply]

16[edit]

My version:

Add four times the first (hundreds) digit to the number formed from the last two digits.

The incorrect version:

Add four times the first digit to the rest.

Perhaps something needs to be done to correct my version in case the number has more than 3 digits, but the previous line adjusts the number formed by the last 3 digits according to the value in the thousands place. — Arthur Rubin (talk) 08:27, 27 May 2010 (UTC)[reply]

I wasn't in advanced math courses throughout middle and high school for nothing, Arthur; my version (your "incorrect version") is in fact mathematically correct -- just confusingly stated. I'll admit, I was just trying to follow the pattern set up by other numbers -- say, 7: "Subtract twice the last digit from the rest," it says, and nothing more. My version probably should have said, "Add four times the first digit to the number formed by the rest," but then you would have to fix all of the other numbers to match it, since it would stick out like a sore thumb. Perhaps a little reminder above/below the table would clear that up... —Preceding unsigned comment added by Black Yoshi (talk • contribs) 13:43, 27 May 2010 (UTC)[reply]

Your expression is wrong unless the number has exactly 3 digits. Mine is correct if the number has at most 3 digits. However, I'm willing to adjust it to cover at least 3 digits, if you insist. — Arthur Rubin (talk) 14:56, 27 May 2010 (UTC)[reply]

Oh my God, I can't believe I've been so darned wrong about that...in fact, you have to multiply the number formed by all except the last two digits by 4, then add that to the number formed by the last two digits...oh God, I'm so sorry about that...about everything I've done to this article in the past twenty-four hours...note to self: CHECK MATH FIRST -- don't just assume something is true...Black Yoshi (talk) 16:49, 27 May 2010 (UTC)[reply]

The similar wording on the rule for 8 confused me as well. I thought it meant to double every digit aside from the last. An example for a three-or-more-digit number (as opposed to the existing two-digit example) in the table would have made this much clearer. Travis Evans (talk) 18:54, 29 September 2022 (UTC)[reply]

Why there is a bias in editing this page and the murad divisibility rule has been deleted!!! —Preceding unsigned comment added by 87.236.232.97 (talk) 16:08, 17 July 2010 (UTC)[reply]

Code boxes all over the place...[edit]

Okay, what is with all these code boxes in the Divisibility by 7 section? Do they even have a purpose, aside from to make the article appear as though it's been hacked? I've tried to remove them (as that's probably why the {{cleanup}} tag has been added) but I haven't been able to; if someone else can either explain their intended purpose and/or clear them up, that would be great. Black Yoshi (talk) 03:17, 8 January 2011 (UTC)[reply]

It makes the article easier to read by ensuring that there is a fixed width font so text lines up properly. Also single line breaks are not ignored the way they are in normal text on MediaWiki. It can be done without the code boxes, yes (just use BR tags after every line) but I find this way easier both to write and to read.—Soap— 14:29, 25 April 2011 (UTC)[reply]

Divisibility Rules in Lotsa Various Bases[edit]

These are for bases in addition to ten. They are grouped according to ease of use (subjective of course).

Base 10:

2: final digit is {0, 2, 4, 6, 8}.
5: final digit is {0, 5}.
4: penultimate digit is odd (2n+1), final digit is {2, 6}; or penultimate digit is even (2n), final digit is {0, 4, 8}.
3: sum of all digits is a multiple of 3.
9: sum of all digits is a multiple of 9.
6: an even number that passes the divisibility test for 3.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.
8: third last digit is odd (2n+1), last two digits are divisible by 4 but not 8 (8n+4); third last digit is even (2n), last two digits are divisible by 8 (8n).
7: successive subtraction of final three digits from all the other digits yields a multiple of 7.
13: successive subtraction of final three digits from all the other digits yields a multiple of 13.

Base 12:

2: final digit is {0, 2, 4, 6, 8, X}.
3: final digit is {0, 3, 6, 9}.
4: final digit is {0, 4, 8}.
6: final digit is {0, 6}.
8: penultimate digit is odd (2n+1), final digit is 4; or penultimate digit is even (2n), final digit is {0, 8}.
9: penultimate digit is 3n+1, final digit is 6; or penultimate digit is 3n−1, final digit is 3; or penultimate digit is 3n, final digit is {0, 9}.
E: sum of all the digits is a multiple of E.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.
16: final two digits are {00, 16, 30, 46, 60, 76, 90, X6}.
14: final two digits are {00, 14, 28, 40, 54, 68, 80, 94, X8}.
5: successive subtraction of final two digits from all the other digits yields a multiple of 5.
X: an even number that passes the divisibility test for 5.
7: successive subtraction of final three digits from all the other digits yields a multiple of 7.

