CHAPTER 3. Number Theory


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1 CHAPTER 3 Number Theory 1. Factors or not According to Carl Friedrich Gauss ( ) mathematics is the queen of sciences and number theory is the queen of mathematics, where queen stands for elevated and beautiful. Number theory is mainly the study of the system of integers Z = {0, ±1, ±2,...} and the consequences of the fact that division is not always possible within Z. E.g. 15/3 Z but 15/4 / Z. Let us start with the wellnown long division that has rather surprising consequences. Example (= ) (= ) (= 6 223) What is really being done and what has been achieved? On the right some zeros are filled in that are not written on the left. We now see that = 45. In effect, multiples of 223 were subtracted from until 223 could not be subtracted anymore without running into negative numbers. The mathematical content of long division is the following theorem and long division is just an efficient way of finding q and r in the Division Theorem. 1
2 2 3. NUMBER THEORY Theorem 1.2. Division Theorem. Given integers a, b, b > 0, there exist unique integers q and r such that a = qb + r, 0 r < b The quotient q and the remainder r can be found by repeated subtraction of b. The division algorithm is just an efficient way for computing q and r by repeated subtractions. Example 1.3. Let a = and b = 735. Find nonnegative integers q and r such that = 735q + r and 0 r < 735. Answer: By Long Division we find q = 772 and r = 503. Example 1.4. Let a = and b = Find nonnegative integers q and r such that = 7593q + r and 0 r < Answer: By Long Division we find q = 429 and r = 0. Definition 1.5. An integer a is even if a = 2x for some integer x, i.e., r = 0 in Theorem 1.2. An integer b is odd if b = 2x + 1 for some integer x in Z, i.e., r = 1 in Theorem 1.2. Proposition 1.6. Every integer is either even or odd. The product of two odd integers is odd. Proof. Let a and b be odd integers. According to the definition of odd integer there exist integers x and y such that a = 2x + 1 and b = 2y + 1. Then ab = (2x + 1)(2y + 1) = (2x + 1)(2y) + (2x + 1) 1 = 2(y(2x + 1)) + 2x + 1 = 2(y(2x + 1) + x) + 1. Here z = y(2x+1) + x is an integer by the closure properties of Z, hence ab = 2z +1 is an odd number. Theorem is not rational. Proof. (Aristotle) By way of contradiction assume that 2 = a/b where either a or b is odd. Then 2b 2 = a 2, hence a 2 is even and therefore a is even. This means that a = 2a and substituting 2b 2 = 4a 2. Thus b 2 = 2a 2. This says that b is even, a contradiction. Definition 1.8. (1) Let a, f be integers. We say that f is a factor of a if a = f some integer or a = f a for some integer a. (2) Let a and b be given integers. An integer f is a common factor of a and b if f is a factor of a and a factor of b.
3 1. FACTORS OR NOT 3 (3) The greatest common factor of two integers a and b is the largest among the common factors of a and b. The greatest common factor of a and b is denoted by gcf(a, b). Remar 1.9. In the literature it is much more common to say f divides a (or f divides a evenly) than to say that f is a factor of a. However, the first gets confused with other uses of divide and therefore we will avoid its use. Remar Suppose that a and b are positive integers. Then b is a factor of a if and only if in the Division Theorem a = qb + r the remainder r = 0. Therefore instead of saying b is a factor of a it is also said that b divides a evenly. Factoring Rules. Let a be a positive integer given in base 10 representation. Then the following rules are true. (1) 2 is a factor of a if its units digit is even. (2) 3 is a factor of a if 3 is a factor of the sum of the digits of a. (3) 9 is a factor of a if 9 is a factor of the sum of the digits of a. Recall that gcf(a, b) is the largest among the common factors of a and b. Therefore gcf(a, b) can be found in the following way which is instructive but not very efficient. Remar Finding greatest common factors Let a and b be given integers. (1) List the positive factors of both a and b. (2) List the common (positive) factors of a and b. (3) Pic the largest in the list of common factors. Example Finding greatest common factors. Let a = 12 and b = 28. (1) The positive factors of 12 are 1, 2, 3, 4, 6, 12. (2) The positive factors of 28 are 1, 2, 4, 7, 14, 28. (3) The common factors are 1, 2, 4. (4) The greatest common factor is 4. Example Let a = 240 and b = 330. The positive factors 0f 240 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20,24, 30, 40, 48,60, 80, 120, 240, and the positive factors of 330 are 1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55,66,110,165, 330. The positive common factors of a and b are 1, 2, 3, 5, 6, 10, 15, 30, and the greatest common factor of a and b is 30, gcf(240, 330) = 30. The following lemmas will be used over and over again and should be memorized. Lemma Let a, b, c be integers and suppose that c = a + b. If a number f is a factor of two of the three integers a, b, c, then f is a factor of the third.
