Talk:Divisor
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In the rule to find n mod 1001 in order to test wheter n is a multiple of 7, or 13, I had written "divide the number in groups of three digits from right to left". It was changed to "left to right", which isn't what I meant. Dividing 1048576 in groups of three digits from left to right would yield 104'857'6, wouldn't it? Army1987 14:38, 2 May 2005 (UTC)
I wrote this generalization:
"Let C, D and B be natural numbers, being B>D, Let C digits be written in a B base. For any D that complies with (B mod D = 1), if the D modulus of the sumatorium of C digits is 0, this implies that C mod D = 0."
...but I am not a professional mathematician. This theorem is likely to be hundreds of years old. ¿Can some expert tell me who discovered this, or if this is considered "common knowledge" or something. I would be most grateful.
Sincerely
Vicente Aceituno
I don't know who discovered it, but I'm sure it's as old as modular arithmetic. In your statement, you can safely drop the assumption B>D, and you can sharpen it a bit by stating: (sum of digits of C) mod D = C mod D. In particular, the sum of digits of C is divisible by D if and only if C is divisible by D. The proof goes like this: if B ≡ 1 (mod D), then Bj ≡ 1j = 1 (mod D), and so
AxelBoldt 18:58, 17 Sep 2004 (UTC)
Thanks a lot. I am not sure if the sharpening breaks something, as what I wanted to do was to explain why divisibility rules work. Particularly the division by 3 rule, that tantalized me since I was a kid. vaceituno 12:0, 19 Sep 2004 (UTC)
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[edit] Dividing by 13
Article currently says, "A number is divisible by 13 iff the result of adding 4 times the last digit to the original number" ... which doesn't look finished. How should it end?
ZagrebMike 23:15, 22 Nov 2004 (UTC)
Yes, i agree. Who can explain?
...Nobody know
I know, I think. 4 is the multiplicative inverse of 10 mod 13. So (10a + b) | 13 iff (40a + 4b) | 13 iff (a + 4b) | 13. So for example you can reduce determining if 12056629 divides 13 to determining if 1205662 + 4*9 = 1205698 divides 13. — ciphergoth 17:40, July 24, 2005 (UTC)
- However, -9 == 4 mod 13 so the existing rule also works, and multiplying by 9 is probably easier than multiplying by 4 - just calculate 1205662 - 90 + 9. — ciphergoth 17:47, July 24, 2005 (UTC)
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- A Vedic algorithm says that since a divisor of 13 times 3 gives 39. 39 is in the nines family so we add one and drop the units digit. Hence, the Ekādhika or multiplier is 4.
Example: Is 12,056,629 divisible by 13? Start on the right, move column-wise to the left. Multiply the first digit by 4, add the product to the next digit to the left, repeat the process on this result and add that result to the next digit to the left. If you end with either 13 or zero, then, yes, the number is divisible by 13, otherwise, no, not divisible. This method is called digit-wise osculation. It follows the vedic ideal of one-line notation. The lower line is the whole necessary notation. Multipling the units digit by 4 and adding the tens and the next digit is mental math.
1 2 0 5 6 6 2 9 13 3 10 22 14 41 38 YES.
Larry R. Holmgren 03:28, 1 March 2007 (UTC)
[edit] 0
From article:
- ...and every integer is a divisor of 0.
Does 0 divide itself?
no it doesn't, because any division by 0 is undefined Xrchz
- No, 0 does divide itself. a|b <==> exists k: ka = b. Obviously for any k, k0 = 0, so the condition is satisfied. — ciphergoth 08:11, August 2, 2005 (UTC)
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- Yeah, but the problem is the "any k" part. If 0|0, then what does 0 ÷ 0 come out to? You can't make a case for any particular answer, so it's considered "indeterminate." (Division by 0 is indeterminate for 0, and undefined for any other number -- either way, it's not allowed.) Perhaps the definition should properly read: "We say m|n for any integers m and n iff there exists a unique integer k such that n = km." Just to clear it up. --Jay (Histrion) 19:53, 4 August 2005 (UTC)
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- Sadly, I can't find my copy of "the theory of numbers" at the moment, but I'm pretty sure that my definition of divisibility is the one they give, not yours. Using yours would make proofs much longer, because you'd have to prove that k was unique all the time. And it's sad to have an exception to the rule "every integer divides zero". It's better to allow an anomaly in the relationship between a|b and b ÷s; a. — ciphergoth 06:57, August 5, 2005 (UTC)
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- I see your point. --Jay (Histrion) 18:28, 6 August 2005 (UTC)
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- I made a survey of mathematics textbooks in my college library. Including Knuth's Concrete Mathematics and Hardy, G. H. and Wright, E. M. An Introduction to the Theory of Numbers. The only one I could find that permitted 0 divides 0 was Ross and Wright's Discrete Mathematics. However, they then added a note in every proof and theorem that m could not be zero. In short the convention is that 0 does not divide 0 and that is what the article should state.