Base 14:

2: final digit is {0, 2, 4, 6, 8, A, C}.
7: final digit is {0, 7}.
4: penultimate digit is odd (2n+1), final digit is {2, 6, A}; or penultimate digit is even (2n), final digit is {0, 4, 8, C}.
D: sum of all the digits is a multiple of D.
3: sum of digits in even places and sum of digits in odd places differ by a multiple of 3.
5: sum of digits in even places and sum of digits in odd places differ by a multiple of 5.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.
6: an even number that passes the divisibility test for 3.
A: an even number that passes the divisibility test for 5.
8: third last digit is odd (2n+1), last two digits are divisible by 4 but not 8 (8n+4); third last digit is even (2n), last two digits are divisible by 8 (8n).
9: successive subtraction of final three digits from all the other digits yields a multiple of 9.

Base 16:

2: final digit is {0, 2, 4, 6, 8, A, C, E}.
4: final digit is {0, 4, 8, C}.
8: final digit is {0, 8}.
3: sum of all digits is a multiple of 3.
5: sum of all digits is a multiple of 5.
F: sum of all digits is a multiple of F.
6: an even number that passes the divisibility test for 3.
A: an even number that passes the divisibility test for 5.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.
7: sum of all blocks of three digits is a multiple of 7.
9: sum of all blocks of three digits is a multiple of 9.
D: sum of all blocks of three digits is a multiple of D.
E: an even number that passes the divisibility test for 7.

Base 6:

2: final digit is {0, 2, 4}.
3: final digit is {0, 3}.
4: penultimate digit is odd (2n+1), final digit is {2}; or penultimate digit is even (2n), final digit is {0, 4}.
13: last two digits are {00, 13, 30, 43}.
5: sum of all digits is a multiple of 5.
14: an even number that passes the divisibility test for 5.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.
12: third last digit is odd (2n+1), last two digits are {04, 20, 32, 44}; or third last digit is even (2n), last two digits are {00, 12, 24, 40, 52}.

Base 8:

2: final digit is {0, 2, 4, 6}.
4: final digit is {0, 4}.
3: sum of digits in even places and sum of digits in odd places differ by a multiple of 3.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.
6: an even number that passes the divisibility test for 3.
7: sum of all the digits is a multiple of 7.
5: successive subtraction of final two digits from all the other digits yields a multiple of 5.
12: an even number that passes the divisibility test for 5.

Base 11 (a prime base, for comparison):

2: sum of all the digits is a multiple of 2. (Alternating-digits rule can also be applied.)
5: sum of all the digits is a multiple of 5.
X: sum of all the digits is a multiple of X.
3: sum of digits in even places and sum of digits in odd places differ by a multiple of 3.
4: sum of digits in even places and sum of digits in odd places differ by a multiple of 4.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.
6: a number that passes the divisibility tests for both 2 and 3.
7: sum of all blocks of three digits is a multiple of 7.
9: successive subtraction of final three digits from all the other digits yields a multiple of 9.

Base 15 (odd but not prime):

3: final digit is {0, 3, 6, 9, C}.
5: final digit is {0, 5, A}.
9: penultimate digit is 3n+1, final digit is {3, C}; or penultimate digit is 3n−1, final digit is 6; or penultimate digit is 3n, final digit is {0, 9}.
2: sum of all the digits is a multiple of 2. (Alternating-digits rule can also be applied.)
7: sum of all the digits is a multiple of 7.
E: sum of all the digits is a multiple of E.
6: a number that passes the divisibility tests for both 3 and 2.
A: a number that passes the divisibility tests for both 5 and 2.
4: sum of digits in even places and sum of digits in odd places differ by a multiple of 4.
8: sum of digits in even places and sum of digits in odd places differ by a multiple of 8.
11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11. — Preceding unsigned comment added by 46.117.126.163 (talk) 21:26, 7 May 2012 (UTC)[reply]

Divisibility testing via modulo division[edit]

Hello fans of divisibility tests and modular math!

I propose adding some text and examples of divisibility testing using modulo division. (And I did add it already once but it was reverted by Arthur Rubin on the grounds that modulo calculation is not a divisibility test.) But, technically, it is a divisibility test, and its easier to perform both manually and mentally than some of the methods currently in the article, and I honesty believe that modulo division is thoroughly in line with the spirit and techniques of the methods discussed in this article. Acknowledging that there may be a legitimate difference of opinion, I'll try proposing it here to see if others agree and want to include it.

