"미디의 정리(Midy's theorem)"의 두 판 사이의 차이
둘러보기로 가기
검색하러 가기
Pythagoras0 (토론 | 기여) (→메타데이터) |
|||
| (사용자 3명의 중간 판 53개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==개요== | |
| − | * [[ | + | * [['142857의 신비' 해설|'142857의 신비']]에서처럼 142857과 같은 수를 적당한 자리마다 쪼개어 더했을때 9가 많이 나타나는 현상에 대한 일반적인 이해 |
| + | ** 1+8=4+5=2+7=9 | ||
| + | ** 142 + 857=999 | ||
| + | ** 428 + 571=999 | ||
| + | ** 285 + 714=999 | ||
| + | ** 857 + 142=999 | ||
| + | ** 571 + 248=999 | ||
| + | ** 712 + 485=999 | ||
| + | ** 14+28+57=99 | ||
| + | ** 42+85+71=198=2*99 | ||
| + | * 여기서 142857과 같은 수란 [[cyclic numbers]] 를 의미한다 | ||
| + | * 대부분의 성질은 [[순환군]] 을 통하여 이해할 수 있다 | ||
| + | * 더 구체적으로는 <math>\mathbb{Z}_p^{x}</math>에서 10^k 꼴의 원소로 생성되는 부분군과 그 coset 의 원소들의 합을 구하는 문제로 이해할 수 있다 | ||
| − | |||
| − | + | ==순환마디의 길이가 2의 배수일때== | |
| − | < | + | ;정리 |
| + | 소수 p에 대하여, 분수 a/p (<math>1\leq a \leq p-1</math>) 를 십진법 전개할 때 얻어지는 순환마디의 길이가 2n 이고, 순환마디가 <math>a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}</math> 라 하자. 다음이 성립한다 | ||
| + | :<math>1\leq i \leq n</math> 에 대하여, <math>a_{i} + a_{i+n}=9</math> | ||
| + | 또한 <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} = 99\cdots 99</math>(n개의 9) 가 성립한다. | ||
| − | + | ;증명 | |
| − | |||
| − | + | 분수 a/p (<math>1\leq a \leq p-1</math>) 를 생각하자. | |
| − | + | <math>g_k \equiv a10^k \pmod p</math> 를 만족시키는 <math>1\leq g_k \leq p-1</math>, <math>(k=0,1,\cdots,2n-1)</math>를 정의하자. <math>g_0=a</math> 이다. | |
| − | + | 분수 a/p의 순환마디의 길이가 2n이면, <math>10^n \equiv -1 \pmod p</math> 가 성립하므로, <math>g_n=p-a</math> 임을 안다. | |
| − | ( | + | <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_0}{p}=\frac{(g_0+g_n)(10^n-1)}{p}=10^n-1</math> ■ |
| − | + | ||
| − | + | ==예 : 1176470588235294== | |
| − | <math> | + | * p=17 |
| + | * 2/17 = 0.'''1176470588235294'''1176470588235294... | ||
| + | * 11764705 + 88235294 = 99999999 | ||
| + | * 이 경우엔 위의 증명에서 <math>g_k</math> 로 쓰인 수는 2, 3, 13, 11, 8, 12, 1, 10, 15, 14, 4, 6, 9, 5, 16, 7 로 주어진다 | ||
| − | + | ||
| − | + | ||
| − | + | ==순환마디의 길이가 3의 배수일 때== | |
| − | + | ;정리 | |
| + | 소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이고, 순환마디가 <math>a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}a_{2n+1}a_{2n+2}\cdots a_{3n}</math> 라 하자. 다음이 성립한다. | ||
| + | :<math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99</math> | ||
| − | |||
| − | + | ;증명 | |
| − | + | 순환마디의 길이가 3n인 분수 1/p 를 생각하자. | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | + | <math>g_k \equiv 10^k \pmod p</math>, <math>0\leq g_k \leq p-1</math> 라 정의하자. <math>g_0=1</math> 이다. | |
| − | + | <math>g_{2n} \equiv g_n^2 \pmod p</math>, <math>g_n^3 \equiv 1 \pmod p</math> 이므로, <math>g_0+g_n+g_{2n}\equiv 1+g_n+g_n^2=(g_n^3-1)/(g_n-1)\equiv 0 \pmod p</math> 이다. | |
| + | |||
| + | 따라서 <math>g_0+g_n+g_{2n}=p</math> 또는 <math>g_0+g_n+g_{2n}=2p</math>가 성립한다. | ||
| − | + | 그러나 <math>1\leq g_k \leq p-1</math> 이므로 <math>1+g_n+g_{2n}=2p</math>일 수 없다. 따라서 <math>g_0+g_n+g_{2n}=p</math> | |
