Talk:Divisibility rule

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  major revision 

message: 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)
replies: text:  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)
thread_title:  Proofs 

message: 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)
replies: text: 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)

text: 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)
replies: text: 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)

text: There might... -- Rmrfstar 00:05, 27 March 2006 (UTC)
replies: text: 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)

text: 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)
thread_title:  53 

message: 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)
replies: text: I love 'em. -- Rmrfstar 23:51, 14 June 2006 (UTC)
replies: text:  Thanks.  Always nice to get some feedback (esp. positive) -- Walt 12:58, 15 June 2006 (UTC)
thread_title:  Revamping the page 

message: 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)
thread_title:  Cut text - trial divisors 

message: 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)
thread_title:  overall form 

message: 216.167.135.24 20:46, 20 February 2007 (UTC)
replies: text: 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)
replies: text: 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)
thread_title:  Confusing Divisibility Rule for 17 

message: 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)
thread_title:  More on General Divisibility Rules 

message: 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)
thread_title:  General rule 

message: 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)
thread_title:  Divisibility by 13 

message: 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===
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)
thread_title:  A Better Algorithm for "Divisible by 3" Rule 

message: 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)
thread_title: The Algorithm

message: 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)
thread_title:  Massive cleanup needed on Aisle 7459634... 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)

text: 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)
thread_title:  16 

message: 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  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)
replies: text: 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)
thread_title:  Code boxes all over the place... 

message: 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)
thread_title:  Divisibility Rules in Lotsa Various Bases 

message: 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)
thread_title:  high powers of two 

message: I don't see the point of the examples added by  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)
replies: text: 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)
thread_title:  Examples 

message: 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)
thread_title:  Examples (April) 

message: 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)
thread_title:  Calculations 

message: 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)
thread_title:  [[WP:BEBOLD]] does not mean to add boldface 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: Neither will I, though I reserve the right to harumph the lack of edit summaries. Favonian (talk) 12:31, 25 June 2016 (UTC)
thread_title:  My reversions 

message: 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)
thread_title:  Move discussion in progress 

message: 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)
thread_title: [[Draft:Divisibility coefficient rule]]

message: 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)
thread_title:  Divisible rule of 10 

message: 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)
thread_title:  "such that gcd(n, 2, 5) = 1" 

message: 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)
replies: text: 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)
thread_title:  Recently removed list of nonnotable divisibility rules 

message: 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)
replies: text: Bloating the list with more and more examples is not helpful. MrOllie (talk) 18:53, 13 September 2023 (UTC)
replies: text: 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)
replies: text: 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)
replies: text: 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)
replies: text: What is best for this article is for it not to become bloated with redundant examples. MrOllie (talk) 22:10, 13 September 2023 (UTC)

text: (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)
thread_title:  Can I add 31 to 40 After I know the rule for 32