The simple and easy-to-do form of modulo division is where you replace the first two digits of your dividend with their remainder, modulo your 1-digit divisor. The proof is constructed similarly to all other divisibility tests. Note the similarity to the divisibility test for 7 where you subtract twice the last digit from the number formed by all but the last digit. By repeatedly applying a 1-digit reduction to the dividend, you can transform the dividend into a 2 or 1 digit number that is easier to test for divisibility. Modulo division is doing exactly the same thing, the only difference is that the reduction operation is a small modulo as opposed to a multiply-and-subtract. (And of course if you think about it, those operations are very nearly equivalent.)

The proof is straightforward. Using the notation in the modular arithmetic article, where ${\overline {x}}_{n}$ signifies $x{\pmod {n}}$ , it is easy to show that when calculating a remainder, you can reduce a dividend by replacing any number of left side digits of the dividend with those digits modulo the divisor.

Suppose we want to find a number ${\overline {LR}}_{n}$ , where two juxtaposed numbers $L$ and $R=b*L+R$ , and where $b$ is your base raised to a power equal to the number of digits in $R$ .

* ${\overline {b*L+R}}_{n}$
* $\equiv {\overline {b*L}}_{n}+{\overline {R}}_{n}$
* $\equiv {\overline {b}}_{n}*{\overline {L}}_{n}+{\overline {R}}_{n}$	← This is proof of the 3's, 9's and 11's test.
* $\equiv b*{\overline {L}}_{n}+{\overline {R}}_{n}$	← The interesting extra step here is to remove the modulo from the base, this is what allows simply juxtaposing the right side of the dividend onto the remainder of the left side modulo n.

The benefits of this particular method of divisibility testing are, in my opinion, numerous. For one, this single rule works for all divisors, though it is most convenient and mainly amenable to mental calculation for divisors 3-11, and I suggest only including examples for divisors <= 11. It also works in all bases, as the construction above demonstrates.

The fact that the modulo division reduction works left to right, ignores the quotient digits, and only requires remembering one single digit at a time is what makes this technique simpler than performing a complete long division. Although it can be argued that a series of small modulo operations looks identical to performing long division, and that the quotient digits are being explicitly calculated and then discarded thus saving no work, in practice it is possible and quite easy to avoid thinking of the quotient digits at all, but rather to mentally calculate 1 digit remainders from 2 digit dividends by thinking of the nearest multiple of the dividend. I am assuming that the 1-digit multiplication table is memorized.

Here are the examples I added originally. Note they can be written in a single line. Some of the divisibility tests are difficult to write in a single line, for example the ${\overline {LR}}_{7}\equiv L-2*R$ test for divisibility by 7 generally requires multiple lines and ends up looking like backwards long division on paper. I believe that a modulo division test for divisibility by 7 is easier to perform by hand than the other tests, but don't take my word for it-- TRY IT!

16,499,205,854,376 (mod 3): 1 6 ¹4 ²9 ²9 ²2 ¹0 ¹5 ⁰8 ²5 ¹4 ²3 ²7 ⁰6^→0 The remainder is zero; 16,499,205,854,376 is divisible by 3.
1,458 (mod 6): 1 4 ²5 ¹8^→0 The remainder is zero, so 1,458 is divisible by 6.
379,587,768 (mod 7): 3 7 ²9 ¹5 ¹8 ⁴7 ⁵7 ¹6 ²8^→0 The remainder is zero; 379,587,768 is divisible by 7.
34152 (mod 8): 1 5 ⁷2^→0 The remainder is zero.
298,247,532 (mod 11): 2 9 ⁷8 ¹2 ¹4 ³7 ⁴5 ¹3 ²2^→0 The remainder is zero, so 298,247,532 is divisible by 11.

DavidAugustus (talk) 19:11, 1 January 2013 (UTC)[reply]

It's indistinguishable from short division, except that it takes less space as you fail to write down the quotient, and I can find no literature using the word "modulo division", as I noted in regard his addition to the short division article. I agree that that (short division) is simpler than some of the tests in the article — Arthur Rubin (talk) 04:45, 2 January 2013 (UTC)[reply]

I'm probably the wrong person to add the modulo discussion to this article, I tried to stub in what I thought was more well known, hoping it could be sourced by someone else. I hope it'll happen eventually. I think it would be worth mentioning short division in this article in the mean time.

