When the sum equals the product

Size: px

Start display at page:

Download "When the sum equals the product"

Penelope Blankenship
7 years ago
Views:

1 Whe the sum equals the product Leo Kurladchik ad Adrzej Nowicki Departmet of Mathematics ad Computer Sciece, Nicholaus Copericus Uiversity, Toruń, Polad mat.ui.toru.pl), mat.ui.toru.pl) 10 November 1998 The atural umbers 1, 2, 3 have a special property: their sum is equal to their product = The umbers 1, 1, 2, 4 possess the same property: Look also at the followig examples: = = = = Let 2 be a atural umber. We are iterested i the sequeces (x 1,..., x ) of atural umbers such that x 1 + x x = x 1 x 2 x ad x 1 x 2 x. We deote by A() the set of all such sequeces. Moreover, we deote by the cardiality of the set A(), that is, is the umber of all elemets of A(). The sequece (2, 2) is a uique elemet of A(2). Thus a(2) = 1. The above examples deal with the cases whe = 3, 4, 5. Now we will prove that these are all the examples of such forms. Theorem 1. a(3) = 1, a(4) = 1, a(5) = 3. Proof. For = 3 we have: x 1 x 2 x 3 = x 1 + x 2 + x 3 3x 3, so x 1 x 2 3. The (x 1, x 2 ) is oe of the pairs (1, 1), (1, 2), (1, 3). But oly the case (x 1, x 2 ) = (1, 2) is good ad i this case x 3 = 3. Hece the set A(3) has oly oe elemet (1, 2, 3). Let = 4. Sice x 1 x 2 x 3 x 4 = x 1 + x 2 + x 3 + x 4 < 4x 4 (the case x 1 = x 2 = x 3 = x 4 is impossible), we have x 1 x 2 x 3 3. The triple (x 1, x 2, x 3 ) is the oe of the triples (1, 1, 1), (1, 1, 2), (1, 1, 3). But oly the case (x 1, x 2, x 3 ) = (1, 1, 2) is good. The set A(4) has oly oe elemet (1, 1, 2, 4). For = 5 we do the same. First we observe that x 1 x 2 x 3 x 4 4. This implies that (x 1, x 2, x 3, x 4 ) is oe of the sequeces (1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3), (1, 1, 1, 4), (1, 1, 2, 2). The sequeces (1, 1, 1, 1) ad (1, 1, 1, 4) are ot good. There is o umber x 5 for such sequeces. From the remaiig sequeces we obtai all the elemets of A(5): (1, 1, 1, 2, 5), (1, 1, 1, 3, 3) ad (1, 1, 2, 2, 2). Oe ca fid aother proof of Theorem 1 i [1] (pp ). 1

2 Theorem 2. For ay 2 the set A() is oempty. Proof. A() cotais the sequece (1, 1,..., 1, 2, ). Let us assume that the sequece (x 1,..., x ) belogs to A(). The x i x x 1 x 2... x = x 1 + x x x + x + + x = x for all i = 1,..., 1. Therefore, the umbers x 1,..., x 1 are smaller tha + 1. But they determie the umber x. Give x 1,..., x 1, we ca fid the umber x from the equality x 1 x = x x. So we get: Theorem 3. For ay 2 the set A() is fiite. Note several facts cocerig A() ad. Theorem 4. If (x 1,..., x ) A() ad 3, the x 1 x 2 x 1 1. Proof. Observe that all the umbers x 1,..., x are ot equal. Suppose that x 1 = = x = x. The x = x ad so x = 1. But 1 < 1 < 2 for 3, so we have a cotradictio. Therefore, hece x 1 x 2 x 1 1. x 1 x 2 x 1 x = x 1 + x x < x, Theorem 5 ([1] 175). For ay atural s there exists a atural such that > s. Proof. Let = 2 2s + 1 ad let x 1 = x 2 = = x 2 = 1. If j {0, 1, 2,..., s}, the we defie: x 1 = 2 j + 1, x = 2 2s j + 1. Every sequece (x 1,..., x ), of the above form, belogs to A() ad the umber of such sequeces equals s + 1. Theorem 6. If (x 1,..., x ) A(), 2, the x x 2. The equality holds oly i the case whe (x 1,..., x ) = (1, 1,..., 1, 2, ). Proof. Let b deote the umber of uit elemets i (x 1,..., x ) A(). Let k be the umber of o-uit elemets i (x 1,..., x ). We will deote the o-uits elemets by y 1 + 1, y 2 + 1,..., y k + 1, respectively, where 1 y 1 y 2 y k. It is clear that k 2, b + k = ad (1) (y 1 + 1)(y 2 + 1)... (y k + 1) = y 1 + y y k + k + b. Let k = 2. The y 1 y 2 = 1, hece y 1 +y =, yieldig x 1 + +x = + y 1 + y 2 2. The equality is oly i the case whe y 1 = 1, y 2 = 1, that is, oly whe (x 1,..., x ) = (1, 1,..., 1, 2, ). 2

