Talk:Divisibility rule

3821:

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.

3765:

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?

2656:

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.

2892:

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.

2903:. 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. 3769:

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.

1152:

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.

3800:

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.

2896:

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.

785:) 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 10 ≡ 4 (mod 17) is the first 1147:

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.

179: 169: 148: 115: 2001:

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.)

21: 2508:

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.

106: 1488:

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...

3764:

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

3613:

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

3052:

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.

1487:

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

1025:

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--

1015:

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

769:

A few examples will help demonstrate this. Since 10 ≡ 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, 10 ≡ 2 (mod 7) so 43106 gives 431×2 + 06 = 868 which gives 8×2+68 = 84 which

4029:

checking to see if it is even. No-one with any clue is going to check that a large number, say 123456789, is divisible by 3 before noticing that it is odd and cannot be divisible by 6. And unless there are objections I'm going to remove the finding a remainder when dividing by 6 example. This is out

3824:

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

2906:

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

2895:

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

2655:

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

1387:

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

1049:

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

1039:

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...

1020:

Does this mean "doubling the last block" of those 2-digit blocks-of-digits which a person (in reverse order) does "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

421:

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

401:

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

295:

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

3820:

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

3799:

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 *

3772:

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

2891:

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.

2000:

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

1638:

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

1364:

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

1146:

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

584:

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

3874:

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 Knowledge isn't a

2645:

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

2503:

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

1996:

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

1258:

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

1205:

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.

1151:

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

1089:

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

3968:

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

2507:

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

1543:

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

1397:

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

495:

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

3776:

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

3768:

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

1204:

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,

1030:

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

1254:

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'

966:

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

1010:

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.

429:

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 Knowledge. Basically, proofs should be in the same text as the theorem, but proofs should be avoided or made sketches unless they add

904:

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.

2665:

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

1185:

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

2635:

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

602:

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.

1103:(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. 1050:

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).

961:

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

1156:

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

2568:

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!

2911:

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.

1379:

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

1021:

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.

1564:

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.

1044:

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: ( / 2) + (2•20) = + 40 = (-131 / 2) + 40 = -65.5 + 40 = -25.5"... and "-25.5•10 = -255 = -15•17".

1265:

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.

1157:

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."

3825:

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.

589:). 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? — 3777:

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.

3231:

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

455:

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.

3773:

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.

2422: 1473:

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. —

3890:

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.

2835:

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

2646:

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.

2492: 2341: 1120:

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.

1068:

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.

1997:

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.

257:

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).

1434:

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...

4082: 3847:

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".

2566: 1262:

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.

235: 2496:← 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. 406:

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. --

3588:

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.

4129: 2272: 3745: 3614:

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.

3147:

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

2120: 2080:, 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. 2040: 459:

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.

3628:

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.

2607:, 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 946:

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.

2666:

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.

696:

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

2511:

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

2078: 1192:

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...

1090:

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.

2848:

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.

3119:

The reduction simplifies the divisibility check. 9,349,990,820,016,829,983 can be reduced to 011000(3),1000000(1)0012 or further before osculating for a check by 17.

2178: 296:

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.--

4119: 3683:

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.

1259:

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.

3151:

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. —

1031:

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?

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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

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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

4070:

the difference between twice the unit digit of the given number and the remaining part of the given number should be a multiple of 7 or it should be equal to 0.

1613:

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

3215:

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. —

1783:

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).

1698:

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).

1361:

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.

3362:

in January 2014. I think those raw examples, without any indication of calculation (and, in some cases, having incorrect calculations) should be removed. —

119: 639:

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.

590: 487: 3719:

Please consider incorporating material from the above draft submission into this article. Drafts are eligible for deletion after 6 months of inactivity. ~

390:

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 Knowledge do without. —

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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}.

4114: 1859:

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}.

1623:

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.

3691:

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

1186:'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. 1388:

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.

4124: 1727:

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}.

585:

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

541: 1374:

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.

462:

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.

3940:

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)

1083:, 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. 201: 3571:

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 ? —

1035:

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

4139: 4104: 818:, 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. 983:

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? --

3826: 3784: 3350: 1965: 1402: 1276: 869: 620: 280: 3738:

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

3911:

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?

3525:

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

1590: 1235: 1040:

becoming "40"). Plus, the "Example" does NOT remove the resulting decimal... by "rounding" (but rather, through multiplication by 10).

47: 32: 4086: 1342: 42: 2434: 2281: 3810: 3638: 3598: 3502: 3389: 1504: 192: 153: 3997:(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. 3522:

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:

3170:

http://vedicmaths.org/images/PDFs/VM_Journal/A%20Different%20Osculation%20Approach%20to%20Test%20Divisibility%20of%20Numbers-1.pdf

1334:

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 =: -->

3749: 41:

at the time. There may be suggestions below for improving the article. Once these issues have been addressed, the article can be

1075:

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

526:

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. --

378:

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.

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evidence that the paper is published, and I would need to see evidence of reliability of the source. I would like to add that

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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}.

1106:

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.

496:

either long article with sensible chapters (which could be skipped by a reader not interested in them) or perhaps even better

3713: 3466:

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

553: 128: 1683:

4: penultimate digit is odd (2n+1), final digit is {2, 6}; or penultimate digit is even (2n), final digit is {0, 4, 8}.

1424:

For one, what is the table of divisibility rules for 2-20 doing ABOVE the table of contents? It should be below it...

20: 2514: 2573:

16,499,205,854,376 (mod 3): 1 6 4 9 9 2 0 5 8 5 4 3 7 6 The remainder is zero; 16,499,205,854,376 is divisible by 3.

403: 1653: 3375:

I tried removing them myself when they were first added, but was reverted twice and didn't feel like starting an

1844:

4: penultimate digit is odd (2n+1), final digit is {2}; or penultimate digit is even (2n), final digit is {0, 4}.

1179:

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

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8: penultimate digit is odd (2n+1), final digit is 4; or penultimate digit is even (2n), final digit is {0, 8}.

1406: 873: 284: 3518:

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

1969: 1280: 624: 3860: 3852: 1239: 476:

I think 50 is enough. After that, it becomes useless information and Knowledge is not the place for that. --

439:

So do you think there's merit to my contention that we would probably be wisest to remove them altogether? —

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This isn't really resolved. I would still be in favor of deleting the "vedic math" section for 7, as it is

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I request a reason for changes to the examples. There have been many editors making arbitrary changes. —

4025:

I'm going to rewrite this. It's ridiculous to state that a number should be checked for divisibility by 3

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

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I've fixed the first problem (and cleaned up a lot of smaller issues), but the last two are beyond me... -

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which affects this page. Please participate on that page and not in this talk page section. Thank you. —

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second number, then it is divisible by 31. E.g., 713 = 690 + 23. 69/3 = 23, thus 713 is divisible by 31.

260:

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

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

2981: 2907:

or 4 arithmetic operations into a single state transition that involves no calculation, because they're

2675: 2589: 2011: 1628: 1553: 1500: 1443: 1369:

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.

325: 134: 3241:

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

319:

Bah, a statement? Do what you will. All I ever did was crank out original researchy home-made rules. --

178: 3494: 2695:

need the same thought. For example, for 7, there's very little difference between having to remember

3954:

No, adding them one at a time does not help. We simply don't need to give examples for every number,

3780: 3741: 3552:

removed examples in the extended list where the rule is a combination of other rules; for example 22.

3490: 3175: 3123: 1961: 1586: 1492: 1338: 1272: 1255:

works. Let's extend that a little, assuming 'b' equals the last two digits and 'a' everything else.

1231: 1226: 1196: 865: 616: 463: 431: 342:

OK, I moved the material, although I'd still like to talk about a name change for the "new" topic. --

276: 3486:

There is no need to change the examples. Would the divisibility conditions be left as is or changed

3296:

The second method is modified long-division (throwing away 7s in other ways during the process). —

3179: 3127: 3027: 908:

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

105: 3183: 3131: 3031: 1704:

13: successive subtraction of final three digits from all the other digits yields a multiple of 13.