4 4 3. NUMBER THEORY Lemma Let a and b be integers. If f is a factor of a, then f is a factor of ab. The Euclidean Algorithm is a very beautiful and efficient way of finding the greatest common factor of two integers. It is bases on the following fact. Lemma Let a and b be integers and a = bq +r for some integers q and r. Then the common factors of a and b are exactly the same as those of b and r. In particular, gcf(a, b) = gcf(b, r). Example Let a = and b = Then a = 16 b + 755, hence Further, 3476 = , so gcf(56371, 3476) = gcf(3476, 755). gcf(56371, 3476) = gcf(3476, 755) = gcf(755, 456). Further, 755 = , 456 = , 299 = , 157 = , hence gcf(56371, 3476) = gcf(755, 456) = gcf(456, 299) = gcf(299, 157) = gcf(157, 122) = gcf(122, 35). It is clear that 35 has the positive factors 1, 5, 7, 35 and of these only 1 is a factor of 122. Hence gcf(56371, 3476) = 1. Theorem Let a and b be (positive) integers. Then there exist integers u and v such that gcf(a, b) = ua + vb. Consequently, every common factor of a and b is a factor of gcf(a, b). The integers u, v and gcf(a, b) can be found efficiently using the Euclidean Algorithm described below. Example Let a = and b = 347. Then gcf(69321, 347) = gcf(a, b) = ua + vb where gcf(a, b) = 1, u = 36 and v = Proof. (1) = (2) 347 = (3) = (1) 1640 (2) 241 = (4) = (2) 1 (3) 106 = (5) = (3) 2 (4) 29 = (6) = (4) 3 (5) 19 = (7) = (5) (6) 10 = (8) = (6) (7) 9 = (9) = (7) (8) 1 =
5 The long divisions used: 1. FACTORS OR NOT = = = = = = = The same process can be done in a short form that only lists the essential data. action a b (1) (2) (3) = (1) 1640 (2) (4) = (2) 1 (3) (5) = (3) 2 (4) (6) = (4) 3 (5) (7)=(5)(6) (8)=(6)(7) (9)=(7)(8) Example Let a = 3675 and b = 791. Then gcf(a, b) = 7 = Proof. action a b (1) (2) (3) = (1) 4 (2) (4) = (2) 1 (3) (5) = (3) 1 (4) (6) = (4) 4 (5) (7) = (5) 1 (6) (8) = (6) 1 (7) (9) = (7) 2 (8) Definition An integer m is a multiple of the integer a if a is a factor of m. An integer m is a common multiple of the integers a and b if m is a multiple of both a and b. The smallest positive common multiple of a and b is the least common multiple of a and b denoted by lcm(a, b).
6 6 3. NUMBER THEORY Given two fractions a/b and c/d, the least common denominator of the fractions is lcm(b, d). Example Let a = 35 and b = 21. Some multiples of a (there are infinitely many of them) are a = 35, 70, 105, 140, 175,... and some multiples of b are b = 21, 42, 63, 84, 105, 126,... We can now see that lcm(35, 21) = 105. It is easy to see that gcf(35, 21) = 7, and we note the curious fact that gcf(a, b) lcm(a, b) = = 735 = ab. Theorem For any postive integers a and b, gcf(a, b) lcm(a, b) = ab. Remar To find the least common multiple of two integers a and b, we use the Euclidean algorithm to find gcf(a, b) and then compute lcm(a, b) = a b. gcf(a,b) Example (1) lcm(240, 330) = = = (See Example 1.13). 30 (2) lcm(56371, 3476) = = = (See Example 1.17). 1 (3) lcm(569321, 347) = = (See Example 1.19). (4) lcm(3675, 791) = = = (See Example 1.20). 7 Example = = The Fundamental Theorem of Arithmetic Given a positive integer, say 113, it can always be factored as 113 = 1 113, in general a = 1 a = a 1. This is an uninteresting trivial factorization. Definition 2.1. An (positive) integer a is composite if it can be factored as a = b c where neither b nor c is equal to one. In other words, the (positive) integer a is composite if it is the product of two positive integers that are both smaller than a. A positive integer that is not composite  can only be factored trivially  is a prime number or simply a prime. Lemma 2.2. Let p be a prime. Then p has exactly two positive factors, namely 1 and p. Let a be any other integer. Then gcf(a, p) = 1 or gcf(a, p) = p depending on whether p is a factor of a or not.