For the sake of discussion, I'll just add a couple of thoughts. If modulo division and short division are indistinguishable, that implies a remainder can't be calculated without dividing (which I covered above), and it implies that the only factor in determining how easy a calculation is to do is the number of sub-operations involved. For a mental calculation, the number of operations is a factor, but potentially more important is storage; the number of digits you have to keep in memory at any given time.

To do a division in your head of an N-digit dividend with a 1 digit divisor, you have to remember N digits. Most people would balk at the suggestion that they try a 30-digit short division in their head and recite the quotient, and not because its hard to do the steps required, but because its hard to remember a 29 digit quotient. If you don't need a quotient because you're only calculating a remainder, then anyone can do divisibility tests with modulo division for 30 digit numbers in their head, because you only ever have to remember 1 digit at a time. Not having to devote any space or thought to the quotient at all is what makes modulo division easy. If the number of operations were all that mattered, there would be no benefit to short division over long division, but many people consider failing to write down the intermediate multiplication and subtraction operations a valuable time and space saving technique.

It is also the space needed to write a calculation, or the memory needed to remember digits during a mental calculation, and not the number of sub-operations involved, that make divisibility tests that use left-side reductions easier to do than right-side reductions. This is why, with large dividends, short division is easier for mod 7 than subtracting twice the last digit repeatedly, and why modulo division is even easier than summing digits in a 3, 9, or 11 test.

DavidAugustus (talk) 06:44, 5 January 2013 (UTC)[reply]

In modulo division, you may not need any space for the quotient, but you do need the same thought. For example, for 7, there's very little difference between having to remember

1	2	3	4	5	6	7	8	9
7	14	21	28	35	42	49	56	63

and

7	14	21	28	35	42	49	56	63

especially if you expect to get it right.

I may not be the best choice for determinging what's "easy" in mental arithmetic, either, but (modified) summing digits seems easier than formal modulo division for 3, 9, or 11. For the example for 9 in the article

2880 → 2+8+8+0 = 18 → 1+8 = 9

could be replaced by

2880 → (2+8)80 = (10)80 → 180 → (1+8)0 = 90 → (9+0) = 9, which produces the same set of digits as your "modulo division", without requiring memorization of the 9 × table.

In other words, in your notation,

2 8 ^{10 1} 8 ⁹ 0 ^{9 → 0},

without having to remember multiples of 9. Similarly, for 11,

9 1 ⁸ 8 ^{16 5} 0 ⁵ 8 ^{13 2} 2 ⁰

with x representing −x, again without having to subtract multiples of 11 (although remembering them isn't a problem). — Arthur Rubin (talk) 07:58, 5 January 2013 (UTC)[reply]

I agree its hard not to think of quotient digits, if you write them down and cross them out, which is why I don't write it that way. I can only tell you I personally find it easy to avoid thinking of quotients at all, I can't otherwise prove it, though I did already demonstrate its not necessary. You are right that its usually trivially easy to recover the quotient and that the difference is very small, but that doesn't matter; there is a difference, and for purely mental calculations, in my opinion, the difference matters. It matters because the ease of the calculation isn't dominated by the ease of the sub-operations, its more about how many things you have to remember, and how many things you have to mentally juggle. I find that allowing the quotient digits to cross my mind slows me down considerably. A little bit of practice might help you see this, if you're willing to try it.

One can also use methods other than division to produce remainders, and the divisibility tests are proof of that fact, quotient digits are not necessarily a byproduct of the sub-operations involved. That's not normally how I calculate remainders, but it does open the possibility of using right-side reductions as sub-operations in a left-side reduction modulo calculation, in order to avoid having to remember the dividend or any intermediate results. That is, at least for any right-side reductions that actually produce valid non-zero remainders- and I'm not sure which ones do if any, but the 7's test doesn't. This is another small reason to prefer modulo division over a divisibility test- if the number being tested isn't divisible, but you need the nearest number that is, calculating the correct remainder helps.

I qualified my statement that modulo division is easier than other tests when the dividend is large. I'm talking about cases where you have to repeat the digit summing process multiple times, which requires remembering the intermediate steps. Modulo division shines for tests on larger dividends because you're always done in one pass, with only 1 digit of temporary storage during the entire process. I don't think modulo division is easier for small dividends like 2880, but I don't think its harder either. The advantage for small dividends is that there's only one method needed, not a weird rule for each divisor, and the result is always a correct remainder.