1079:, the multiplier. Start on the right, one digit at-a-time, moving to the left. If you finish with 2878:, again without having to subtract multiples of 11 (although remembering them isn't a problem). — 200:

on Knowledge. If you would like to participate, please visit the project page, where you can join

3653: 2585:

298,247,532 (mod 11): 2 9 8 2 4 7 5 3 2 The remainder is zero, so 298,247,532 is divisible by 11.

2045: 2005: 1920:

9: successive subtraction of final three digits from all the other digits yields a multiple of 9.

1786:

9: successive subtraction of final three digits from all the other digits yields a multiple of 9.

1748:

7: successive subtraction of final three digits from all the other digits yields a multiple of 7.

1701:

7: successive subtraction of final three digits from all the other digits yields a multiple of 7.

728: 663: 184: 3441:

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

3356: 820:

In the "We can generalize" paragraph, next-to-last sentence, the following alteration was made (

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5: successive subtraction of final two digits from all the other digits yields a multiple of 5.

1742:

5: successive subtraction of final two digits from all the other digits yields a multiple of 5.

1135:********************************************************************************************** 3988: 3931: 3880: 3572: 3560: 3470: 3406: 3376: 3363: 3317: 3297: 3216: 3152: 3041: 3015: 2879: 2620: 1993: 1958:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

1610: 1583:

Why there is a bias in editing this page and the murad divisibility rule has been deleted!!!

1566: 1532: 1475: 486:

I agree, 50 is a good place to stop. There might even be a case for stopping much earlier. —

4050: 4035: 4002: 3959: 3896: 3801: 3629: 3619: 3589: 3511: 3380: 3072:

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

2579:

379,587,768 (mod 7): 3 7 9 5 8 7 7 6 8 The remainder is zero; 379,587,768 is divisible by 7.

2145: 1911:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

1876:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

1856:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

1818:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

1774:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

1733:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

1695:

11: sum of digits in even places and sum of digits in odd places differ by a multiple of 11.

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2: sum of all the digits is a multiple of 2. (Alternating-digits rule can also be applied.)

1896:

2: sum of all the digits is a multiple of 2. (Alternating-digits rule can also be applied.)

3316:

I don't see how the tests for 32, 64, 128, 256, and 512 given are in any way practical...

1955:

8: sum of digits in even places and sum of digits in odd places differ by a multiple of 8.

1952:

4: sum of digits in even places and sum of digits in odd places differ by a multiple of 4.

1908:

4: sum of digits in even places and sum of digits in odd places differ by a multiple of 4.

1905:

3: sum of digits in even places and sum of digits in odd places differ by a multiple of 3.

1873:

3: sum of digits in even places and sum of digits in odd places differ by a multiple of 3.

1771:

5: sum of digits in even places and sum of digits in odd places differ by a multiple of 5.

1768:

3: sum of digits in even places and sum of digits in odd places differ by a multiple of 3.

440: 423: 391: 3003: 1461:

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

1431:: 53, 67, 73, 89, 97...and so many more between 91 and 989 that I can't even pin down... 1335:

135 + 2(4) = 143 = 13 * 11. They prove this rule as part of Ring Theory in university.

1312:

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

3983:

What is best for this article is for it not to become bloated with redundant examples.

3875:

WP:HOWTO site, such lists of examples are off topic here anyway, so I have removed it.

3724: 3293:

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

2616: 2608: 2604: 1989: 1646: 1427:

And the table of rules for numbers beyond 20 is missing a ton of numbers, specifically

705: 648: 3680:

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

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No parenthesis (unless needed for other reasons, as in some of the expressions for 7)

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It seems also to be the creation of the editor (board name = raajesh; editor name =

3006:

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

3984: 3927: 3876: 3011: 1428: 970: 930: 911: 804: 604: 364: 3648:

Neither will I, though I reserve the right to harumph the lack of edit summaries.

2417:{\displaystyle \equiv {\overline {b}}_{n}*{\overline {L}}_{n}+{\overline {R}}_{n}} 363:

to "properties". If that sticks there then maybe that is another potential name.

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Is 9,349,990,820,016,829,983 divisible by 17? As above for a divisor of 17 the

885: 545: 515: 297: 197: 3541:; I haven't standarized those yet; and add the last digit to twice the rest as 3379:, so I wikified them a bit and left it. If you want to remove them go ahead. 1223:

it is easily, see general divisibility test,... for examples: 121=120+1--: -->

379: 174: 3056:

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

3720: 1641: 4090: 4054: 4039: 4006: 3992: 3978: 3963: 3949: 3935: 3920: 3900: 3884: 3864: 3834: 3815: 3792: 3753: 3728: 3707: 3657: 3643: 3623: 3603: 3578: 3566: 3476: 3412: 3394: 3369: 3325: 3303: 3222: 3191: 3158: 3139: 3047: 3021: 3014:, but some relationship to reality is required, and that is not evident. — 2989: 2885: 2683: 2626: 2597: 1973: 1659: 1632: 1598: 1574: 1557: 1538: 1508: 1481: 1447: 1410: 1350: 1302:

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

1284: 1243: 1217: 1209: 1199: 1130: 1112: 1004: 987: 973: 933: 924: 914: 893: 877: 807: 628: 607: 571: 557: 530: 518: 480: 466: 434: 382:

made a request for the subpages, maybe we should wait for the decision. --

367: 346: 329: 314: 300: 288: 2899:

I qualified my statement that modulo division is easier than other tests

2691:

In modulo division, you may not need any space for the quotient, but you

2576:

1,458 (mod 6): 1 4 5 8 The remainder is zero, so 1,458 is divisible by 6.

1385:

And now for the random-number example; is 235418983207893 divisible by 3?

360: 1988:

I propose adding some text and examples of divisibility testing using

310:

to make a statement also, as he did quite a bit here a while back. --

3969:

digit times 3 from the rest 37: add blocks of 3 from right to left )

1182:--Changed the fonts and presentation to make it readable. 2007.06.08 793:< 10) but lends itself to fast calculations in other cases where 613:

Well 50 is a good area for now, but for biggners it should be 20

3677:

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

2504:

also works in all bases, as the construction above demonstrates.

3469:

I don't want to make cosmetic changes without justification. —

3422:

The examples have different styles of calculations; for example

2487:{\displaystyle \equiv b*{\overline {L}}_{n}+{\overline {R}}_{n}} 2336:{\displaystyle \equiv {\overline {b*L}}_{n}+{\overline {R}}_{n}} 2200:

is your base raised to a power equal to the number of digits in

1949:

A: a number that passes the divisibility tests for both 5 and 2.

1946:

6: a number that passes the divisibility tests for both 3 and 2.

1914:

6: a number that passes the divisibility tests for both 2 and 3.

758:

digits and repeat if necessary. If the result is a multiple of

2860:

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

3926:

Bloating the list with more and more examples is not helpful.

1739:

14: final two digits are {00, 14, 28, 40, 54, 68, 80, 94, X8}.

851:

and add the product to the last (or more precisely, smallest)

754:

and add the product to the last (or more precisely, smallest)

430:

significant value in understanding the article. That's my 2c.

99: 929:

Thanks. Always nice to get some feedback (esp. positive) --

264:

basic ideas, few rules, generalizations of the ideas/concepts

3247:

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

1888:

12: an even number that passes the divisibility test for 5.

1853:

14: an even number that passes the divisibility test for 5.

1879:

6: an even number that passes the divisibility test for 3.

1830:

E: an even number that passes the divisibility test for 7.

1815:

A: an even number that passes the divisibility test for 5.

1812:

6: an even number that passes the divisibility test for 3.

1780:

A: an even number that passes the divisibility test for 5.

1777:

6: an even number that passes the divisibility test for 3.

1745:

X: an even number that passes the divisibility test for 5.

1736:

16: final two digits are {00, 16, 30, 46, 60, 76, 90, X6}.

1692:

6: an even number that passes the divisibility test for 3.

1326:

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

1016:

operation, rounding any decimal end result as necessary."

2004:

The proof is straightforward. Using the notation in the

1992:. (And I did add it already once but it was reverted by 3609:

I think the original editor confused everyone including

1917:

7: sum of all blocks of three digits is a multiple of 7.