7 2. THE FUNDAMENTAL THEOREM OF ARITHMETIC 7 Example 2.3. The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97 are all prime numbers. The number 123 is composite because 123 = The number 3127 is composite because 3127 = The number 1111 is composite because 1111 = Proposition 2.4. Let n be a positive integer and suppose that n = a b where a and b are again positive integers. If b n, then a n. Proof. If b n and also a > n then a b > n n = n while a b = n by hypothesis. Corollary 2.5. If the positive integer p has no factors that are less than or equal to p then p is a prime number. Proof. If p had a factor larger than n then it would have a factor n and these have all been eliminated. Corollary 2.6. If the positive integer p has no prime factors which are less than or equal to p then p is a prime number. Proof. If p has no prime factors n then there is no composite factor n either, since such a factor would in turn contain a prime factor of n. Theorem 2.7. There exist infinitely many primes. Proof. This is an example of a proof by contradiction. We mae an assumption, then derive an impossibility from it and conclude that the assumption was false. Assume that there are only finitely many primes. List them in increasing order: (2.8) 2, 3, 5, 7, 11, 13, 17,..., P, so P is the largest prime. Study the number n obtained by multiplying all the primes and adding 1: n = p P + 1. Now 2 is not a factor of n since 2 is not a factor of 1; 3 is not a factor of n since 3 is not a factor of 1; 5 is not a factor of n since 5 is not a factor of 1; in general, p is not a factor or n because p is not a factor of 1. So n is either a prime itself different from any one in the list (2.8) or it is a product of primes, each of which is not contained in the list (2.8). This means that the list (2.8) did not contain all the primes after all, so the assumption that there are only finitely many primes is false. Therefore, there are infinitely many primes. Theorem 2.9. Fundamental Theorem of Arithmetic. Every integer a > 1 can be factored uniquely as a = p n 1 1 pn 2 2 pn,
8 8 3. NUMBER THEORY n i 1, p 1 < p 2 < < p, p i primes. We will indicate a proof later. For the moment we observe that every integer a > 1 is a product of primes for the following reason. If a is itself a prime it is considered a product of primes. If a is composite, then a = a 1 a 2 and both a 1 and a 2 are smaller than a. Now loo at a 1 and a 2. Factor them if they are composite, and eep going until you arrive at primes. The process have to come to a halt because integers cannot become smaller and smaller forever. Example = = (11 49) (17 17) = 11 (7 7) = = Before getting into the uniqueness proof, let us see what it can do for us. The uniqueness says that however we arrive at a factorization into a product of primes, the result is always the same. Example Let a = Now suppose that b is a factor of a and a = b c. Let b = p n 1 1 p n and c = p m 1 1 p m, where p 1 < p 2 < < p, and 0 n i, m i, be the prime factorizations of b and c as in Theorem 2.9. Now a = b c = p n 1+m 1 1 p n +m. By the uniqueness of the prime factorization we must have a = = p n 1+m 1 1 p n +m. This means that p 1 = 2, n 1 + m 1 = 3, p 2 = 5, n 2 + m 2 = 2, p 3 = 7, and n 3 + m 3 = 1. We conclude that = 3, n 1 3, n 2 2 and n 3 1. Using this we can list the positive factors of a = = 1400 (there are = 24 of them) as follows: = 1, = 7, = 5, = 35, = 25, = 175, = 2, = 14, = 10, = 70, = 50, = 700, = 4, = 28, = 20, = 140, = 100, = 700, = 8, = 56, = 40, = 280, = 200, = Corollary Let a = p n 1 1 pn 2 2 pn, n i 1, p 1 < p 2 < < p, p i primes. Then the positive factors of a are exactly the integers f = p i 1 1 p i 2 2 p i, where 0 i j n i. There are (n 1 + 1)(n 2 + 1) (n + 1) such factors.