Here's how I'm really thinking about this, and the way that I first learned how easy calculating modulos really can be. You can make a state machine for divisibility that operates on 1 digit of dividend at a time, read from left to right. Below is the table for divisibility by 3. This collapses 3 or 4 arithmetic operations into a single state transition that involves no calculation, because they're baked into the table. You could memorize this table, and that's probably the easiest divisibility test there is, even easier than summing digits. I don't think that's very practical, but the closer you can approximate this mentally, the easier testing divisibility becomes.

mod 3
Cur. state → Input ↓	0	1	2
0,3,6,9	0	1	2
1,4,7	1	2	0
2,5,8	2	0	1

DavidAugustus (talk) 17:42, 5 January 2013 (UTC)[reply]

Vedic math forum[edit]

A recently added link:

Osculation Method Simplified Osculation Technique - Vedic Maths to check Divisibility by 7, 13 and 17.

is not a "simplified osculation technique", but I can't figure out what it actually may be. It's not a reliable, or even a credible, source; that's not required for external links, but some relationship to reality is required, and that is not evident. — Arthur Rubin (talk) 15:02, 20 August 2013 (UTC)[reply]

It seems also to be the creation of the editor (board name = raajesh; editor name = Raajesh23 (talk · contribs) ) — Arthur Rubin (talk) 15:09, 20 August 2013 (UTC)[reply]

The link explains how the Osculation process adopted in Vedic Maths to check divisibility of a number can be further simplified. In the case of 17 to check divisibility of 9,349,990,820,016,829,983 for example, we can reduce 9,349,990,820,016,829,983 using a 17-point circle and do the divisibility. Let us take a smaller example to understand this.

To check divisibility of 91 by 7, if we use Osculation, we Osculate 91 by 5.

The steps are;

9 1

14

14 => 5 * 1 + 9

In the modified approach, we can reduce 91 using a 7-point circle like the one shown below:

91 -> 21

21 is divisible by 7 and so 91 is.

Another example;

11459

By Osculation;

1 1 4 5 9

35 46 9 50

As 35 is divisible by 7, we say 11459 is also divisible by 7.Instead, we can reduce 11459 first. We get

=> 10003 | 14 is replaced by 00, 59 by 03
=> 03003 | 10 is replaced by 03
=> 0203 | 30 is replaced by 2
=> 63 | 20 is replaced by 6

63 is divisible by 7 and the number 11459 also.

The reduction simplifies the divisibility check. 9,349,990,820,016,829,983 can be reduced to 011000(3),1000000(1)0012 [bracket indicate bar digits] or further before osculating for a check by 17. — Preceding unsigned comment added by Raajesh23 (talk • contribs) 05:36, 21 August 2013‎ (UTC)[reply]

We still need a reliable source for the method, and that it has anything to do with Vedic math or "Osculation". A forum is not a reliable source, except insofar as the poster is a published expert. I have doubts about the "Vedic" section of "divisibility by 7", but, at least, it purports to be from a book, which I cannot prove is flaky at this time. — Arthur Rubin (talk) 05:51, 21 August 2013 (UTC)[reply]

Here is a published white paper on the concept:

http://vedicmaths.org/images/PDFs/VM_Journal/A%20Different%20Osculation%20Approach%20to%20Test%20Divisibility%20of%20Numbers-1.pdf — Preceding unsigned comment added by Raajesh23 (talk • contribs) 04:51, 12 September 2013 (UTC)[reply]

I don't see direct evidence that the paper is published, and I would need to see evidence of reliability of the source. I would like to add that some examples of "Vedic mathematics" fall under Category:Pseudomathematics; I don't know whether this topic of "Vedic mathematics" fits in that area, but we need to be careful about what sources are real-world reliable and what are only reliable in context. — Arthur Rubin (talk) 14:40, 18 October 2013 (UTC)[reply]

This isn't really resolved. I would still be in favor of deleting the "vedic math" section for 7, as it is more complicated than the third method of the table, add 5 times the last digit to the rest of the number. At best, it's a folding of that method, so it should be reported as a sentence (possibly plus tabular example; the primary example here is misaligned.)

It would be cleaner to write:

Is 438,722,025 divisible by seven?  Multiplier = 5.
 4  3  8  7  2  2  0  2  5
42 37 46 37  6 40 37 27

In any case, the steps need to be explained, something like:

Sample calculation step
v	w
V = v + W₁ + 5 × W₀	W (calculated earlier)= W₁W₀

If they cannot be explained from a reliable source, then it's too confusing to include.