1827:

D: sum of all blocks of three digits is a multiple of D.

1824:

9: sum of all blocks of three digits is a multiple of 9.

1821:

7: sum of all blocks of three digits is a multiple of 7.

1319:" it should be written "The answer must be divisible by 500:

split it in serveral separate articles - for instance:

4030:

of topic, and we don't do this for the other examples.

77: 3870:

Recently removed list of nonnotable divisibility rules

2619:) is simpler than some of the tests in the article — 1527:

3 digits. However, I'm willing to adjust it to cover

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Where it is written "The answer must be divisible by

958:

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

677:

if and only if the sum of its digits is divisible by

1093:2 0 9 8 6 5 85 116 89 86 66 YES. 267:

a somewhat comprehensive list of divisibilitiy rules

196:, a collaborative effort to improve the coverage of 51:

of the decision if they believe there was a mistake.

1985:Hello fans of divisibility tests and modular math! 1220:i still don't get how you do divisibility rule 11! 3053:Let us take a smaller example to understand this. 2560: 2486: 2416: 2335: 2266: 2212: 2192: 2172: 2134: 2114: 2072: 2034: 3555:changed the example for 999 so it's less obvious. 1161:Thanking you, in advance, for your prompt reply, 1054:Thanking you, in advance, for your prompt reply, 689:=9 given above are special cases of this result ( 3559:If there are any objections, please comment. — 3116:63 is divisible by 7 and the number 11459 also. 2561:{\displaystyle {\overline {LR}}_{7}\equiv L-2*R} 1365:is better explained by this series of examples: 734:). Now rather than summing the digits, we take 4130:Knowledge level-5 vital articles in Mathematics 3907:Can I add 31 to 40 After I know the rule for 32 3335:I don't see the point of the examples added by 3165:Here is a published white paper on the concept: 2426:← This is proof of the 3's, 9's and 11's test. 1609:Okay, what is with all these code boxes in the 359:I changed the visible part of the pipe over at 4065:Add the rule for the divisibility rule for 7. 1523:3 digits. Mine is correct if the number has 1224:12-1=11; another example, 143,...140+3=: --> 814:The above was changed by another editor. Per 8: 2832:especially if you expect to get it right. 1943:E: sum of all the digits is a multiple of E. 1940:7: sum of all the digits is a multiple of 7. 1902:X: sum of all the digits is a multiple of X. 1899:5: sum of all the digits is a multiple of 5. 1882:7: sum of all the digits is a multiple of 7. 1765:D: sum of all the digits is a multiple of D. 1730:E: sum of all the digits is a multiple of E. 1357:A Better Algorithm for "Divisible by 3" Rule 831:Now rather than summing the digits, we take 3207:examples of "Vedic mathematics" fall under 2582:34152 (mod 8): 1 5 2 The remainder is zero. 1794:2: final digit is {0, 2, 4, 6, 8, A, C, E}. 1469:Add four times the first digit to the rest. 1420:Okay, seriously, this article is a mess... 3778: 3739: 3699:There is a move discussion in progress on 1416:Massive cleanup needed on Aisle 7459634... 1336: 863: 770:is easily noted as being a multiple of 7. 711:. Then we first find a pair of integers ( 614: 142: 56: 15: 2534: 2519: 2516: 2478: 2468: 2458: 2448: 2436: 2408: 2398: 2388: 2378: 2368: 2358: 2352: 2327: 2317: 2307: 2289: 2283: 2258: 2234: 2231: 2205: 2185: 2147: 2127: 2106: 2091: 2088: 2052: 2047: 2026: 2016: 2013: 1847:13: last two digits are {00, 13, 30, 43}. 1666:Divisibility Rules in Lotsa Various Bases 1548:-- don't just assume something is true... 1000:20:46, 20 February 2007 (UTC) From: Dann 762:then the original number is divisible by 3668:Simpler 25 Rule & missing 29 example 2914: 2780: 2697: 1981:Divisibility testing via modulo division 1892:Base 11 (a prime base, for comparison): 1850:5: sum of all digits is a multiple of 5. 1809:F: sum of all digits is a multiple of F. 1806:5: sum of all digits is a multiple of 5. 1803:3: sum of all digits is a multiple of 3. 1756:2: final digit is {0, 2, 4, 6, 8, A, C}. 1689:9: sum of all digits is a multiple of 9. 1686:3: sum of all digits is a multiple of 3. 4120:Knowledge vital articles in Mathematics 3103:10003 | 14 is replaced by 00, 59 by 03 2267:{\displaystyle {\overline {b*L+R}}_{n}} 1544:past twenty-four hours...note to self: 144: 103: 4083:2001:4456:C7E:1400:2405:E396:8C79:2D65 4135:C-Class vital articles in Mathematics 1712:2: final digit is {0, 2, 4, 6, 8, X}. 773:Note that there is no unique triple ( 7: 4061:100% working divisibility rule for 7 3462:Spaced additive operators ("+", "−") 2115:{\displaystyle {\overline {LR}}_{n}} 1299:224: it is divisible by 2 and by 7. 190:This article is within the scope of 3714:Draft:Divisibility coefficient rule 3079:21 is divisible by 7 and so 91 is. 2035:{\displaystyle {\overline {x}}_{n}} 950:Proof using trial divisors For 21: 352:how about "Divisibility test"? -- 133:It is of interest to the following 2852:In other words, in your notation, 1928:3: final digit is {0, 3, 6, 9, C}. 1677:2: final digit is {0, 2, 4, 6, 8}. 1175:More on General Divisibility Rules 1142:Thank you for posting your reply. 993:Confusing Divisibility Rule for 17 954:Choose a number n such that n: --> 14: 4145:Mid-priority mathematics articles 3746:2409:4042:4E9A:7618:0:0:544B:9000 2083:Suppose we want to find a number 2061: 210:Knowledge:WikiProject Mathematics 4115:Knowledge level-5 vital articles 1605:Code boxes all over the place... 