9 2. THE FUNDAMENTAL THEOREM OF ARITHMETIC 9 Example How many positive factors does 120 posses? We factor 120 = There are = 16 factors. Example (1) The number a = has = 315 different positive factors. (2) The number 512 = 2 9 has 10 different positive factors. (3) The number = has 64 different positive factors. We come to a crucial result. Theorem Let a, b, c be integers. If a is a factor of bc and gcf(a, b) = 1, then a is a factor of c. Proof. It is given that gcf(a, b) = 1. By Theorem 1.18 there are integers u, v such that 1 = ua + vb. Multiplying the equation by c we obtain c = uac + vab. Now a is a factor of uac trivially, and c is a factor of ab, so vab by hypothesis. By Lemma 1.14 a is a factor of c. Corollary Let p, p 1, p 2 primes. If p is a factor of p 1 p 2, then p = p 1 or p = p 2. Corollary Let p, p 1,...,p n be primes. If p is a factor of p 1 p 2 p n, then p = p 1 or p = p 2, or... or p = p n. Corollary 2.17 is the result that is needed to prove the uniqueness part of the Fundamental Theorem of Arithmetic. We illustrate the formal proof by an example. Example We find that = Somebody else come up with a mysterious factorization (2.19) = p n 1 1 p n 2 2 p n, p 1 < p 2 < < p. Therefore we have that = p n 1 1 p n 2 2 p n Tae a factor p i of the right hand side which is then also a factor of the left hand side. By Corollary 2.17 p i = 2 or p i = 3 or p i = 5 or p i = 7. Hence (2.19) must loo lie = 2 n 1 3 n 2 5 n 3 7 n 4 and we have = 2 n 1 3 n 2 5 n 3 7 n 4
10 10 3. NUMBER THEORY Now 2 is a factor of the left hand side so it is a factor of the right hand side by Corollary This means that n 1 1 and we can cancel 2 to get = 2 n n 2 5 n 3 7 n 4 The prime 2 is still a factor of the left hand side, so of the right hand side, and we must have n This means we can cancel another 2 and obtain = 2 n n 2 5 n 3 7 n 4. Now there is no 2 on the left so there cannot be a 2 on the right and we conclude that n 1 2 = 0 or n 1 = 2. We next loo at the prime 3 that appears on the left. It must appear on the right also, so n 2 1. Canceling the 3 we have = 3 n n 3 7 n 4. We conclude that n 2 = 1 because no 3 appears on the left and have now = 5 n 3 7 n 4. We have a five on the left, so n 3 1 enabling us to cancel a 5 and get = 5 n n 4. We still have a five on the left, so n enabling us to cancel another 5 to get 5 7 = 5 n n 4. We still have a five on the left, so n enabling us to cancel another 5 to get 7 = 5 n n 4. We finally get n 3 = 3 and n 4 = 1 showing that the mysterious factorization is identical with ours. Proposition Common Factors. Suppose a = p n 1 1 pn 2 2 pn pn is the factorization of the integer a into a product of primes, and b = p m 1 1 p m 2 2 p m p m is the factorization of the integer b into a product of primes, then the common factors of a and b are all integers f = p E 1 1 pe 2 2 pe p E where E i is less than or equal to both n i and m i. Therefore the greatest common factor of a and b is gcf(a, b) = p e 1 1 p e 2 2 p e p e where e i is the lesser of n i and m i. Proposition Multiples. (1) The positive multiples of an integer a are the integers a b where b is any positive integer.
11 2. THE FUNDAMENTAL THEOREM OF ARITHMETIC 11 (2) If a = p n 1 1 pn 2 2 pn pn is the factorization of a into a product of primes, then maing the exponents n i larger produces multiples, also bringing in additional primes produces multiples. (3) The multiples of a are all integers of the form where m i n i and s i 0. m = p m 1 1 pm 2 2 pm p m qs 1 1 qs 2 2 qs 3 3 Proposition Common Multiples. Suppose a = p n 1 1 p n 2 2 p n p n is the factorization of the integer a into a product of primes, and b = p m 1 1 p m 2 2 p m p m is the factorization of the integer b into a product of primes, then the common multiples of a and b are all integers m = p M 1 1 p M 2 2 p M p M qs 1 1 qs 2 2 qs 3 3 where M i is greater or equal to both n i and m i and s i 0. Therefore the least common multiple of a and b is where M i is the greater of n i and m i. lcm(a, b) = p M 1 1 p M 2 2 p M p M Theorem Let a and b be positive integers. Then Example Let and let Then and and hence lcm(a, b) gcf(a, b) = a b. a = b = lcm(a, b) = gcf(a, b) = lcm(a, b) gcf(a, b) = ( )( ) = ab.