| − | < | + | <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_{2n}}{p}+\frac{g_{2n} 10^n-g_{0}}{p}=\frac{(g_0+g_n+g_{2n})(10^n-1)}{p}=10^n-1</math> ■ |
| − | * p= | + | * 일반적으로 소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이라고 하자. 분수 a/p (<math>1\leq a \leq p-1</math>) 또는 (p-a)/p (<math>1\leq a \leq p-1</math>) 의 순환소수전개를 생각하자. 둘 중의 하나는 <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99</math> 다른 하나는, <math>a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=2* 99\cdots 99</math> 를 만족한다 |
| − | |||
| − | * | ||
| − | + | ==예 : 052631578947368421== | |
| − | + | * p=19 | |
| + | ** 1/19=0.'''052631578947368421'''052... | ||
| + | ** 52631+578947+368421=999999 | ||
| + | * p=7 | ||
| + | ** 3/7 = 0.4285714286... | ||
| + | ** 42+ 85+71=198 | ||
| + | ** 4/7 = 0.5714285714 | ||
| + | ** 57+14+28=99 | ||
| − | + | ||
| − | + | ||
| − | + | ==역사== | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| + | * 1836년 미디 | ||
* http://www.google.com/search?hl=en&tbs=tl:1&q= | * http://www.google.com/search?hl=en&tbs=tl:1&q= | ||
| − | * [[ | + | * [[수학사 연표]] |
| − | + | ||
| − | + | ||
| − | + | ==관련된 항목들== | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
* [[합동식 (모듈로 modulo 연산)]] | * [[합동식 (모듈로 modulo 연산)]] | ||
| − | + | ||
| − | |||
| − | + | ==매스매티카 파일 및 계산 리소스== | |
| − | * | + | * https://docs.google.com/leaf?id=0B8XXo8Tve1cxYWNhZDQwN2MtNjMyMS00ZDc2LTgzZTQtMzFmMTQxYWFmZWM0&sort=name&layout=list&num=50 |
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | + | ||
| − | + | ==사전 형태의 자료== | |
| − | |||
| − | |||
* http://ko.wikipedia.org/wiki/ | * http://ko.wikipedia.org/wiki/ | ||
* [http://en.wikipedia.org/wiki/Midy%27s_theorem http://en.wikipedia.org/wiki/Midy's_theorem] | * [http://en.wikipedia.org/wiki/Midy%27s_theorem http://en.wikipedia.org/wiki/Midy's_theorem] | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | + | ||
| − | + | ==리뷰, 에세이, 강의노트== | |
| − | + | * [http://www.jstor.org/stable/3621748 A Curious String of Nines] | |
| − | |||
| − | * [http://www.jstor.org/stable/3621748 A Curious String of Nines] | ||
** Hans Liebeck, | ** Hans Liebeck, | ||
** <cite>The Mathematical Gazette</cite>, Vol. 85, No. 504 (Nov., 2001), pp. 431-438 | ** <cite>The Mathematical Gazette</cite>, Vol. 85, No. 504 (Nov., 2001), pp. 431-438 | ||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | |||
| − | + | ==관련논문== | |
| − | * | + | * Lewittes, Joseph. 2006. “Midy’s Theorem for Periodic Decimals”. <em>math/0605182</em> (5월 7). http://arxiv.org/abs/math/0605182 |
| − | * | + | * A. Gupta and B. Sury, [http://www.emis.de/journals/INTEGERS/papers/f19/f19.pdf Decimal expansion of 1/p and subgroup sums], Integers: Electronic Journal Of Combinatorial Number Theory 5 (2005), |
| − | * | + | * Brian D. Ginsberg [http://www.jstor.org/stable/4146879 Midy's (Nearly) Secret Theorem: An Extension after 165 Years], <cite>The College Mathematics Journal</cite>, Vol. 35, No. 1 (Jan., 2004), pp. 26-30 |