1081:17 (or a multiple of 17) or zero 404:W:WikiProject_Mathematics/Proofs 213:Template:WikiProject Mathematics 177: 167: 146: 113: 104: 19: 3855:) 11:15, 16 January 2023 (UTC) 3701:Talk:Divisibility (ring theory) 2122:, where two juxtaposed numbers 1867:2: final digit is {0, 2, 4, 6}. 1797:4: final digit is {0, 4, 8, C}. 1715:3: final digit is {0, 3, 6, 9}. 1296:It is divisible by 2 and by 7. 1026:be a pointless waste-of-time). 860:digits and repeat if necessary. 839:digits) and multiply the first 742:digits) and multiply the first 230:This article has been rated as 4125:C-Class level-5 vital articles 4007:22:12, 13 September 2023 (UTC) 3993:22:10, 13 September 2023 (UTC) 3979:22:05, 13 September 2023 (UTC) 3964:21:58, 13 September 2023 (UTC) 3950:21:51, 13 September 2023 (UTC) 3936:18:53, 13 September 2023 (UTC) 3921:18:47, 13 September 2023 (UTC) 3192:04:51, 12 September 2013 (UTC) 2066: 2055: 1575:18:54, 29 September 2022 (UTC) 878:07:45, 11 November 2014‎ (UTC) 37:nominee, but did not meet the 1: 4091:10:13, 2 September 2024 (UTC) 3729:16:40, 27 November 2020 (UTC) 3514:does not mean to add boldface 3437:29: 261: 1×3 = 3; 3 + 26 = 29 3426:16b: 1168: 11 × 4 + 68 = 112. 3395:16:04, 24 February 2014 (UTC) 3370:13:32, 24 February 2014 (UTC) 3235:It would be cleaner to write: 2840:2880 → 2+8+8+0 = 18 → 1+8 = 9 2053: 1924:Base 15 (odd but not prime): 1005:20:46, 20 February 2007 (UTC) 894:08:58, 11 November 2014 (UTC) 719:) that solves the congruence 204:and see a list of open tasks. 4140:C-Class mathematics articles 4105:Former good article nominees 3865:12:47, 15 January 2023 (UTC) 3843:"such that gcd(n, 2, 5) = 1" 3326:12:43, 18 October 2013 (UTC) 3223:14:40, 18 October 2013 (UTC) 3140:05:36, 21 August 2013‎ (UTC) 3106:03003 | 10 is replaced by 03 2603:It's indistinguishable from 2529: 2473: 2453: 2403: 2383: 2363: 2322: 2302: 2253: 2101: 2073:{\displaystyle x{\pmod {n}}} 2021: 1931:5: final digit is {0, 5, A}. 1838:2: final digit is {0, 2, 4}. 1718:4: final digit is {0, 4, 8}. 1531:3 digits, if you insist. — 1269:Best of luck, George Wunder 1218:03:22, 24 October 2007 (UTC) 1064:Hello Danny. I read a book, 504:basics including a few rules 426:07:57, August 9, 2005 (UTC) 386:15:26, August 7, 2005 (UTC) 3695:Move discussion in progress 3445:16b: 1168: 11×4 + 68 = 112. 3252: 3159:05:51, 21 August 2013 (UTC) 3048:15:09, 20 August 2013 (UTC) 3022:15:02, 20 August 2013 (UTC) 2990:17:42, 5 January 2013 (UTC) 2886:07:58, 5 January 2013 (UTC) 2684:06:44, 5 January 2013 (UTC) 2627:04:45, 2 January 2013 (UTC) 2598:19:11, 1 January 2013 (UTC) 1633:03:17, 8 January 2011 (UTC) 988:03:51, 4 January 2007 (UTC) 507:comprehensive list of rules 443:22:09, August 9, 2005 (UTC) 410:02:44, August 9, 2005 (UTC) 394:22:24, August 7, 2005 (UTC) 356:15:30, August 7, 2005 (UTC) 4161: 3835:20:42, 13 April 2022 (UTC) 3816:18:55, 13 April 2022 (UTC) 3793:15:20, 13 April 2022 (UTC) 3754:06:35, 20 March 2022 (UTC) 3477:19:44, 19 April 2015 (UTC) 3413:05:15, 19 April 2015 (UTC) 3304:20:26, 19 April 2015 (UTC) 3209:Category:Pseudomathematics 3109:0203 | 30 is replaced by 2 2901:when the dividend is large 2006:modular arithmetic article 1660:14:29, 25 April 2011 (UTC) 1639:both to write and to read. 1411:02:34, 30 April 2010 (UTC) 1398:number is divisible by 3. 955:0. (2n)*10+n=21n, 21n/21=n 593:08:23, 27 March 2006 (UTC) 572:00:05, 27 March 2006 (UTC) 558:12:10, 17 April 2007 (UTC) 531:22:26, 17 April 2007 (UTC) 519:12:11, 17 April 2007 (UTC) 490:22:23, 26 March 2006 (UTC) 481:00:59, 25 March 2006 (UTC) 467:22:14, 9 August 2005 (UTC) 435:15:40, 9 August 2005 (UTC) 347:20:23, 6 August 2005 (UTC) 330:23:06, 17 April 2007 (UTC) 315:22:24, 17 April 2007 (UTC) 301:16:53, 17 April 2007 (UTC) 45:. Editors may also seek a 4055:19:25, 23 July 2024 (UTC) 4040:22:58, 29 June 2024 (UTC) 3901:20:05, 28 July 2023 (UTC) 3885:19:24, 24 July 2023 (UTC) 3760:Divisibility rules for 11 3708:13:31, 9 April 2017 (UTC) 3658:12:31, 25 June 2016 (UTC) 3644:06:32, 25 June 2016 (UTC) 3624:22:34, 24 June 2016 (UTC) 3604:22:07, 24 June 2016 (UTC) 3579:16:39, 7 March 2016 (UTC) 3567:16:32, 7 March 2016 (UTC) 3505:) 17:44, January 2, 2016‎ 3277:W (calculated earlier)= W 1870:4: final digit is {0, 4}. 1841:3: final digit is {0, 3}. 1800:8: final digit is {0, 8}. 1759:7: final digit is {0, 7}. 1721:6: final digit is {0, 6}. 1680:5: final digit is {0, 5}. 1599:16:08, 17 July 2010 (UTC) 1309:364: (3 × 2) + 64 = 70." 1244:00:44, 2 March 2008 (UTC) 1210:17:36, 22 June 2007 (UTC) 1131:18:36, 3 March 2007 (UTC) 1113:11:23, 3 March 2007 (UTC) 942:Cut text - trial divisors 934:12:58, 15 June 2006 (UTC) 925:23:51, 14 June 2006 (UTC) 629:23:12, 29 July 2015 (UTC) 289:17:35, 8 April 2009 (UTC) 273:Proofs & techniques 229: 162: 141: 59: 55: 33:Mathematics good articles 3254:Sample calculation step 3112:63 | 20 is replaced by 6 1558:16:49, 27 May 2010 (UTC) 1539:14:56, 27 May 2010 (UTC) 1509:13:43, 27 May 2010 (UTC) 1482:08:27, 27 May 2010 (UTC) 1448:21:27, 25 May 2010 (UTC) 1351:19:56, 25 May 2016 (UTC) 1285:17:55, 28 May 2008 (UTC) 1200:01:55, 9 June 2007 (UTC) 1195:no problem for the font 974:01:46, 2 June 2006 (UTC) 915:23:57, 29 May 2006 (UTC) 808:00:18, 29 May 2006 (UTC) 608:00:10, 29 May 2006 (UTC) 368:00:06, 29 May 2006 (UTC) 338:Split from Divisor topic 236:project's priority scale 4075:Example: 798 (8x2=16) 3429:21: 168: 16 − (8×2) = 0 3211:; I don't know whether 2999:A recently added link: 2611:article. I agree that 2173:{\displaystyle R=b*L+R} 1974:21:26, 7 May 2012 (UTC) 1465:The incorrect version: 270:proofs & techniques 193:WikiProject Mathematics 4110:C-Class vital articles 3482:Divisibility condition 3451:29: 261: 26 + 1×3 = 29 2562: 2488: 2418: 2337: 2268: 2214: 2194: 2174: 2136: 2116: 2074: 2036: 1519:unless the number has 801:are relatively small. 