The second method is modified long-division (throwing away 7s in other ways during the process). — Arthur Rubin (talk) 20:26, 19 April 2015 (UTC)[reply]

high powers of two[edit]

I don't see how the tests for 32, 64, 128, 256, and 512 given are in any way practical... Double sharp (talk) 12:43, 18 October 2013 (UTC)[reply]

Examples[edit]

I don't see the point of the examples added by 69.120.155.201 (talk · contribs · WHOIS) in January 2014. I think those raw examples, without any indication of calculation (and, in some cases, having incorrect calculations) should be removed. — Arthur Rubin (talk) 13:32, 24 February 2014 (UTC)[reply]

I tried removing them myself when they were first added, but was reverted twice and didn't feel like starting an edit war, so I wikified them a bit and left it. If you want to remove them go ahead. Black Yoshi (Yoshi! | Yoshi's Eggs) 16:04, 24 February 2014 (UTC)[reply]

Examples (April)[edit]

I request a reason for changes to the examples. There have been many editors making arbitrary changes. — Arthur Rubin (talk) 05:15, 19 April 2015 (UTC)[reply]

Calculations[edit]

The examples have different styles of calculations; for example

16b: 1168: 11 × 4 + 68 = 112.
21: 168: 16 − (8×2) = 0

but

29: 261: 1×3 = 3; 3 + 26 = 29

I think we should agree on a single style for all the expressions. For these, I would suggest:

16b: 1168: 11×4 + 68 = 112.
21: 168: 16 − 8×2 = 0
29: 261: 26 + 1×3 = 29

No parenthesis (unless needed for other reasons, as in some of the expressions for 7)
Unspaced "×"
Spaced additive operators ("+", "−")

Also, all the math formulas use \cdot or juxtaposition for multiplication; I think we should change them all to \times.

I don't want to make cosmetic changes without justification. — Arthur Rubin (talk) 19:44, 19 April 2015 (UTC)[reply]

Divisibility condition[edit]

There is no need to change the examples. Would the divisibility conditions be left as is or changed — Preceding unsigned comment added by 47.20.7.225 (talk • contribs) 17:44, January 2, 2016‎

WP:BEBOLD does not mean to add boldface[edit]

Returning to this article after a long absence, I found a lot of unnecessary bolding. While I fixed that, I

adjusted commas so that number of 4 or less digits do not have commas, and numbers of 5 or more digits have commas every 3 digits:
adjusted some of the formulas to leave the number in the same order; for example "add 2 times the last 2 digits to the rest" is written r + t × 2 (or r + 2 × t; I haven't standarized those yet; and add the last digit to twice the rest as r × 2 + d.
removed examples in the extended list where the rule is a combination of other rules; for example 22.
changed the example for 999 so it's less obvious.

If there are any objections, please comment. — Arthur Rubin (talk) 16:32, 7 March 2016 (UTC)[reply]

So I made some of the changes I suggested last year, as I had to make significant adjustments. Any objection to fixing the rest of them ? — Arthur Rubin (talk) 16:39, 7 March 2016 (UTC)[reply]

My reversions[edit]

To clarify, the reason I keep reverting those edits us because very similar edits have been reverted by other editors in the past and I see no reason why they shouldn't be reverted now. Black Yoshi (Yoshi! | Yoshi's Eggs) 22:07, 24 June 2016 (UTC)[reply]

I think the original editor confused everyone including user:Favonian by removing a large block and re-inserting it earlier in the article, and only later fixed the look of the table, instead of doing all of this in 1 edit. I was also confused at first before I realized that he was improving the article by gathering the 20-30 block with the 1-20 block because they both are continuous. Beyond 30, there are more holes (34, 38, 40 are not mentioned...). Of course it would be more helpful if everyone was using the edit summaries to explain their edits. Dhrm77 (talk) 22:34, 24 June 2016 (UTC)[reply]

I realize now that it is an improvement, and I won't revert it again. I can't say the same for those who were reverting the prior edits in the first place, but I will back off. Black Yoshi (Yoshi! | Yoshi's Eggs) 06:32, 25 June 2016 (UTC)[reply]

Neither will I, though I reserve the right to harumph the lack of edit summaries. Favonian (talk) 12:31, 25 June 2016 (UTC)[reply]

Simpler 25 Rule & missing 29 example[edit]

25 Rule[edit]

I think that the 25 rule could be a tiny bit simpler than it is.

