Just the Factors, Ma am HAROLD B. REITER.

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Just the Factors, Ma am HAROLD B REITER UNIVERSITY OF NORTH CAROLINA CHARLOTTE http://wwwmathunccedu/~hbreiter The purpose of this note is to find and study a method for determining and counting all the positive integer divisors of a positive integer Let N be a given positive integer We say d is a divisor of N and write d N if N/d is a positive integer Thus, for example, 2 6 Denote by D N the set of all positive integer divisors of N For example D 6 = {, 2,, 6} There are four parts to this note In the first part, we count the divisors of a given positive integer N based on its prime factorization In the second part, we construct all the divisors, and in the third part we discuss the geometry of the D N In part four, we discuss applications to contest problems Counting the divisors of N First consider the example N = 72 To find the number of divisors of 72, note that the prime factorization of 72 is given by 72 = 2 2 Each divisor d of 72 must be of the form d = 2 i j where 0 i and 0 j 2 Otherwise, 2 2 /d could not be an integer, by the Fundamental Theorem of Arithmetic (the theorem that guarantees the unique factorization into primes of each positive integer) So there are 4 choices for the exponent i and choices for j Hence there are 4 = 2 divisors of 72 Reasoning similarly, we can see that for any integer N = p e p e 2 2 p e k k, the number of divisors is Π k i=(e i + ) = (e + )(e 2 + ) (e k + ) 2 Constructing the divisors of N In part we found the number of members of D N for any positive integer N In this part, we seek the list of divisors themselves Again we start with the prime factorization of N Suppose N = p e p e 2 2 p e k k If k = 2 the listing is straightforward In this case, we build a table by first listing the powers of p across the top of the table and the powers of p 2 down the side,

thus obtaining an (e + ) (e 2 + ) matrix of divisors Again we use N = 72 as an example Notice that each entry in the table is the product of its row label and its column label 2 0 2 2 2 2 0 2 4 8 6 2 24 2 9 8 6 72 What do we when the number of prime factors of N is more than 2? If k = we can construct e + matrices of divisors, one for each power of p For example, if N = 60 = 2 2 5, we construct one 4 matrix for 5 0 and one for 5 The result is a 4 2 matrix of divisors The two 4 matrices are shown below 5 0 2 0 2 2 2 2 0 2 4 8 6 2 24 2 9 8 6 72 and 5 2 0 2 2 2 2 0 5 0 20 40 5 0 60 20 2 45 90 80 60 For larger values of k we can create multiple copies of the matrix associated with the number N = p e p e 2 2 p e k k The geometry of D N To investigate the geometry of D N, we first explore the relation divides Recall that a b means that a and b are positive integers for which b/a is an integer The relation has several important properties, three of which are crucial to our discussion Reflexive For any positive integer a, a a 2 Antisymmetry For any pair of positive integers a, b, if a b and b a, then a = b Transitivity For any three positive integers, a, b, c, if a b and b c, then a c These properties are easy to prove The first says that each integer is a divisor of itself; that is, a/a is an integer The second says that no two different integers can be divisors of one another This is true since a larger integer can never be a divisor of a smaller one The third property follows from the arithmetic b/a c/b = c/a together with the property that the product of two positive integers is a positive integer Any set S with a relation defined on it that satisfies all three of the properties above is called a partially ordered set, or a poset A branch of discrete mathematics studies the properties of posets, (S, ) Each finite poset has a unique directed graph representation This pictorial representation is what we mean by the geometry of D N To construct the directed graph of a poset (S, ), draw a vertex (dot) for each member of S Then connect 2

two vertices a and b with a directed edge (an arrow) if a b Of course, in our case D N this means we connect a to b if a b The case D 6 is easy to draw: 6 2 Fig The digraph of D 6 The circles at each of the four vertices are called loops They are included as directed edges because each number is a divisor of itself The reader should imagine that all the non-loop edges are upwardly directed The directed edge from to 6 indicates that 6 But since we know that the vertices of D 6 satisfy all three properties required of a poset, we can leave off both (a) the loops, which are implied by the reflexive property, and (b) the edges that are implied by the transitivity condition The slimmed down representation, called the Hasse diagram, is much easier to understand It captures all the essential information without cluttering up the scene The Hasse diagram of D 6 is shown below 6 2 Fig 2, the Hasse diagram of D 6 In general, the Hasse diagram for D N has only those non-loop edges which are