12 12 3. NUMBER THEORY 3. Exercises Exercise 3.1. For each of the following values of a and b find the unique integers q and r such that a = qb + r, 0 r < b. (1) a = 723, b = 23. (2) a = 1582, b = 231. (3) a = , b = (4) a = 12345, b = (5) a = 0, b = 13. (6) a = 365, 904, b = Exercise 3.2. Decide whether or not the following numbers have factors of 2, 3, 4, 5 and , , , 29480, , Exercise 3.3. For the following integers a and b list the positive factors and find the greatest common factor gcf(a, b). (1) a = 115, b = 225. (2) a = 1111, b = 333. (3) a = 237, b = (4) a = 17 23, b = Exercise 3.4. Find the greatest common factor of the following integers a and b using the method of Euclid. (1) a = 543, b = 113. (2) a = 4563, b = 981. (3) a = 451, b = 85. (4) a = 1111, b = (5) a = 12345, b = Exercise 3.5. Find the greatest common factor of the following integers a and b and find integers u, v such that gcf(a, b) = ua + vb. (1) a = 543, b = 113. (2) a = 4563, b = 981. (3) a = 451, b = 85. (4) a = 1111, b = (5) a = 12345, b = Exercise 3.6. Find the least common multiple of the following integers a and b by listing common multiples and looing for the least one among these. (1) a = 543, b = 113. (2) a = 56, b = 98. (3) a = 45, b = 85.
13 (4) a = 111, b = 111. (5) a = 123, b = EXERCISES 13 Exercise 3.7. Find the least common multiple of the following integers a and b using Theorem (1) a = 543, b = 112. (2) a = 4563, b = 981. (3) a = 451, b = 85. (4) a = 1111, b = (5) a = 12345, b = Exercise 3.8. Compute the following sums and differences using least common denominators. 28 (1) (2) (3) (4) (5) Exercise 3.9. Which ones of the following numbers are prime and which ones are composite? 211, 373, 453, 565, 463, 371, 637, 343, , Exercise How many positive factors do the following integers possess? 24, 67, 69, 2445, 1111, 999, 5 6, p 7 where p is a prime, 7 n, p n where p is a prime. Exercise Using your calculator find integers q and r such that = q r, 0 r < Exercise Let x and y be (unnown) integers that are related by the equality x 130 = 75y. Why are the following statements true? (1) 5 is a factor of x. (2) 3 is not a factor of x. (3) If y has a factor of 13, then x has a factor of 13.
14 14 3. NUMBER THEORY (4) If y is even, then x is also even. (5) If y is odd, then x is also odd. (6) If gcf(13, y) = 1, then 13 cannot be a factor of x. Exercise Verify the following statements. (1) odd odd = odd. (2) even even = even. (3) even odd = even. (4) odd + odd = even. (5) even + even = even. Exercise Let a = b where b is some unnown positive integer such that gcf(2 5 19, b) = 1. Which of the following numbers are factors of a and which ones are not? 16, 80, 95, 361, 100, 76. Exercise The integer 378 has 16 different positive factors and a partial list of factors is 1, 2, 3, 6, 7, 9, 14, 18, 21, 63, 378. Complete the list. 4. Solution to exercises 3.1 Double chec your answers by computing qb + r. It has to come to be a has factors 3 and 5, has factor 2, has factor 5, has factors 4 and 5, has a factor 2, has a factor of gcf(115, 225) = 5, 115 = 5 23, 225 = gcf(1111, 333) = 1, 1111 = , 333 = gcf(237, 1659) = 237, 237 = 3 79, 1659 = gcf(17 23, 17 19) = gcf(543, 113) = 1; gcf(4563, 981) = 9; gcf(451, 85) = 1; gcf(1111, 11111) = 1; gcf(12345, 1234) = In the following the triples (,, ) are the rows of the tables in the text. (543, 1, 0) 4(113, 0, 1) = (91, 1, 4) (113, 0, 1) (91, 1, 4) = (22, 1, 5) (91, 1, 4) 4 (22, 1, 5) = (3, 5, 24) (22, 1, 5) 7 (3, 5, 24) = (1, 36, 173). Chec: = 1.