Current Rule: Divisible if the number formed by the last 2 digits is divisible by 25.

My Rule: Divisible if the last 2 digits are 25, 50, 75, or 00 (if there is a hundred preceding the zeros.) example: 1575: 75 is the last two digits, so its divisible.

29 Example[edit]

The example for the 2nd way of finding 29 ([last 2 digits] x 9 + [remainder of number])is missing, and I have no idea of what it could be. Let's debate on it and see what it might be!

Move discussion in progress[edit]

There is a move discussion in progress on Talk:Divisibility (ring theory) which affects this page. Please participate on that page and not in this talk page section. Thank you. —RMCD bot 13:31, 9 April 2017 (UTC)[reply]

Draft:Divisibility coefficient rule[edit]

Please consider incorporating material from the above draft submission into this article. Drafts are eligible for deletion after 6 months of inactivity. ~Kvng (talk) 16:40, 27 November 2020 (UTC)[reply]

Divisible rule of 10[edit]

If a number has 0 in the units place than it is divisible by 10 A) 11233450 it is divisible by 10 because it is units place there is 0 — Preceding unsigned comment added by 2409:4042:4E9A:7618:0:0:544B:9000 (talk) 06:35, 20 March 2022 (UTC)[reply]

Divisibility rules for 11[edit]

I made a small edit to the first 2 rules for 11 in the table but it got reverted. Just wondering why? The first was for "Form the alternating sum of the digits, or equivalently sum(odd) - sum(even)." I added the text "from right to left" after "digits". While technically you don't necessarily have to start from the right for the divisibility test to be correct, you do if you want to claim "or equivalently sum(odd) - sum(even)." If there are an even number of digits and you start from the left, that is sum(even) - sum(odd) and thus not equivalent. Further, I'd argue this is "more correct" because it will always give you the correct remainder mod 11. Finally, this language is used in the 7 and 13 test descriptions where it talks about doing the alternating sum of digits in blocks of 3. But order doesn't matter for those rules either if all you are interested in is the divisibility and not the actual remainder so not sure why we say it there but not here. Shouldn't we be consistent throughout the article?

The second edit was for "Add the digits in blocks of two from right to left. The result must be divisible by 11." Here I removed "from right to left" because we are doing only addition and addition is commutative, so it makes absolutely no difference whatsoever what direction we start from. In fact, in the example given it shows the addition being done from left to right, contradicting the description! I can kinda sorta see the rationale for reverting my first edit above for the alternating sum since, like I said, if all you're interested in is the divisibility and not the remainder it doesn't matter. But it makes absolutely no sense to require the "from right to left" language here when it truly doesn't matter at all but then delete it from the alternating sum description when it actually does make some difference. It's completely backwards.

Also, my edits didn't include this but I kind of feel like there should be an entry in the 11 section for the alternating sum in blocks of 3 (as described in the 7 and 13 sections) which also works for 11. I realize we already have the alternating sum of single digits which is easier. But I think there is some utility to knowing the blocks of 3 rule for 11 too. This is because if you do this alternating sum calculation for a number you can use the result to test the divisibility for 7, 11 and 13 all in one calculation.

Also, for whatever it's worth, in general for 11, you can do the alternating sum on any odd sized blocking and the simple addition on any even sized blocking. For example, breaking the number into blocks of 5 or 7 or 9 etc. and doing the alternating sum also works. And breaking the number into blocks of 4 or 6 or 8, etc. and simply adding also works. Granted, not really all that useful so not sure if it makes sense to include it as a rule, but kind of interesting. Possibly deserves a mention somewhere in the article. — Preceding unsigned comment added by 50.238.167.160 (talk) 15:20, 13 April 2022 (UTC)[reply]

Because adding the numbers in blocks of two must be done from right to left; from left to right, it won't work unless the number has exactly an even number of digits, whereas doing the alternating sum doesn't matter in terms of which direction you move. For example, take a number like 142,318 (11 * 12,938). It has an even number of digits, therefore it doesn't matter which direction you move when you add the blocks of two numbers. A number like 152,332,213 (11 * 13,848,383), for example, has an odd number of digits, so you must start from the right-hand side: 13 + 22 + 33 + 52 + 1 = 121 = 11 * 11, whereas 15 + 23 + 32 + 21 + 3 makes 94, which is not divisible by 11. Therefore, in general, when adding in blocks of two, at least to check for divisibility by 11, you MUST start from the right-hand side for it to work. Black Yoshi (Yoshi! | Yoshi's Eggs) 18:55, 13 April 2022 (UTC)[reply]