| + | * M. Shrader-Frechette, Complementary Rational Numbers, Math. Mag., 51 (1978) 90–98. | ||
| + | * E. Midy, De quelques proprietes des nombres et des fractions decimals periodiques, Nantes, (1836), 21 pages. | ||
| − | + | == 노트 == | |
| − | + | ===말뭉치=== | |
| + | # In 1836, E. Midy proved that if p is a prime greater than 5, and the period of 1/p is 2 1.<ref name="ref_472101d4">[http://math.colgate.edu/~integers/h3/h3.pdf Integers: electronic journal of combinatorial number theory 7 (2007), #a03]</ref> | ||
| + | # (The results of Midy and Ginsberg follow quickly from this).<ref name="ref_472101d4" /> | ||
| + | # E. Midy, De quelques proprietes des nombres et des fractions decimals periodiques, Nantes, (1836), 6.<ref name="ref_472101d4" /> | ||
| + | # This theorem, which we will examine in this paper, be- came known as Midys Theorem due to a pamphlet published by E. Midy in Nantes, France, 1836.<ref name="ref_07047442">[http://sand.truman.edu/~dgarth/klinecapstone.pdf A graphical analysis of midy’s theorem]</ref> | ||
| + | # In 1836 E. Midy published at Nantes, France, a pamphlet of twenty-one pages on some topics in number theory with applications to decimals.<ref name="ref_758326cf">[http://emis.impa.br/EMIS/journals/INTEGERS/papers/h2/h2.pdf Integers: electronic journal of combinatorial number theory 7 (2007), #a02]</ref> | ||
| + | # Note that (iii) explains the dierence between 1/77 where k = 3 and gcd(77, 103 which has the Midy property (1), and 1/803 where k = 4 and gcd(803, 104 which (1) fails.<ref name="ref_758326cf" /> | ||
| + | # Midy himself considered the case of period length 3k, but he focused on the sums ai + ai+k + ai+2k, 1 k, which do not give smooth results.<ref name="ref_758326cf" /> | ||
| + | # Section 2 contains the main results as we study the Midy property in a more general setting.<ref name="ref_758326cf" /> | ||
| + | # This is known as Midy’s theorem.<ref name="ref_1d2f78ea">[https://www.johndcook.com/blog/2018/11/19/midys-theorem/ Midy's theorem: fractions with prime denominator]</ref> | ||
| + | # If k is any divisor of the period of the decimal expansion of a/p (where p is again a prime), then Midy's theorem can be generalised as follows.<ref name="ref_5a20be13">[https://en.wikipedia.org/wiki/Midy%27s_theorem Midy's theorem]</ref> | ||
| + | # To prove the original Midy's theorem, take the special case where h = 2.<ref name="ref_12344c07">[https://handwiki.org/wiki/Midy%27s_theorem Midy's theorem]</ref> | ||
| + | # Now this is even in number therefore 1+8 = 9 which again shows the validity of Midy’s theorem.<ref name="ref_84efc06e">[https://www.geeksforgeeks.org/midys-theorem/ Midy's theorem]</ref> | ||
| + | # Given numerator and denominator, the task is to find if the resultant floating point number follows Midy’s theorem or not.<ref name="ref_84efc06e" /> | ||