587:multiplicative inverse 3448:21: 168: 16 − 8×2 = 0 2844:could be replaced by 2563: 2489: 2419: 2338: 2269: 2215: 2195: 2175: 2137: 2117: 2075: 2037: 544:comment was added by 120:level-5 vital article 39:good article criteria 3734:Divisible rule of 10 2515: 2435: 2351: 2282: 2230: 2204: 2184: 2146: 2126: 2087: 2046: 2012: 216:mathematics articles 85:Good article nominee 3255: 2917: 1515:Your expression is 658:is an integer with 3312:high powers of two 3253: 3091:1 1 4 5 9 2915: 2558: 2484: 2414: 2333: 2264: 2210: 2190: 2170: 2132: 2112: 2070: 2062: 2054: 2032: 1330:Divisibility by 13 900:Revamping the page 567:There might... -- 185:Mathematics portal 129:content assessment 60:Article milestones 3814: 3795: 3783:comment added by 3756: 3744:comment added by 3642: 3602: 3507: 3493:comment added by 3393: 3288: 3287: 3195: 3178:comment added by 3143: 3126:comment added by 3096: 3082:Another example; 3004:Osculation Method 2978: 2977: 2830: 2829: 2776: 2775: 2532: 2500: 2499: 2476: 2456: 2406: 2386: 2366: 2325: 2305: 2256: 2213:{\displaystyle R} 2193:{\displaystyle b} 2135:{\displaystyle L} 2104: 2024: 1964:comment added by 1611:Divisibility by 7 1589:comment added by 1512: 1495:comment added by 1353: 1341:comment added by 1287: 1275:comment added by 1246: 1234:comment added by 1215:I am the man 2859 1207:Larry R. Holmgren 1128:Larry R. Holmgren 1110:Larry R. Holmgren 1066:Vedic Mathematics 880: 868:comment added by 819: 635:Text from Divisor 631: 619:comment added by 561: 279:comment added by 250: 249: 246: 245: 242: 241: 98: 97: 94: 93: 27:Divisibility rule 4152: 4071: 4021:Divisibilty by 6 3804: 3632: 3592: 3506: 3487: 3401:Examples (April) 3383: 3361: 3360: 3341: 3256: 3229:more complicated 3194: 3172: 3142: 3120: 3094: 2995:Vedic math forum 2918: 2873: 2864:9 1 8 0 8 2 2781: 2698: 2567: 2565: 2564: 2559: 2539: 2538: 2533: 2528: 2520: 2493: 2491: 2490: 2485: 2483: 2482: 2477: 2469: 2463: 2462: 2457: 2449: 2423: 2421: 2420: 2415: 2413: 2412: 2407: 2399: 2393: 2392: 2387: 2379: 2373: 2372: 2367: 2359: 2342: 2340: 2339: 2334: 2332: 2331: 2326: 2318: 2312: 2311: 2306: 2301: 2290: 2273: 2271: 2270: 2265: 2263: 2262: 2257: 2252: 2235: 2223: 2222: 2219: 2217: 2216: 2211: 2199: 2197: 2196: 2191: 2179: 2177: 2176: 2171: 2141: 2139: 2138: 2133: 2121: 2119: 2118: 2113: 2111: 2110: 2105: 2100: 2092: 2079: 2077: 2076: 2071: 2069: 2041: 2039: 2038: 2033: 2031: 2030: 2025: 2017: 1976: 1656: 1651: 1644: 1622: 1616: 1601: 1546:CHECK MATH FIRST 1511: 1489: 1270: 1229: 813: 681:. The rules for 673:is divisible by 539: 291: 218: 217: 214: 211: 208: 187: 182: 181: 171: 164: 163: 158: 150: 143: 126: 117: 116: 109: 108: 100: 80: 57: 23: 16: 4160: 4159: 4155: 4154: 4153: 4151: 4150: 4149: 4095: 4094: 4069: 4063: 4023: 3909: 3872: 3845: 3762: 3736: 3717: 3697: 3689: 3675: 3670: 3586: 3516: 3488: 3484: 3420: 3403: 3342: 3337: 3336: 3333: 3314: 3284: 3280: 3274: 3270: 3242: 3173: 3121: 3088:By Osculation; 3059:The steps are; 2997: 2869: 2521: 2518: 2513: 2512: 2467: 2447: 2433: 2432: 2397: 2377: 2357: 2349: 2348: 2316: 2291: 2288: 2280: 2279: 2236: 2233: 2228: 2227: 2202: 2201: 2182: 2181: 2144: 2143: 2124: 2123: 2093: 2090: 2085: 2084: 2044: 2043: 2015: 2010: 2009: 1990:modulo division 1983: 1959: 1668: 1654: 1647: 1642: 1620: 1614: 1607: 1584: 1490: 1455: 1418: 1395: 1359: 1332: 1252: 1225:14-3=11;...etc 1177: 1136: 1107: 1094: 995: 981: 944: 920:I love 'em. -- 902: 637: 540:—The preceding 474: 464:Oleg Alexandrov 432:Oleg Alexandrov 408:Yannick Gingras 384:Yannick Gingras 376: 354:Yannick Gingras 340: 274: 255: 215: 212: 209: 206: 205: 183: 176: 156: 127:on Knowledge's 124: 114: 76: 12: 11: 5: 4158: 4156: 4148: 4147: 4142: 4137: 4132: 4127: 4122: 4117: 4112: 4107: 4097: 4096: 4062: 4059: 4058: 4057: 4022: 4019: 4018: 4017: 4016: 4015: 4014: 4013: 4012: 4011: 4010: 4009: 3995: 3908: 3905: 3904: 3903: 3871: 3868: 3844: 3841: 3840: 3839: 3838: 3837: 3827:50.238.167.160 3822: 3785:50.238.167.160 3761: 3758: 3735: 3732: 3716: 3711: 3696: 3693: 3688: 3685: 3674: 3671: 3669: 3666: 3665: 3664: 3663: 3662: 3661: 3660: 3585: 3582: 3557: 3556: 3553: 3550: 3523: 3515: 3509: 3483: 3480: 3464: 3463: 3460: 3457: 3453: 3452: 3449: 3446: 3439: 3438: 3431: 3430: 3427: 3419: 3416: 3402: 3399: 3398: 3397: 3339:69.120.155.201 3332: 3329: 3313: 3310: 3309: 3308: 3307: 3306: 3294: 3286: 3285: 3282: 3278: 3275: 3272: 3268: 3264: 3263: 3260: 3251: 3250: 3249: 3248: 3240: 3239: 3238: 3237: 3236: 3233: 3167: 3166: 3162: 3161: 3113: 3110: 3107: 3104: 3095:35 46 9 50 3012:external links 3008: 3007: 2996: 2993: 2980: 2976: 2975: 2972: 2969: 2966: 2962: 2961: 2958: 2955: 2952: 2948: 2947: 2944: 2941: 2938: 2934: 2933: 2930: 2927: 2924: 2922: 2890: 2874:representing − 2866: 2865: 2858: 2857: 2850: 2849: 2842: 2841: 2828: 2827: 2822: 2817: 2812: 2807: 2802: 2797: 2792: 2787: 2774: 2773: 2768: 2763: 2758: 2753: 2748: 2743: 2738: 2733: 2727: 2726: 2723: 2720: 2717: 2714: 2711: 2708: 2705: 2702: 2689: 2688: 2687: 2686: 2670: 2669: 2668: 2667: 2660: 2659: 2658: 2657: 2650: 2649: 2648: 2647: 2640: 2639: 2638: 2637: 2630: 2629: 2617:short division 2609:short division 2605:short division 2587: 2586: 2583: 2580: 2577: 2574: 2557: 2554: 2551: 2548: 2545: 2542: 2537: 2531: 2527: 2524: 2502: 