3 Let k 3. The the equality (1) implies: y y k y 1 y 2 + y 2 y y k y 1 < (y 1 + 1)(y 2 + 1)... (y k + 1) (y y k ) =. Therefore x x = y y k + < 2. The above theorem was offered as a problem at the Polish Mathematical Olympiad i Theorem 7. Let (x 1,..., x ) A(), 2. Deote by b the umber of uit elemets i (x 1,..., x ). The b 1 [log 2 ]. The equality holds, for example, i the case whe is of the form 2 s s (where s 2) ad (x 1,..., x ) = (1,..., 1, 2, 2,..., 2). }{{} s Proof. Theorem 6 implies that 2 b x 1 x = x x 2. Hece b log 2 (2) = 1 + log 2 ad so b 1 [log 2 ]. The remaiig part of this theorem is obvious. Theorem 8. If is eve ad (x 1,..., x ) A() the the umber x x is divisible by 4. Proof. Suppose that all the umbers x 1,..., x are odd. The we have a eve umber of odd umbers. The sum x x is the a eve umber, ad the product x 1 x is odd. Therefore, at least oe of the umbers x 1,..., x is eve. This meas that the product is eve ad cosequetly the sum is also eve. This implies that we have at least two eve umbers. Thus the product, which is equal to the sum, is divisible by

4 The tables, obtaied by a computer programme, preset the umbers for We see, for example, that a(50) = 4, a(100) = 5. The set A(50) has exactly 4 elemets. We ca prove that every sequece (x 1,..., x 50 ) belogig to A(50) is such that x 1 = x 2 = = x 47 = 1 ad (x 48, x 49, x 50 ) is oe of the triples: (1, 2, 50), (1, 8, 8), (2, 2, 17), (2, 5, 6). The set A(100) has exactly 5 elemets. Every elemet is of the form (x 1,..., x 100 ), where x 1 = x 2 = = x 95 = 1 ad (x 96, x 97, x 98, x 99, x 100 ) is oe of the sequeces: (1, 1, 1, 2, 100), (1, 1, 1, 4, 34), (1, 1, 1, 10, 12), (1, 1, 4, 4, 7), (2, 2, 3, 3, 3). Usig a computer we ca prove that a(1997) = 20, a(1998) = 8, a(1999) = 16, a(2000) = 10. We see, lookig at the above tables, that 24 is the maximal two-digit umber such that = 1. There exist exactly 3 atural three-digit umbers with the property = 1. They are: 114, 174 ad 444. The authors do ot kow the aswer to the followig questio: Is there a atural such that = 1 ad > 444? Now we preset some facts cocerig the case = 1. Theorem 9. Let > 2. If = 1 the 1 is prime. Proof. Suppose that 1 is ot prime. The = ab + 1 for some atural a, b with 2 a b. The the two sequeces (1, 1,..., 1, 2, ) ad (1, 1,..., 1, a + 1, b + 1) are differet ad they belog to A(). As a cosequece of the above theorem we get Theorem 10. If 4 ad = 1, the 2. Note also the followig Theorem 11. If 5 ad = 1, the 3. Proof. Theorem 9 implies that is ot of the form 3k + 1. If = 3k + 2 the the set A() has two differet sequeces (1,..., 1, 2, ) ad (1, 1,..., 1, 2, 2, k + 1). From the above facts we obtai Theorem 12. If = 1 ad 5, the 6. Note also the followig Theorem 13. If = 1 ad > 100, the is of the form either 7k or 7k + 2 or 7k + 3 or 7k + 6 (k 14). 4

5 Proof. The set A() cotais the sequece (1,..., 1, 2, ). If = 7k + 1 or 7k + 4 or 7k + 5, the A() cotais also (1, 1,..., 1, 8, k + 1), (1, 1,..., 1, 2, 4, k + 1), (1, 1,..., 2, 2, 2, k + 1), respectively. Theorem 14. If = 1 ad > 100, the is of the form 30k or 30k +24 (k 3). Proof. Sice 6 (Theorem 12), the umber has oe of the forms 30k, 30k + 6, 30k + 12, 30k + 18 or 30k If = 30k + 6, the 1 is ot prime; a cotradictio with Theorem 9. We kow that the set A() always cotais the sequece (1,..., 1, 2, ). I the case whe = 30k + 12 or = 30k + 18, the set A() cotais also respectively. (1, 1,..., 1, 2, 2, 2, 2, 2k + 1), (1, 1,..., 1, 2, 3, 6k + 4), It follows from the above facts that if > 100 ad = 1 the the umber has oe of the forms 210k, 210k + 24, 210k + 30, 210k + 84, 210k + 90, 210k + 114, 210k or 210k We proved (see Theorem 14) that if = 1 ad 5 the is of the form 30k or 30k + 24 (k 0). We thik, however, that the case = 30k does ot hold. Cojecture 1. If 5 ad = 1, the is of the form 30k Cojecture 2. If > 100 ad = 1, the = 114 or = 174 or = 444. Refereces [1] W. Sierpiński, Number Theory, Part II, (i Polish), PWN, Warszawa

1. MATHEMATICAL INDUCTION

1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1