TESTS FOR DIVISIBILITY BY PRIME NUMBERS

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1 Bulletin of the Marathwada Mathematical Society Vol. 9, No. 1, June 2008, Pages TESTS FOR DIVISIBILITY BY PRIME NUMBERS G S. Kawale Department of Mathematics, Vasantrao Naik Mahavidyalaya, Aurangabad Dist. Aurangabad (M.S.), INDIA. Abstract The purpose of this paper is to give two general tests for divisibility of a given positive integer by any given prime number and compare their usefulness as far as the number of steps required for testing the divisibility is concerned. These tests include the tests developed in [2, 3, 4] as particular cases. 1 INTRODUCTION During the last five decades several authors have, developed many tests of divisibility of a given positive integer by some suitable primes including 19,23,7 [1, 2, 3, 4, 5] In [1], the author discussed such tests in detail, but nowhere he tried to give a mathematical justification for these tests. In [4], the author has produced a mathematical justification for such tests. In this paper a short review of two different tests of divisibility of a given positive integer by any given prime number considered in [4], is taken and it is shown that the tests given in [2, 3, 4] are the particular cases. Further a comparison between these tests is discussed by taking into consideration the number of steps required for testing the divisibility of a given number, provided no calculator or a computer is used for doing the calculations. 2 NOTATIONS AND PRELIMINARIES Let N be the set of positive integers and S be the subset of N defined by : S = {z : (z,) = 1 }, where (b, c) represents the g.c.d. of b and c. Definition 2.1 An integer b is said to be divisible by an integer a, (not zero), if there is an integer x such that b = ax, and we write a b. For example 7 28, etc. Notation: If n N, consists of r + 1 digits a 0,a 1,...,a r, then n can be expressed as n = r.a r + r 1 a r a 1 + a 0. We write this number as follows. n = a r a r 1...a 1 a 0. (2.1) 55

2 56 G S. Kawale If n stands for the integer, then the relation between n and n is as follows. n = a r a r 1...a 2 a 1 (2.2) n = n a 0, (2.3) For example if n = 3512,n = 351, then 351 = , i..e. the relation (2.3) holds. Positive and Negative oscillators: Let d S, then there exists one and only one integer c {1,3,7,9} such that the product cd ends with 9. When this 9 is dropped the remaining integer when added to 1 gives us another integer say P(d). This integer P(d) is called the positive osculator of d. If d S then it can be observed that P(d) = c.d + 1 (2.4) or c.d =.P(d) 1 (2.5) For example, If d = 23, then P(d) is obtained by considering the product 23 3 = 69, and dropping 9 and adding 1. Thus P(23) = = 7. i.e. 7 is the positive osculator of 23, and can be obtained by using the above mentioned formula (2.4). Thus P(23) = = 7. Negative osculator of d is defined as the integer N(d) given by N(d) = d P(d). (2.6) For example, N(23) = 23 P(23) = 23 7 = 16. Now if d S and d = a k a k 1...a 1 a 0 with a 0 = 1, then it can be verified that N(d) = d 1 = a k a k 1...a 1. For example if d = 51 then N(51) = 51 1 = 5. Note that P(51) = 46. Here P(51) > N(51). If d does not end with 1, then N(d) = N(c.d) where c { 1, 3, 7, 9 } is such that the product c.d ends with 1. For example, N(7) = N(7 3) = N(21) = 2. In the following table values of P(d) and N(d) are given for some integers d satisfying (d,) = 1.