15 4. SOLUTION TO EXERCISES 15 (4563, 1, 0) 4(981, 0, 1) = (639, 1, 4) (981, 0, 1) (639, 1, 4) = (342, 1, 5) (639, 1, 4) (342, 1, 5) = (297, 2, 9) (342, 1, 5) (297, 2, 9) = (45, 3, 14) (297, 2, 9) 6 (45, 3, 14) = (27, 20, 93) (45, 3, 14) (27, 20, 93) = (18, 23, 107) (27, 20, 93) (18, 23, 107) = (9, 43, 200) Chec: = 9. (451, 1, 0) 5(85, 0, 1) = (26, 1, 5) (85, 0, 1) 3 (26, 1, 5) = (7, 3, 16) (26, 1, 5) 3 (7, 3, 16) = (5, 10, 53) (7, 3, 16) (5, 10, 53) = (2, 13, 69) (5, 10, 53) 2 (2, 13, 69) = (1, 36, 191) (11111, 1, 0) 10(1111, 0, 1) = (1, 1, 10) (12345, 1, 0) 10(1234, 0, 1) = (5, 1, 10) (1234, 0, 1) 246 (5, 1, 10) = (4, 246, 2461) (5, 1, 10) (4, 246, 2461) = (1, 247, 2471) Chec: = These problems are very time consuming and unpleasant. The lesson is that with more mathematics life gets much easier. The answers given are obtained in the advanced fashion. 3.7 gcf(543, 112) = 1, lcm(543, 112) = / gcf(543, 112) = , gcf(4563, 981) = 9; lcm(4563, 981) = / gcf(4563, 981) = gcf(451, 85) = 1; = gcf(1111, 11111) = 1; lcm(1111, 11111) = = gcf(12345, 1234) = 1; lcm(12345, 1234) = = = = = = = = 211 is prime; 373 = 373 is prime; 453 = is composite; 565 = is composite; 463 = 463 is prime; 371 = 7 53 is composite 637 = is composite; 343 = 7 3 is composite; = is comp[osite; = 4026 = is composite = So q = 269 and r = = (1) 5 is a factor of x because 5 is a factor of 75y and 130.
16 16 3. NUMBER THEORY (2) 3 is not a factor of x because if it were, then 3 would be a factor of x and 75, hence of 75y and x, hence of 130 which is not true. (3) If y has a factor of 13, then x has a factor of 13. True. (4) If y is even, then x is also even. True. (5) If y is odd, then x is also odd. True. (6) If gcf(13, y) = 1, then 13 cannot be a factor of x. True. Assume to the contrary that 13 is a factor of x. Then 13 is a factor of x and 130 hence of 75y. Since gcf(13, y) = 1, we now that 13 is not a factor of y so it would have to be a factor of 75, which is false (1) odd odd = odd. This was done in class. (2) even even = even. Tae two even numbers x and y. Being even they are of the form x = 2x, y = 2y. Hence xy = 2x 2y = 4x y, so xy even has a factor 4 which is more than having a factor 2. (3) even odd = even. Let x be even and y be odd. Then x = 2x and y = 2y + 1. Hence xy = 2x y which is even. (4) odd + odd = even. True. (5) even + even = even. True Let a = b where b is some unnown positive integer such that Since gcf(2 5 19, b) = 1 any integer having only factors 2, 5, and 19 must be a factor of = So 16 : Yes, 80 Yes, 95 Yes, 361 = 19 2 No, 100 Yes, 76 = 4 19 Yes Loo for the complementary factors 378/1 = 378, 378/2 = 189, 378/3 = 126, 378/6 = 63, 378/7 = 54, 378/9 = 42, 378/14 = 27, 378/18 = 21. Hence the complete list is 1, 2, 3, 6, 7, 9, 14, 18, 21, 27, 42, 54, 63, 126, 189, 378.
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