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- This issue is similar to wheather 1 is prime. In many imprecise definitions of primality it is--and historically it was considered to be--but to define 1 to be prime is not useful and force you to add an exception for it to almost every theorem involving prime numbers.
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- If no one has further information on this topic I will edit the definition accordingly. --Elfan 02:26, 7 March 2006 (UTC)
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- I shall have to find my copy of Niven and Zuckerman's "The Theory of Numbers" to contradict you on this. Suffice to say at this stage that the article is now inconsistent with itself, since the definition has not been changed. You'll need to update the definition to reflect this convention... — ciphergoth 16:26, 24 March 2006 (UTC)
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- Fixed consistency. Do Niven and Zuckerman actually use 0|0 anywhere? --Elfan 18:19, 26 March 2006 (UTC)
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[edit] iff?
Being a Dane, I wonder if(f) this is a word missing in my vocabulary :-) [[1]]
- Shall we say iff. It is math lingo. Oleg Alexandrov 15:09, 22 Apr 2005 (UTC)
[edit] Divisibility rule
should I think be reinstated as a separate article - it's not central to the concept of "divisor" and there is enough material for a separate article. — ciphergoth 20:13, July 24, 2005 (UTC)
- I'd certainly vote for that. --Jay (Histrion) 19:33, 4 August 2005 (UTC)
- my vote goes for that too -- Yannick Gingras 17:36, August 6, 2005 (UTC)
- Yes please. The way it stands now, this takes way too much space in the article, and now with the proofs thrown in. By the way, according to Wikipedia:WikiProject_Mathematics/Proofs proofs should not be that proeminent in articles, especially if they are long, as the recently added proof in the divisor article. Oleg Alexandrov 17:52, 6 August 2005 (UTC)
- I agree - a new article "Divisibilty rules", and link to that. Bubba73 18:14, August 6, 2005 (UTC)
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- For some reason, though, I feel "rules" is the wrong term. Would "checks," "shortcuts," or "tricks" be better? Or maybe "criteria," which has a meaning closer to "rules" but seems a bit more appropriate? --Jay (Histrion) 20:08, 6 August 2005 (UTC)
- OK, I moved the material to Divisibility rule, although I'd still like to talk about a name change for the "new" topic. --Jay (Histrion) 20:23, 6 August 2005 (UTC)
I just found this page. I read a book, Vedic Mathematics, 1965, 1978. In chapters 29 and 30, pages 273-295, the author gives an algorithm for divisibility called osculation. It gets complicated as there are several methods: digit-wise, whole number, positive and negative, complex, and multiplex osculation. It can be done mentally or with some notation of the results of an addition and multiplication. It does not require division! The positive osculator plus the negative osculator equals the divisor, P+Q=D. Example: is the number 5,293,240,096 divisible by 139? Round up to fourteen tens. The osculator (multiplier) is 14. Start on the right, moving column-wise to the left. 14x6=86. Then adding to the tens column gives 86+9=93. Osculate 93. 3x14+9+0=51. Next column, osculating 51 gives 1x14+5+0=19. Osculating 19 gives 14x9+1+4=131. Then osculate 131, 14x1+13+2=29. Next osculate 29, 14x9+2+3=131. Next osculate 131, 1x14+13+9=36. Osculate 36, 6x14+3+2=89. Osculate 89, 9x14+8+5=139. Yes, the number is divisible by 139. When we end on either zero or the divisor 139 (or a multiple of 139), then, yes, the number is divisible by 139.
5 2 9 3 2 4 0 0 9 6 139 89 36 131 29 131 19 51 93 YES.
More later. Larry R. Holmgren 03:08, 1 March 2007 (UTC)
[edit] Problems
Looks like some technical problem in red text...I have no idea what to do about it. Gert2 03:35, 4 January 2007 (UTC)
- Nevermind...I fixed it. Gert2 03:39, 4 January 2007 (UTC)