3 Tests for Divisibility by Prime Numbers 57 Table : 2.1 d P(d) N(d) Positively osculated numbers: If n is given by n = a r a r 1... a 1 a 0 and d S, then the integer n + P(d)a 0 is called positively osculated number of n (osculated by d), and is denoted by O d (n), where n = a r a r 1... a 1. For example if n = 138 and d = 23, then P(23) = 7 and n = 13. Hence O 23 (138) = = 69. Also O 23 (69) = = 69. Negatively osculated Numbers: Let n and n be as above and d be any integer in S. Then the integer n a 0 N(d) is called negatively osculated number with respect to d. We shall denote this number by O d (n). For example if d = 7 and n = 364, then n = 36,N(d) = N(7) = N(21) = 2 and a 0 = 4. Hence O 7 (364) = = 36 8 = 28. Similarly O 17 (561) = = 51,because, n = 56,N(17) = N(51) = 5 and a 0 = 1. The concepts about the positive and negative osculators and osculated numbers given here are the same as those given in [1], and [4] but explained in a different way. Lastly we state a theorem, which is used in the next section. Theorem 2.1 If a, b, c are non-zero integers such that c ab and (b, c) = 1, then c a. For the proof see [6]. 3 TESTS OF DIVISIBILITY In this section two tests of divisibility are discussed and a comparison is made between them. We begin with the following lemmas. Lemma 3.1 If d S, then d [.P(d) 1]. Proof: The stated result follows immediately by the relation (2.5). As an illustration if d = 23, then P(23) = 7 and.p(d) 1 = 7 1 = 69. Clearly Lemma 3.2 If d S then d [.N(d) + 1]. Proof: We have by using the definition of N(d),.N(d) + 1 = (d P(d)) + 1 = d.p(d) + 1 =.d [.P(d) 1]. The result now follows from Lemma 3.1.

4 58 G S. Kawale As an example if d = 17, then N(d) = d P(d) = = 5, Also N(d) + 1 = = 51, clearly We are now in a position to state and prove the following theorem related to divisibility of a given positive integer n by another positive integer d. Theorem 3.1 (Test 1) : Let n and n be as in (2.1) and (2.2) respectively. For d S, let P(d) and O d (n) represent the positive osculator of d and positively osculated number w.r.t. d respectively. Then d n iff d O d (n). Proof By the definition of O d (n) we have O d (n) = n + P(d)a 0. Hence using (2.3), we have.o d (n) = n + P(d)a 0 i.e.,.o d (n) = n a 0 +.P(d)a 0 = n + a 0 (.P(d) 1). The proof now follows from Lemma 3.1 and Theorem 2.1. Theorem 3.2 (Test 2): Let n and n be given by (2.1) and (2.2) respectively, for d S, let N(d) and O d (n) stand for negative osculator of d and negatively osculated number w.r.t. d. Then d n iff d O d (n). Proof: By definition of O d (n), we have O d (n) = n N(d)a 0. Hence using (2.3), we have,.o d (n) = n N(d)a 0, i.e..o d (n) = n a 0(N(d) + 1). - The proof now follows from Lemma 3.2 and Theorem 2.1. Let us illustrate the tests considered in the above theorems and compare which test is more convenient. Example: Let d = 23 and n = 805. Then P(d) = P(23) = P(69) = 7, and N(d) = 23 7 = 16. Further O d (n) = O 23 (805) = = 115. and O d (n) = O 23(805) = = 0. Since 115 is divisible by 23 (The divisibility of 115 by 23 can be tested by using Test 1), we conclude by Test 1, that 805 is divisible by 23. Further 0 is divisible by 23 and hence 805 is divisible by 23 by Test 2. Here Test 2 is more useful than Test 1, since within one step the divisibility of 805 by 23 is tested by the use of Test 2, while Test 1 requires application of Test 1 two times. The tests given above can also be used to test the divisibility of a given integer by a number which is a product of two or more primes. The following example illustrates this point. Let n = and d 1 = 119,d 2 = 201. Here d 1 = 17 7 and d 2 = Both d 1 and d 2 are prdoucts of two primes. We observe that P(d 1 ) = = 12 and N(d 2 ) = 20. Also N(d 1 ) = = 7 and P(d 2 ) = = 181.