2498: 2497: 2494: 2481: 2475: 2472: 2466: 2461: 2455: 2452: 2446: 2443: 2440: 2428: 2427: 2424: 2411: 2405: 2402: 2396: 2391: 2385: 2382: 2376: 2371: 2365: 2362: 2356: 2344: 2343: 2330: 2324: 2321: 2315: 2310: 2304: 2300: 2297: 2294: 2287: 2275: 2274: 2261: 2255: 2251: 2248: 2245: 2242: 2239: 2209: 2189: 2169: 2166: 2163: 2160: 2157: 2154: 2151: 2131: 2109: 2103: 2099: 2096: 2068: 2065: 2060: 2057: 2051: 2029: 2023: 2020: 1982: 1979: 1978: 1977: 1966:46.117.126.163 1956: 1953: 1950: 1947: 1944: 1941: 1938: 1935: 1932: 1929: 1922: 1921: 1918: 1915: 1912: 1909: 1906: 1903: 1900: 1897: 1890: 1889: 1886: 1883: 1880: 1877: 1874: 1871: 1868: 1861: 1860: 1857: 1854: 1851: 1848: 1845: 1842: 1839: 1832: 1831: 1828: 1825: 1822: 1819: 1816: 1813: 1810: 1807: 1804: 1801: 1798: 1795: 1788: 1787: 1784: 1781: 1778: 1775: 1772: 1769: 1766: 1763: 1760: 1757: 1750: 1749: 1746: 1743: 1740: 1737: 1734: 1731: 1728: 1725: 1722: 1719: 1716: 1713: 1706: 1705: 1702: 1699: 1696: 1693: 1690: 1687: 1684: 1681: 1678: 1667: 1664: 1663: 1662: 1606: 1603: 1582: 1580: 1579: 1578: 1577: 1562: 1561: 1560: 1471: 1470: 1463: 1462: 1454: 1451: 1436: 1435: 1432: 1425: 1417: 1414: 1403:123.243.217.67 1394: 1391: 1390: 1389: 1386: 1382: 1381: 1376: 1375: 1371: 1370: 1358: 1355: 1331: 1328: 1277:69.115.168.129 1251: 1248: 1189: 1176: 1173: 1171: 1168: 1164: 1160: 1155: 1150: 1145: 1141: 1138:Hello, Larry. 1134: 1125: 1124: 1123: 1122: 1105: 1092: 1070: 1069: 1060: 1053: 1048: 1043: 1038: 1034: 1029: 1024: 1019: 1014: 1009: 1002:216.167.135.24 998:216.167.135.24 994: 991: 980: 977: 964: 963: 959: 956: 949: 943: 940: 939: 938: 937: 936: 901: 898: 897: 896: 883: 882: 881: 870:121.244.182.76 861: 647:is written in 643:If an integer 642: 636: 633: 621:137.186.196.22 611: 610: 599: 598: 597: 596: 595: 594: 577: 576: 575: 574: 538: 536: 535: 534: 533: 512: 511: 508: 505: 498: 497: 492: 491: 473: 470: 453: 452: 451: 450: 449: 448: 447: 446: 445: 444: 414: 413: 412: 411: 396: 395: 375: 372: 371: 370: 357: 344:Jay (Histrion) 339: 336: 335: 334: 333: 332: 293: 292: 281:198.189.251.26 271: 268: 265: 254: 253:major revision 251: 248: 247: 244: 243: 240: 239: 228: 222: 221: 219: 202:the discussion 189: 188: 172: 160: 159: 151: 139: 138: 132: 110: 96: 95: 92: 91: 88: 81: 73: 72: 69: 66: 62: 61: 53: 52: 24: 13: 10: 9: 6: 4: 3: 2: 4157: 4146: 4143: 4141: 4138: 4136: 4133: 4131: 4128: 4126: 4123: 4121: 4118: 4116: 4113: 4111: 4108: 4106: 4103: 4102: 4100: 4093: 4092: 4088: 4084: 4079: 4076: 4073: 4072: 4066: 4060: 4056: 4052: 4048: 4044: 4043: 4042: 4041: 4037: 4033: 4028: 4020: 4008: 4004: 4000: 3996: 3994: 3990: 3986: 3982: 3981: 3980: 3976: 3972: 3971:Helpe30 wikis 3967: 3966: 3965: 3961: 3957: 3953: 3952: 3951: 3947: 3943: 3942:Helpe30 wikis 3939: 3938: 3937: 3933: 3929: 3925: 3924: 3923: 3922: 3918: 3914: 3913:Helpe30 wikis 3906: 3902: 3898: 3894: 3889: 3888: 3887: 3886: 3882: 3878: 3869: 3867: 3866: 3862: 3858: 3857:Jonathan 2357 3854: 3850: 3849:Jonathan 2357 3842: 3836: 3832: 3828: 3823: 3819: 3818: 3817: 3812: 3808: 3803: 3798: 3797: 3796: 3794: 3790: 3786: 3782: 3774: 3770: 3766: 3759: 3757: 3755: 3751: 3747: 3743: 3733: 3731: 3730: 3726: 3722: 3715: 3712: 3710: 3709: 3706: 3702: 3694: 3692: 3686: 3684: 3681: 3678: 3672: 3667: 3659: 3655: 3651: 3647: 3646: 3645: 3640: 3636: 3631: 3627: 3626: 3625: 3621: 3617: 3612: 3611:user:Favonian 3608: 3607: 3606: 3605: 3600: 3596: 3591: 3584:My reversions 3583: 3581: 3580: 3577: 3574: 3569: 3568: 3565: 3562: 3554: 3551: 3548: 3544: 3540: 3536: 3532: 3528: 3524: 3521: 3520: 3519: 3513: 3510: 3508: 3504: 3500: 3496: 3492: 3481: 3479: 3478: 3475: 3472: 3467: 3461: 3458: 3455: 3454: 3450: 3447: 3444: 3443: 3442: 3436: 3435: 3434: 3428: 3425: 3424: 3423: 3417: 3415: 3414: 3411: 3408: 3400: 3396: 3391: 3387: 3382: 3378: 3374: 3373: 3372: 3371: 3368: 3365: 3358: 3355: 3352: 3349: 3346: 3340: 3330: 3328: 3327: 3323: 3319: 3311: 3305: 3302: 3299: 3295: 3292: 3291: 3290: 3289: 3276: 3266: 3265: 3261: 3258: 3257: 3246: 3245: 3244: 3243: 3234: 3230: 3226: 3225: 3224: 3221: 3218: 3214: 3210: 3206: 3202: 3198: 3197: 3196: 3193: 3189: 3185: 3181: 3177: 3171: 3164: 3163: 3160: 3157: 3154: 3150: 3146: 3145: 3144: 3141: 3137: 3133: 3129: 3125: 3117: 3114: 3100: 3097: 3092: 3089: 3086: 3083: 3080: 3077: 3073: 3070: 3066: 3063: 3060: 3057: 3054: 3050: 3049: 3046: 3043: 3039: 3036: 3033: 3029: 3024: 3023: 3020: 3017: 3013: 3005: 3002: 3001: 3000: 2994: 2992: 2991: 2987: 2983: 2982:DavidAugustus 2973: 2970: 2967: 2964: 2963: 2959: 2956: 2953: 2950: 2949: 2945: 2942: 2939: 2936: 2935: 2931: 2928: 2925: 2921:Cur. state → 2920: 2919: 2913: 2910: 2904: 2902: 2897: 2893: 2888: 2887: 2884: 2881: 2877: 2872: 2863: 2862: 2861: 2855: 2854: 2853: 2847: 2846: 2845: 2839: 2838: 2837: 2833: 2826: 2823: 2821: 2818: 2816: 2813: 2811: 2808: 2806: 2803: 2801: 2798: 2796: 2793: 2791: 2788: 2786: 2783: 2782: 2779: 2772: 2769: 2767: 2764: 2762: 2759: 2757: 2754: 2752: 2749: 2747: 2744: 2742: 2739: 2737: 2734: 2732: 2729: 2728: 2724: 2721: 2718: 2715: 2712: 2709: 2706: 2703: 2700: 2699: 2696: 2694: 2685: 2681: 2677: 2676:DavidAugustus 2674: 2673: 2672: 2671: 2664: 2663: 2662: 2661: 2654: 2653: 2652: 2651: 2644: 2643: 2642: 2641: 2634: 2633: 2632: 2631: 2628: 2625: 2622: 2618: 2614: 2610: 2606: 2602: 2601: 2600: 2599: 2595: 2591: 2590:DavidAugustus 2584: 2581: 2578: 2575: 2572: 2571: 2570: 2555: 2552: 2549: 2546: 2543: 2540: 2535: 2525: 2522: 2509: 2505: 2495: 2479: 2470: 2464: 2459: 2450: 2444: 2441: 2438: 2430: 2429: 2425: 2409: 2400: 2394: 2389: 2380: 2374: 