| + | # Write( "Midy's theorem holds!" ); } else { Console.<ref name="ref_84efc06e" /> | ||
| + | # This is to provide the necessary machinery for the proof of Midy's theorem, as well as for completeness.<ref name="ref_b7c9355f">[https://digitalcommons.unl.edu/mathfacpub/48/ A THEOREM ON REPEATING DECIMALS]</ref> | ||
| + | # In Extended Midy’s Theorem, the repeating portion is divided into m digits, then their sum is a multiple of 10m - 1.<ref name="ref_cedb9ba6">[https://www.tutorialspoint.com/extended-midy-s-theorem-in-cplusplus Extended Midy's theorem in C++]</ref> | ||
| + | # 2352 9411 7647 Extended Midy's theorem holds!<ref name="ref_cedb9ba6" /> | ||
| + | # We then use our method to give an elementary proof of Midy’s theorem on repeating decimals.<ref name="ref_35609a5a">[https://www.semanticscholar.org/paper/A-SIMPLE-PROOF-OF-MIDY%27S-THEOREM-Hamarsheh-Jaradat/edbcb21bb9dd4e861e5512f819a0702c14c51791 [PDF] A SIMPLE PROOF OF MIDY'S THEOREM]</ref> | ||
| + | # In Extended Midy’s theorem if we divide the repeating portion of a/p into m digits, then their sum is a multiple of 10m -1.<ref name="ref_c9954aeb">[https://www.geeksforgeeks.org/extended-midys-theorem/ Extended Midy's theorem]</ref> | ||
| + | # Write( "Denominator is not prime, " + "thus Extended Midy's theorem " + "is not applicable" ); return ; } int l = str.<ref name="ref_c9954aeb" /> | ||
| + | # Write( "Extended Midy's " + "theorem holds!" ); else Console.<ref name="ref_c9954aeb" /> | ||
| + | # Write( "Extended Midy's " + "theorem doesn't hold!" ); } else if (l % 2 != 0) { Console.<ref name="ref_c9954aeb" /> | ||
| + | # The 2-block property for primes is known as Midy’s theorem (1836).<ref name="ref_8162793e">[https://www.jstor.org/stable/10.4169/002557010x479974 Repeating Decimals: A Period Piece on JSTOR]</ref> | ||
| + | ===소스=== | ||
| + | <references /> | ||
| − | + | == 메타데이터 == | |
| − | * [ | + | ===위키데이터=== |
| − | * | + | * ID : [https://www.wikidata.org/wiki/Q856158 Q856158] |
| − | * | + | ===Spacy 패턴 목록=== |
| + | * [{'LOWER': 'midy'}] | ||
| + | * [{'LOWER': 'midy'}, {'LOWER': "'s"}, {'LEMMA': 'theorem'}] | ||
2021년 2월 22일 (월) 23:25 기준 최신판
개요
- '142857의 신비'에서처럼 142857과 같은 수를 적당한 자리마다 쪼개어 더했을때 9가 많이 나타나는 현상에 대한 일반적인 이해
- 1+8=4+5=2+7=9
- 142 + 857=999
- 428 + 571=999
- 285 + 714=999
- 857 + 142=999
- 571 + 248=999
- 712 + 485=999
- 14+28+57=99
- 42+85+71=198=2*99
- 여기서 142857과 같은 수란 cyclic numbers 를 의미한다
- 대부분의 성질은 순환군 을 통하여 이해할 수 있다
- 더 구체적으로는 \(\mathbb{Z}_p^{x}\)에서 10^k 꼴의 원소로 생성되는 부분군과 그 coset 의 원소들의 합을 구하는 문제로 이해할 수 있다
순환마디의 길이가 2의 배수일때
- 정리
소수 p에 대하여, 분수 a/p (\(1\leq a \leq p-1\)) 를 십진법 전개할 때 얻어지는 순환마디의 길이가 2n 이고, 순환마디가 \(a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}\) 라 하자. 다음이 성립한다 \[1\leq i \leq n\] 에 대하여, \(a_{i} + a_{i+n}=9\) 또한 \(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} = 99\cdots 99\)(n개의 9) 가 성립한다.