5 Tests for Divisibility by Prime Numbers 59 lt is clear that the efforts required to multiply a number by 12 are less than the efforts required to multiply the same number by 7, provided no calculator or computer is used in doing some intermediate calculations. Hence as far as divisibility of by d 1 is concerned, Test 1 is more convenient than Test 2. Since n is a five digit number we have to apply Test 1, three times. The procedure of testing the divisibility is given below (7 is dropped) +84 (7 12 is added) (9 is dropped) +8 (9 12 is added) 83 3 (3 is dropped) +36 (3 12 is added) 119 Now 119, is divisible by 119. Hence is also divisible by 119. i.e is divisible by 7 as well as by 17. For d 2 = 201,N(d 2 ) = 20 < 181 = P(d 2 ). Hence an application of Test 2 is more convenient than the application of Test 1, though the number of steps required for both the tests is the same. The procedure of testing the divisibility is as follows (drop 7 ) 140 (7 20 is subtracted) (5 is dropped) 0 (5 20 is subtracted) 60 3 (3 is dropped) 60 (3 20 is subtracted) 00 But 0 is divisible by 201. Hence is also divisible by 201,i.e is divisible by 67 as well as by 3. By taking d 1 = 7 17 = 119 and P(d 1 ) = 12 and applying the procedure of osculation as illustrated in the above example for the integer n = 1652, we get the following steps.

6 60 G S. Kawale (drop 2 ) +24 (2 12 is added) 18 9 (9 is dropped) +8 (9 12 is added) 12 6 (6 is dropped) +72 (6 12 is added) is divisible by 7 but not by 17. Hence the given integer 1652 is divisible by 7 but not by 17. Similarly if n = 5338, we observe that 5338 is divisible by 17 but not by 7. Remark 3.1:The tests of divisibility considered in [2, 3, 4] are particular cases of the Theorems 3.1 and 3.2 proved above. For example the theorem of [2] is a particular case of Theorem 3.1, if we take d = 23 and n = abcd. The multiplying factor 7 considered in [2] for the divisibility of n by 23 is the same as the positive osculator of 23. Remark 3.2: For d S, if p(d) < N(d) then Test 1 is more convenient than Test 2. As particular cases of Theorem 3.1 and Theorem 3.2, we state below two results for the divisibility of four digit number n by 29 and 31. Theorem 3.3 Let n = abcd. Then 29 n iff 29 (abc + 3d). Proof: n = abcd = (abc) + d = (abc + 3d) 29d. This relation along with Theorem 2.1 shows that 29 n iff 29 (abc + 3d). Hence the proof. Theorem 3.4 Let n = abcd. Then 31 abcd iff 31 (abc 3d). Proof: n = abcd = (abc) + d = (abc 3d) + 31d. Hence by theorem 2.1, we conclude that 31 n iff 31 (abc 3d). The procedure of testing the divisibility of an integer n, by a given d S can also be explained with the help of an algorithm. For example if d = 29 and n is given by (2.1) then the algorithm is as follows. Step 1 : Let n = a r a r 1...a 2 a 1 a 0 and n = a r a r 1... a 2 a 1. Step 2 : Multiply a 0 by 3 and add the product to n and call the result as n 1. Step 3 : Take n and n, as n 1 and n 1. Step 4 : Go to step 2. Step 5 : If n, at the end of step 3 is divisible by 29, then conclude that the given number n is also divisible by 29 and stop. 4 ACKNOWLEDGMENT I am thankful to Dr. S. R. Joshi for his guidance during the preparation of the paper. I am also thankful to the referee for making some valuable suggestions, which helped me to modify the paper.

7 Tests for Divisibility by Prime Numbers 61 References [1] J.S.S., Bharati - Krishna Tirtha, Vedic Mathematics, Motilal Banarasidas Ideological Publishers Delhi - 7, [2] N.S. Birajdar, Tests for divisibility by some prime numbers, Bull.Marathwada Math.Soc., 2,(2001). [3] M.Humphreys,Tests for divisibility by 19, Math. Gaz., 82, (1998), [4] S.R. Joshi, On tests of divisibility,mathematics Education, (Siwan), 15, 2, (1981).B 1 B 2. [5] J. Kashangaki,A test for divisibility by 7, Math. Gaz., (2002). [6] I.Niven and H.S. Zukermann, An Introduction to the Theory of Numbers, (Third Edition), Wiley Eastern limited, New Delhi 1976, Chapter 1.

NONLINEAR CONGRUENCES IN THE THEORY OF NUMBERS

Bulletin of the Marathwada Mathematical Society Vol. 12, No. 2, December 2011, Pages 24-31 NONLINEAR CONGRUENCES IN THE THEORY OF NUMBERS S.R.Joshi 8, Karmayog, Tarak Housing Society, Opp to Ramkrishna