2369: 2360: 2354: 2346: 2345: 2328: 2319: 2313: 2308: 2298: 2295: 2292: 2285: 2277: 2276: 2259: 2249: 2246: 2243: 2240: 2237: 2225: 2224: 2221: 2207: 2187: 2180:, and where 2167: 2164: 2161: 2158: 2155: 2152: 2149: 2129: 2107: 2097: 2094: 2081: 2063: 2058: 2049: 2027: 2018: 2007: 2002: 1998: 1995: 1991: 1986: 1980: 1975: 1971: 1967: 1963: 1957: 1954: 1951: 1948: 1945: 1942: 1939: 1936: 1933: 1930: 1927: 1926: 1925: 1919: 1916: 1913: 1910: 1907: 1904: 1901: 1898: 1895: 1894: 1893: 1887: 1884: 1881: 1878: 1875: 1872: 1869: 1866: 1865: 1864: 1858: 1855: 1852: 1849: 1846: 1843: 1840: 1837: 1836: 1835: 1829: 1826: 1823: 1820: 1817: 1814: 1811: 1808: 1805: 1802: 1799: 1796: 1793: 1792: 1791: 1785: 1782: 1779: 1776: 1773: 1770: 1767: 1764: 1761: 1758: 1755: 1754: 1753: 1747: 1744: 1741: 1738: 1735: 1732: 1729: 1726: 1723: 1720: 1717: 1714: 1711: 1710: 1709: 1703: 1700: 1697: 1694: 1691: 1688: 1685: 1682: 1679: 1676: 1675: 1674: 1671: 1665: 1661: 1658: 1657: 1652: 1650: 1645: 1637: 1636: 1635: 1634: 1630: 1626: 1619: 1612: 1604: 1602: 1600: 1596: 1592: 1591:87.236.232.97 1588: 1576: 1572: 1568: 1563: 1559: 1555: 1551: 1547: 1542: 1541: 1540: 1537: 1534: 1530: 1526: 1522: 1518: 1514: 1513: 1510: 1506: 1502: 1498: 1494: 1486: 1485: 1484: 1483: 1480: 1477: 1468: 1467: 1466: 1460: 1459: 1458: 1452: 1450: 1449: 1445: 1441: 1433: 1430: 1429:prime numbers 1426: 1423: 1422: 1421: 1415: 1413: 1412: 1408: 1404: 1399: 1393:The Algorithm 1392: 1384: 1383: 1378: 1377: 1373: 1372: 1368: 1367: 1366: 1362: 1356: 1354: 1352: 1348: 1344: 1340: 1329: 1327: 1324: 1322: 1318: 1313: 1310: 1307: 1305: 1300: 1297: 1294: 1291: 1288: 1286: 1282: 1278: 1274: 1267: 1263: 1260: 1256: 1249: 1247: 1245: 1241: 1237: 1236:81.203.145.59 1233: 1228: 1221: 1219: 1216: 1212: 1211: 1208: 1202: 1201: 1198: 1193: 1190: 1187: 1183: 1180: 1174: 1172: 1169: 1166: 1162: 1158: 1153: 1148: 1143: 1139: 1133: 1132: 1129: 1119: 1118: 1117: 1116: 1115: 1114: 1111: 1104: 1102: 1098: 1091: 1088: 1084: 1082: 1078: 1074: 1067: 1063: 1062: 1061: 1058: 1055: 1051: 1046: 1041: 1036: 1032: 1027: 1022: 1017: 1012: 1007: 1006: 1003: 999: 992: 990: 989: 986: 978: 976: 975: 972: 968: 960: 957: 953: 952: 951: 947: 941: 935: 932: 928: 927: 926: 923: 919: 918: 917: 916: 913: 909: 906: 899: 895: 891: 887: 884: 879: 875: 871: 867: 862: 859: 858: 855: 850: 846: 842: 838: 834: 830: 829: 827: 823: 817: 812: 811: 810: 809: 806: 802: 800: 796: 792: 788: 784: 780: 776: 771: 767: 765: 761: 757: 753: 749: 745: 741: 737: 733: 730: 726: 722: 718: 714: 710: 709: 703: 699: 694: 692: 688: 684: 680: 676: 672: 668: 665: 661: 657: 653: 652: 646: 640: 634: 632: 630: 626: 622: 618: 609: 606: 601: 600: 592: 588: 583: 582: 581: 580: 579: 578: 573: 570: 566: 565: 564: 563: 562: 559: 555: 551: 547: 543: 532: 529: 525: 524: 523: 522: 521: 520: 517: 509: 506: 503: 502: 501: 494: 493: 489: 485: 484: 483: 482: 479: 471: 469: 468: 465: 460: 457: 442: 438: 437: 436: 433: 428: 427: 425: 422:divide 12. — 420: 419: 418: 417: 416: 415: 409: 405: 400: 399: 398: 397: 393: 389: 388: 387: 385: 381: 373: 369: 366: 362: 358: 355: 351: 350: 349: 348: 345: 337: 331: 328: 327: 323: 318: 317: 316: 313: 309: 305: 304: 303: 302: 299: 290: 286: 282: 278: 272: 269: 266: 263: 262: 261: 258: 252: 237: 233: 227: 224: 223: 220: 203: 199: 195: 194: 186: 180: 175: 173: 170: 166: 165: 161: 155: 152: 149: 145: 140: 136: 130: 122: 121: 111: 107: 102: 101: 89: 87: 86: 82: 79: 78:June 19, 2006 75: 74: 70: 67: 64: 63: 58: 54: 50: 49: 44: 40: 36: 35: 34: 28: 25: 22: 18: 17: 4080: 4077: 4074: 4068: 4067: 4064: 4026: 4024: 3910: 3873: 3846: 3811:Yoshi's Eggs 3779:— Preceding 3775: 3771: 3767: 3763: 3740:— Preceding 3737: 3718: 3698: 3690: 3682: 3679: 3676: 3639:Yoshi's Eggs 3599:Yoshi's Eggs 3587: 3573:Arthur Rubin 3570: 3561:Arthur Rubin 3558: 3546: 3542: 3538: 3534: 3530: 3526: 3517: 3489:— Preceding 3485: 3471:Arthur Rubin 3468: 3465: 3459:Unspaced "×" 3440: 3432: 3421: 3418:Calculations 3407:Arthur Rubin 3404: 3390:Yoshi's Eggs 3364:Arthur Rubin 3353: 3347: 3334: 3318:Double sharp 3315: 3298:Arthur Rubin 3232:misaligned.) 3228: 3217:Arthur Rubin 3212: 3204: 3200: 3199:I don't see 3174:— Preceding 3168: 3153:Arthur Rubin 3148: 3122:— Preceding 3118: 3115: 3101: 3098: 3093: 3090: 3087: 3084: 3081: 3078: 3074: 3071: 3067: 3064: 3061: 3058: 3055: 3051: 3042:Arthur Rubin 3034: 3025: 3016:Arthur Rubin 3009: 2998: 2979: 2908: 2905: 2900: 2898: 2894: 2889: 2880:Arthur Rubin 2875: 2870: 2867: 2859: 2851: 2843: 2834: 2831: 2784: 2777: 2730: 2692: 2690: 2621:Arthur Rubin 2612: 2588: 2510: 2506: 2501: 2082: 2003: 1999: 1994:Arthur Rubin 1987: 1984: 1960:— Preceding 1923: 1891: 1862: 1833: 1789: 1751: 1707: 1672: 1669: 1648: 1640: 1608: 1581: 1567:Travis Evans 1545: 1533:Arthur Rubin 1528: 1524: 1520: 1516: 1476:Arthur Rubin 1472: 1464: 1457:My version: 1456: 1437: 1419: 1400: 1396: 1363: 1360: 1343:184.105.71.2 1337:— Preceding 1333: 1325: 1320: 1316: 1314: 1311: 1308: 1303: 1301: 1298: 1295: 1292: 1289: 1268: 1264: 1261: 1257: 1253: 1250:General rule 1222: 1213: 1203: 1194: 1191: 1188: 1184: 1181: 1178: 1170: 1167: 1163: 1159: 1154: 1149: 1144: 1140: 1137: 1126: 1108: 1100: 1097:Example two: 1096: 1095: 1087:The process: 1086: 1085: 1080: 1076: 1073:Example one: 1072: 1071: 1065: 1059: 1056: 1052: 1047: 1042: 1037: 1033: 1028: 1023: 1018: 1013: 1008: 996: 982: 979:overall form 969: 965: 948: 945: 910: 907: 903: 864:— Preceding 856: 853: 852: 848: 844: 840: 836: 832: 825: 821: 803: 798: 794: 790: 786: 782: 778: 774: 772: 768: 763: 759: 755: 751: 747: 743: 739: 735: 731: 724: 720: 716: 712: 707: 701: 697: 695: 690: 686: 682: 678: 674: 670: 666: 659: 655: 650: 644: 641: 638: 615:— Preceding 612: 537: 513: 499: 475: 461: 458: 454: 377: 341: 320: 294: 259: 256: 232:Mid-priority 231: 191: 157:Mid‑priority 135:WikiProjects 118: 84: 83: 48:reassessment 46: 31: 30: 26: 3802:Black Yoshi 3630:Black Yoshi 3590:Black Yoshi 3495:47.20.7.225 3381:Black Yoshi 3069:5 * 1 + 9 2856:2 8 8 0 , 1625:Black Yoshi 1585:—Preceding 1550:Black Yoshi 1497:Black Yoshi 1491:—Preceding 1440:Black Yoshi 1401:- Ricketts 1271:—Preceding 1230:—Preceding 835:(which has 738:(which has 322:Cryptic C62 308:Cryptic C62 275:—Preceding 207:Mathematics 198:mathematics 154:Mathematics 43:renominated 4099:Categories 4081:63/7=9 ✔️ 4078:79-16=63 3687:29 Example 2042:signifies 1380:mentioned. 