- 증명
분수 a/p (\(1\leq a \leq p-1\)) 를 생각하자.
\(g_k \equiv a10^k \pmod p\) 를 만족시키는 \(1\leq g_k \leq p-1\), \((k=0,1,\cdots,2n-1)\)를 정의하자. \(g_0=a\) 이다.
분수 a/p의 순환마디의 길이가 2n이면, \(10^n \equiv -1 \pmod p\) 가 성립하므로, \(g_n=p-a\) 임을 안다.
\(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_0}{p}=\frac{(g_0+g_n)(10^n-1)}{p}=10^n-1\) ■
예 : 1176470588235294
- p=17
- 2/17 = 0.11764705882352941176470588235294...
- 11764705 + 88235294 = 99999999
- 이 경우엔 위의 증명에서 \(g_k\) 로 쓰인 수는 2, 3, 13, 11, 8, 12, 1, 10, 15, 14, 4, 6, 9, 5, 16, 7 로 주어진다
순환마디의 길이가 3의 배수일 때
- 정리
소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이고, 순환마디가 \(a_1a_2\cdots a_{n} a_{n+1}a_{n+2}\cdots a_{2n}a_{2n+1}a_{2n+2}\cdots a_{3n}\) 라 하자. 다음이 성립한다. \[a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99\]
- 증명
순환마디의 길이가 3n인 분수 1/p 를 생각하자.
\(g_k \equiv 10^k \pmod p\), \(0\leq g_k \leq p-1\) 라 정의하자. \(g_0=1\) 이다.
\(g_{2n} \equiv g_n^2 \pmod p\), \(g_n^3 \equiv 1 \pmod p\) 이므로, \(g_0+g_n+g_{2n}\equiv 1+g_n+g_n^2=(g_n^3-1)/(g_n-1)\equiv 0 \pmod p\) 이다.
따라서 \(g_0+g_n+g_{2n}=p\) 또는 \(g_0+g_n+g_{2n}=2p\)가 성립한다.
그러나 \(1\leq g_k \leq p-1\) 이므로 \(1+g_n+g_{2n}=2p\)일 수 없다. 따라서 \(g_0+g_n+g_{2n}=p\)
\(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=\frac{g_0 10^n-g_n}{p} + \frac{g_{n} 10^n-g_{2n}}{p}+\frac{g_{2n} 10^n-g_{0}}{p}=\frac{(g_0+g_n+g_{2n})(10^n-1)}{p}=10^n-1\) ■
- 일반적으로 소수 p에 대하여, 분수 1/p 를 십진법 전개할 때 얻어지는 순환마디의 길이가 3n 이라고 하자. 분수 a/p (\(1\leq a \leq p-1\)) 또는 (p-a)/p (\(1\leq a \leq p-1\)) 의 순환소수전개를 생각하자. 둘 중의 하나는 \(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}= 99\cdots 99\) 다른 하나는, \(a_1a_2\cdots a_{n} + a_{n+1}a_{n+2}\cdots a_{2n} +a_{2n+1}a_{2n+2}\cdots a_{3n}=2* 99\cdots 99\) 를 만족한다
예 : 052631578947368421
- p=19
- 1/19=0.052631578947368421052...
- 52631+578947+368421=999999
- p=7
- 3/7 = 0.4285714286...
- 42+ 85+71=198
- 4/7 = 0.5714285714
- 57+14+28=99
역사
관련된 항목들
매스매티카 파일 및 계산 리소스
사전 형태의 자료
리뷰, 에세이, 강의노트
- A Curious String of Nines
- Hans Liebeck,
- The Mathematical Gazette, Vol. 85, No. 504 (Nov., 2001), pp. 431-438
관련논문
- Lewittes, Joseph. 2006. “Midy’s Theorem for Periodic Decimals”. math/0605182 (5월 7). http://arxiv.org/abs/math/0605182
- A. Gupta and B. Sury, Decimal expansion of 1/p and subgroup sums, Integers: Electronic Journal Of Combinatorial Number Theory 5 (2005),
- Brian D. Ginsberg Midy's (Nearly) Secret Theorem: An Extension after 165 Years, The College Mathematics Journal, Vol. 35, No. 1 (Jan., 2004), pp. 26-30
- M. Shrader-Frechette, Complementary Rational Numbers, Math. Mag., 51 (1978) 90–98.