Okay, I see now what you are getting at. Even though for an odd number of digits you could just add a leading 0, as a matter of practicality it makes sense to start from the right so that you don't need to count the number of digits beforehand if the number is long (though the number would have to be pretty darn long to not be able to tell by just looking at it, esp. if it has commas). And I see now other places in the article where that is consistent. Basically, I guess the article is specifying going from right to left any time a rule requires using blocks of multiple digits. So fine, but I feel like this language is very confusing because it makes it sound like there is a mathematical reason to do it as opposed to simply a practical one. This confusion is further reinforced because in some cases there really is a mathematical reason to start from the right, namely all the rules involving an alternating sum. If you want to preserve the remainder in those cases you definitely need to start from right. I guess if it were up to me, I wouldn't write the article this way but I can see it's at least consistent for the most part and too much to change. However, there is some inconsistency in some of the examples for alternating sums. For example for the alternating sum rules for 7, 13, 73, 77 and 137 the example illustrates starting from the right. But for 91, 101 and 143 it does not. The examples for those show starting from the left even though the description says start from the right.

All that said, I feel like my edit for alternating sum for 11 should stand. It is just wrong (or at least very confusing) to say that the alternating sum, without specifying right to left, is always the equivalent of sum(odd) - sum(even). It is not and even the given example shows that. Either my edit should stay or else the "or equivalently..." clause should be removed. Or possibly it should be rewritten something like "or equivalently sum(odd) - sum(even) or sum(even) - sum(odd)." But I don't care enough to bother with it any more. 50.238.167.160 (talk) 20:42, 13 April 2022 (UTC)[reply]

"such that gcd(n, 2, 5) = 1"[edit]

The gcd(n, 2, 5) will always equal 1 because 2 and 5 are prime. I believe it was intended that n is not divisible by 2 or 5. In that case it should be "such that gcd(n, 2) = 1 and gcd(n, 5) = 1".Jonathan 2357 (talk) 11:15, 16 January 2023 (UTC) Jonathan 2357 (talk) 12:47, 15 January 2023 (UTC)[reply]

Recently removed list of nonnotable divisibility rules[edit]

Despite the the article section title's claim that the rules were notable, the list was largely unsourced and had no coherent inclusion criteria. Any effort to trim it back to just the sourced stuff would be doomed, because it will just become bloated again in short order. Since Wikipedia isn't a WP:HOWTO site, such lists of examples are off topic here anyway, so I have removed it. MrOllie (talk) 19:24, 24 July 2023 (UTC)[reply]

I agree with this removal. This section is just a coatrack with no inclusion criteria other than supposed notability (which apparently means any example an editor wants to include). There's no discernable end to this. Meters (talk) 20:05, 28 July 2023 (UTC)[reply]

Can I add 31 to 40 After I know the rule for 32[edit]

I know 31, 33, 34, 35, 36, 37, 38, 39, and 40 but not 32 (Because it's a power of a prime number not on there) so can I add 31-40 later? Helpe30 wikis (talk) 18:47, 13 September 2023 (UTC)[reply]

Bloating the list with more and more examples is not helpful. MrOllie (talk) 18:53, 13 September 2023 (UTC)[reply]

maybe just doing 1 at a time is better like 31 then 32 then so on until 40 (P.S. Ill give you 31 subtract three times last digit from the rest) Helpe30 wikis (talk) 21:51, 13 September 2023 (UTC)[reply]

No, adding them one at a time does not help. We simply don't need to give examples for every number, Meters (talk) 21:58, 13 September 2023 (UTC)[reply]

I know I feel like 40's a great middle ground stopping place for this that's why I talked about 40 as the end of my edits because I felt like 3 more unique rules while stopping right before the next means that I contributed with 31 and 37 (now I need 32 but here's the others 31: subtract the last digit times 3 from the rest 37: add blocks of 3 from right to left ) Helpe30 wikis (talk) 22:05, 13 September 2023 (UTC)[reply]

What is best for this article is for it not to become bloated with redundant examples. MrOllie (talk) 22:10, 13 September 2023 (UTC)[reply]

(ec)Just drop it please. We have previously removed specific cases for numbers higher than 30. It's not that we don't know rules. Meters (talk) 22:12, 13 September 2023 (UTC)[reply]