not implied by transitivity, that is, those edges from a to b for which b is a prime number multiple of a The Hasse diagrams of D 72, D 0, and D 60 are shown below 72 24 6 8 2 8 4 6 9 2 Fig The Hasse diagram of D 72 Notice that each prime divisor of 0 can be considered a direction, multiplication by 2 moves us to the left ( ), by moves us upward ( ) and, by 5 moves us to the right ( ) Also note that if a and b are divisors of 0 then a b if and only if there is a sequence of upwardly directed edges starting at a and ending at b For example, 0 and (, ), (, 5), (5, 0) are all directed edges in the digraph of D 0 On the other hand, we say 2 and 5 are incomparable because neither divides the other, and indeed there is no upwardly directed sequence of edges from either one to the other 4

0 6 0 5 2 5 Fig 4 The Hasse diagram of D 0 What would the divisors of 60 look like if we build such a diagram for them? Try to construct it before you look at D 60 60 2 20 0 4 6 0 5 2 5 Fig 5 The Hasse diagram of D 60 Consider the same (lattice/hasse) diagram for the divisors of 20 We can draw this in several ways The first one (Fig 6a) places each divisor of 20 at a level determined by its number of prime divisors The second one (Fig 6b) emphasizes the degrees of freedom These two diagrams are representations of a four dimensional cube, not surprising since the Hasse diagram for D 0 is a three-dimensional cube A mathematical way to say the two digraphs are the same is to say they are isomorphic This means that they have the same number of vertices and the same 5

number of edges and that a correspondence between the vertices also serves as a correspondence between the edges Note that the digraphs in 6a and 6b have the required number of vertices (6) and the required number of edges (2) Can you find an N such that the Hasse diagram of D N is a representation of a five-dimensional cube Such a digraph must have 2 5 = 2 vertices, and 2 2 + 6 = 80 edges 20 0 42 70 05 6 0 5 4 2 5 2 5 7 Fig 6a The Hasse diagram of D 20 6

42 4 20 05 2 70 5 7 6 2 0 5 0 5 Fig 6b The Hasse diagram of D 20 How can the geometry help us do number theory? One way to use the geometry is in the calculation of the GCF and LCM of two members of D N Note that each element d of D N generates a downward cone of divisors and an upward cone of multiples We can denote these cones by F (d) and M(d) respectively Then the GCF (d, e) = max{f (d) F (e)} and LCM(d, e) = min{m(d) M(e)} 4 Problems from competitions The following problems come from MathCounts and the American Mathematics Competitions Find the number of three digit divisors of 600 2 How many positive integers less than 50 have an odd number of positive integer divisors? (The Locker Problem) A high school with 000 lockers and 000 students tries the following experiment All lockers are initially closed Then student number opens all the lockers Then student number 2 closes the even numbered 7

lockers Then student number changes the status of all the lockers numbered with multiples of This continues with each student changing the status of all the lockers which are numbered by multiples of his or her number Which lockers are closed after all the 000 students have done their jobs? 4 If N is the cube of a positive integer, which of the following could be the number of positive integer divisors of N? (A) 200 (B) 20 (C) 202 (D) 20 (E) 204 5 Let N = 69 5 + 5 69 4 + 0 69 + 0 69 2 + 5 69 + How many positive integers are factors of N? (A) (B) 5 (C) 69 (D) 25 (E) 26 6 A teacher rolls four dice and announces both the sum S and the product P Students then try to determine the four dice values a, b, c, and d Find an ordered pair (S, P ) for which there is more than one set of possible values 7 How many of the positive integer divisors of N = 2 5 7 have exactly 2 positive integer divisors? 8 How many ordered pairs (x, y) of positive integers satisfy xy + x + y = 99? 9 How many positive integers less than 400 have exactly 6 positive integer divisors? 0 The product of four distinct positive integers, a, b, c, and d is 8! The numbers also satisfy What is d? ab + a + b = 9 () bc + b + c = 99 (2) How many multiples of 0 have exactly 0 divisors? I am grateful to Howard Groves (Uniter Kingdom) for this problem 2 Find the number of odd divisors of! 8