- E. Midy, De quelques proprietes des nombres et des fractions decimals periodiques, Nantes, (1836), 21 pages.
노트
말뭉치
- In 1836, E. Midy proved that if p is a prime greater than 5, and the period of 1/p is 2 1.[1]
- (The results of Midy and Ginsberg follow quickly from this).[1]
- E. Midy, De quelques proprietes des nombres et des fractions decimals periodiques, Nantes, (1836), 6.[1]
- This theorem, which we will examine in this paper, be- came known as Midys Theorem due to a pamphlet published by E. Midy in Nantes, France, 1836.[2]
- In 1836 E. Midy published at Nantes, France, a pamphlet of twenty-one pages on some topics in number theory with applications to decimals.[3]
- Note that (iii) explains the dierence between 1/77 where k = 3 and gcd(77, 103 which has the Midy property (1), and 1/803 where k = 4 and gcd(803, 104 which (1) fails.[3]
- Midy himself considered the case of period length 3k, but he focused on the sums ai + ai+k + ai+2k, 1 k, which do not give smooth results.[3]
- Section 2 contains the main results as we study the Midy property in a more general setting.[3]
- This is known as Midy’s theorem.[4]
- If k is any divisor of the period of the decimal expansion of a/p (where p is again a prime), then Midy's theorem can be generalised as follows.[5]
- To prove the original Midy's theorem, take the special case where h = 2.[6]
- Now this is even in number therefore 1+8 = 9 which again shows the validity of Midy’s theorem.[7]
- Given numerator and denominator, the task is to find if the resultant floating point number follows Midy’s theorem or not.[7]
- Write( "Midy's theorem holds!" ); } else { Console.[7]
- This is to provide the necessary machinery for the proof of Midy's theorem, as well as for completeness.[8]
- In Extended Midy’s Theorem, the repeating portion is divided into m digits, then their sum is a multiple of 10m - 1.[9]
- 2352 9411 7647 Extended Midy's theorem holds![9]
- We then use our method to give an elementary proof of Midy’s theorem on repeating decimals.[10]
- In Extended Midy’s theorem if we divide the repeating portion of a/p into m digits, then their sum is a multiple of 10m -1.[11]
- Write( "Denominator is not prime, " + "thus Extended Midy's theorem " + "is not applicable" ); return ; } int l = str.[11]
- Write( "Extended Midy's " + "theorem holds!" ); else Console.[11]
- Write( "Extended Midy's " + "theorem doesn't hold!" ); } else if (l % 2 != 0) { Console.[11]
- The 2-block property for primes is known as Midy’s theorem (1836).[12]
소스
- ↑ 1.0 1.1 1.2 Integers: electronic journal of combinatorial number theory 7 (2007), #a03
- ↑ A graphical analysis of midy’s theorem
- ↑ 3.0 3.1 3.2 3.3 Integers: electronic journal of combinatorial number theory 7 (2007), #a02
- ↑ Midy's theorem: fractions with prime denominator
- ↑ Midy's theorem
- ↑ Midy's theorem
- ↑ 7.0 7.1 7.2 Midy's theorem
- ↑ A THEOREM ON REPEATING DECIMALS
- ↑ 9.0 9.1 Extended Midy's theorem in C++
- ↑ [PDF A SIMPLE PROOF OF MIDY'S THEOREM]
- ↑ 11.0 11.1 11.2 11.3 Extended Midy's theorem
- ↑ Repeating Decimals: A Period Piece on JSTOR
메타데이터
위키데이터
- ID : Q856158
Spacy 패턴 목록
- [{'LOWER': 'midy'}]
- [{'LOWER': 'midy'}, {'LOWER': "'s"}, {'LEMMA': 'theorem'}]