1227:Donquimico 1197:Donquimico 847:digits by 750:digits by 591:ciphergoth 488:ciphergoth 441:ciphergoth 424:ciphergoth 392:ciphergoth 380:User:Linas 90:Not listed 3512:WP:BEBOLD 3267:V = v + W 3180:Raajesh23 3149:published 3128:Raajesh23 3075:91 -: --> 3068:14 =: --> 3028:Raajesh23 1790:Base 16: 1752:Base 14: 1708:Base 12: 1673:Base 10: 962:addition. 123:is rated 3781:unsigned 3742:unsigned 3705:RMCD bot 3650:Favonian 3533:× 2 (or 3503:contribs 3491:unsigned 3377:edit war 3351:contribs 3331:Examples 3188:contribs 3176:unsigned 3136:contribs 3124:unsigned 3038:contribs 2937:0,3,6,9 2008:, where 1962:unsigned 1863:Base 8: 1834:Base 6: 1587:unsigned 1529:at least 1505:contribs 1493:unsigned 1339:unsigned 1273:unsigned 1232:unsigned 1101:Ekādhika 1077:Ekādhika 985:Rmrfstar 922:Rmrfstar 866:unsigned 669:), then 617:unsigned 569:Rmrfstar 554:contribs 542:unsigned 528:Rmrfstar 478:Rmrfstar 312:Rmrfstar 277:unsigned 3985:MrOllie 3928:MrOllie 3877:MrOllie 3673:25 Rule 3271:+ 5 × W 2923:Input ↓ 1618:cleanup 1525:at most 1521:exactly 1290:Quote: 822:removed 685:=3 and 402:but on 361:Divisor 234:on the 125:C-class 68:Process 4047:Meters 4032:Meters 4027:before 3999:Meters 3956:Meters 3893:Meters 3807:Yoshi! 3635:Yoshi! 3616:Dhrm77 3595:Yoshi! 3576:(talk) 3564:(talk) 3545:× 2 + 3537:+ 2 × 3474:(talk) 3410:(talk) 3386:Yoshi! 3367:(talk) 3301:(talk) 3220:(talk) 3201:direct 3156:(talk) 3111:=: --> 3108:=: --> 3105:=: --> 3102:=: --> 3085:11459 3045:(talk) 3040:) ) — 3019:(talk) 2965:2,5,8 2951:1,4,7 2916:mod 3 2883:(talk) 2624:(talk) 1536:(talk) 1479:(talk) 1165:Danny 1057:Danny 886:DMacks 816:WP:TPO 789:where 693:=10). 654:, and 546:Kmhkmh 516:Kmhkmh 510:proofs 374:Proofs 298:Kmhkmh 131:scale. 71:Result 29:was a 4045:done 3357:WHOIS 3062:9 1 2909:baked 2868:with 2636:time. 1517:wrong 826:added 706:base 662:≡ 1 ( 649:base 112:This 4087:talk 4051:talk 4036:talk 4003:talk 3989:talk 3975:talk 3960:talk 3946:talk 3932:talk 3917:talk 3897:talk 3881:talk 3861:talk 3853:talk 3831:talk 3789:talk 3750:talk 3725:talk 3721:Kvng 3654:talk 3620:talk 3499:talk 3433:but 3345:talk 3322:talk 3213:this 3205:some 3184:talk 3132:talk 3032:talk 2986:talk 2778:and 2680:talk 2613:that 2594:talk 2142:and 1970:talk 1649:Soap 1629:talk 1595:talk 1571:talk 1554:talk 1501:talk 1444:talk 1407:talk 1347:talk 1293:"14 1281:talk 1240:talk 971:Walt 931:Walt 912:Walt 890:talk 874:talk 805:Walt 797:and 625:talk 605:Walt 550:talk 365:Walt 326:Talk 285:talk 65:Date 3076:21 3065:14 2059:mod 1323:". 828:): 766:. 729:mod 704:in 664:mod 226:Mid 4101:: 4089:) 4053:) 4038:) 4005:) 3991:) 3977:) 3962:) 3948:) 3934:) 3919:) 3899:) 3883:) 3863:) 3833:) 3809:| 3791:) 3752:) 3727:) 3656:) 3637:| 3622:) 3597:| 3529:+ 3501:• 3388:| 3324:) 3262:w 3190:) 3186:• 3138:) 3134:• 2988:) 2974:1 2960:0 2946:2 2932:2 2825:63 2820:56 2815:49 2810:42 2805:35 2800:28 2795:21 2790:14 2771:63 2766:56 2761:49 2756:42 2751:35 2746:28 2741:21 2736:14 2725:9 2722:8 2719:7 2716:6 2713:5 2710:4 2707:3 2704:2 2701:1 2693:do 2682:) 2596:) 2553:∗ 2547:− 2541:≡ 2530:¯ 2474:¯ 2454:¯ 2445:∗ 2439:≡ 2431:* 2404:¯ 2384:¯ 2375:∗ 2364:¯ 2355:≡ 2347:* 2323:¯ 2303:¯ 2296:∗ 2286:≡ 2278:* 2254:¯ 2241:∗ 2226:* 2220:. 2159:∗ 2102:¯ 2022:¯ 1972:) 1631:) 1621:}} 1615:{{ 1597:) 1573:) 1556:) 1507:) 1503:• 1453:16 1446:) 1409:) 1349:) 1321:14 1306:. 1283:) 1242:) 1121:OK 892:) 876:) 781:, 777:, 723:≡ 715:, 700:| 627:) 556:) 552:• 514:-- 472:53 324:· 287:) 4085:( 4049:( 4034:( 4001:( 3987:( 3973:( 3958:( 3944:( 3930:( 3915:( 3895:( 3879:( 3859:( 3851:( 3829:( 3813:) 3805:( 3787:( 3748:( 3723:( 3652:( 3641:) 3633:( 3618:( 3601:) 3593:( 3549:. 3547:d 3543:r 3539:t 3535:r 3531:t 3527:r 3497:( 3392:) 3384:( 3359:) 3354:· 3348:· 3343:( 3320:( 3283:0 3281:W 3279:1 3273:0 3269:1 3259:v 3182:( 3130:( 3035:· 3030:( 2984:( 2971:0 2968:2 2957:2 2954:1 2943:1 2940:0 2929:1 2926:0 2876:x 2871:x 2785:7 2731:7 2678:( 2615:( 2592:( 2556:R 2550:2 2544:L 2536:7 2526:R 2523:L 2480:n 2471:R 2465:+ 2460:n 2451:L 2442:b 2410:n 2401:R 2395:+ 2390:n 2381:L 2370:n 2361:b 2329:n 2320:R 2314:+ 2309:n 2299:L 2293:b 2260:n 2250:R 2247:+ 2244:L 2238:b 2208:R 2188:b 2168:R 2165:+ 2162:L 2156:b 2153:= 2150:R 2130:L 2108:n 2098:R 2095:L 2067:) 2064:n 2056:( 2050:x 2028:n 2019:x 1968:( 1655:— 1643:— 1627:( 1593:( 1569:( 1552:( 1499:( 1442:( 1405:( 1345:( 1317:7 1304:7 1279:( 1238:( 888:( 872:( 857:n 854:k 849:k 845:n 843:- 841:m 837:m 833:a 824:/ 799:k 795:n 791:k 787:n 783:d 779:k 775:n 764:d 760:d 756:k 752:k 748:n 746:- 744:m 740:m 736:a 732:d 727:( 725:k 721:b 717:k 713:n 708:b 702:a 698:d 691:b 687:d 683:d 679:d 675:d 671:n 667:d 660:b 656:d 651:b 645:n 623:( 560:. 548:( 283:( 238:. 137::

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