DOCSLIB.ORG

Generalized Perfect Numbers Connected with Arithmetic Functions

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Generalized Perfect Numbers Connected with Arithmetic Functions Math. Sci. Lett. 3, No. 3, 249-253 (2014) 249 Mathematical Sciences Letters An International Journal http://dx.doi.org/10.12785/msl/030318 Generalized Perfect Numbers Connected with Arithmetic Functions Azizul Hoque∗ and Himashree Kalita Department of Mathematics, Gauhati University, Guwahati, 781014, India Received: 13 Apr. 2014, Revised: 24 May 2014, Accepted: 26 May 2014 Published online: 1 Sep. 2014 Abstract: In this paper, we obtain some new results for perfect numbers and generalized perfect numbers connected with the relationship among arithmetic functions σ, φ and ψ. These arithmetic functions and their compositions play vital role in this work. Keywords: Perfect number, Superperfect number, k-hyperperfect number, Super-hyperperfect number, Arithmetic functions. 1 Introduction been checked without success [2]. Recently T. Goto, Y. Ohno [3], D. Ianucci [5,6], K. G. Hare [4] established For a natural number n, we denote the sum of positive several results on odd perfect numbers. A positive integer σ n is called superperfect number if σ(σ(n)) = 2n. The divisors of n by (n) = ∑d|n d and the sum of proper positive divisors of n by ρ(n) = σ(n) − n. A natural notion of these numbers was introduced by D. ρ Suryanarayana [11] in 1969. Even superperfect numbers number n is called a perfect number if (n) = n or p−1 p equivalently σ(n)= 2n. The first few perfect numbers are are of the form 2 , where 2 − 1 is a Mersenne prime. 6, 28, 496, 8128..... (Sloanes A000396 [13]). Euclid The first few superperfect numbers are 2, 4, 16, 64, 4096, discovered that the first four perfect numbers are 65536, 262144..... It is not known whether there are any generated by the formula 2n−1(2n − 1) about 300 B.C. odd superperfect numbers. An odd superperfect number n [8]. He also noticed that 2n 1 is a prime number for would have to be a square number such that either n or − σ every instance, and in Proposition IX.36 of Elements (n) is divisible by at least three distinct primes [7]. gave the proof, that the discovered formula gives an even perfect number whenever 2n − 1 is prime. In order for 2n − 1 to be a prime, n must itself to be a prime. A A positive integer n is called k-hyperperfect number if p k + 1 k − 1 Mersenne prime is a prime number of the form 2 − 1, σ(n)= n + . On can remark that a number is where p must also be a prime number. Any even perfect k k perfect iff it is 1-hyperperfect. The concept of number n is of the form n = 2p−1(2p − 1), where 2p − 1 is a Mersenne prime. Perfect numbers are intimately k-hyperperfect number was given by Minoli and Bear connected with these primes, since there is a concrete [10] and they also conjecture that there are k-hyperperfect numbers for every k. All hyperperfect numbers less than one-to-one association between even perfect numbers and 11 Mersenne primes. The fact that Euclids formula gives all 10 have been computed by J.S. Craine [9]. Bege and Fogarasi [1] introduced the concept of super-hyperperfect possible even perfect numbers was proved by Euler two millennia after the formula was discovered. There are number. A positive integer n is called super-hyperperfect k + 1 k − 1 only 48 known Mersenne primes (2013 [12]) and hence number if σ(σ(n)) = n + . They have k k only 48 even perfect numbers are known. There is a conjectured that all super-hyperperfect numbers are of the conjecture that there are infinitely many perfect numbers. 3p − 1 The search for new ones is to keep on going by search form 3p−1, where p and are primes. For any 2 program via the Internet; named GIMPS (Great Internet natural number n, Eulers phi-function and Dedekinds Mersenne Prime Search). It is not known if any odd 1 300 Arithmetic function are given by φ(n)= n∏ (1 − ) perfect number exists, although numbers up to 10 have p|n p ∗ Corresponding author e-mail: [email protected] c 2014 NSP Natural Sciences Publishing Cor. 250 A. Hoque, H. Kalita: Generalized Perfect Numbers Connected with... 1 and ψ(n) = n∏ (1 + ) respectively, where p runs Hence the theorem follows. p|n p through the distinct prime divisors of n. Corollary 2.4. If k > 1,1 + 2 + 4 + 8 + ... + k = 2k − 1, where 2k − 1 is a prime and k(2k − 1) is a perfect number, then 2 Main Results (2φ)n ◦(φ ◦ψ)(k(2k−1)) = (φ ◦2ψ)n ◦(φ ◦ψ)(k(2k−1)), Theorem 2.1. If k > 1,1 + 2 + 4 + 8 + ... + k = 2k − 1, where 2k − 1 is a prime and k(2k − 1) is a perfect number, n ≥ 0 then (φ ◦ ψ ◦ σ)(k(2k − 1)) = 2(φ ◦ ψ)(k(2k − 1)). Proof: It is clear that (k,2k − 1)= 1, and k is even and of Theorem 2.5. If k > 1,1 + 2 + 4 + 8 + ... + k = 2k − 1, the form 2n,n ≥ 1. Since both φ and ψ are multiplicative where 2k − 1 is a prime and k(2k − 1) is a perfect number, ψn n−1 2 functions, then (k(2k − 1)) = 3.2 k , n ≥ 1. (φ ◦ ψ ◦ σ)(k(2k − 1)) = φ(ψ(σ(k(2k − 1)))) Proof: It is clear that (k,2k − 1)= 1, and k is even and of m ψ = φ(ψ(2k(2k − 1))) the form 2 , m ≥ 1. Since is multiplicative function, = φ(ψ(2k)ψ(2k − 1)) ψ(k(2k − 1)) = ψ(k)ψ(2k − 1)= 3k2 = φ(2k(1 + 1/2)2k) 2 = 2.2k (1 − 1/2) Thus the result is true for n = 1. = 2k2 (1) Suppose the result is true for n > 1. Then ψn(k(2k − 1)) = 3.2n−1k2 Also Now (φ ◦ ψ)(k(2k − 1)) = φ(ψ(k(2k − 1))) = φ(ψ(k)ψ(2k − 1)) ψn+1(k(2k − 1)) = ψ(3.2n−1k2) = φ(k(1 + 1/2)2k) = ψ(3.22m+n−1), = φ(3k2) = ψ(3)ψ(22m+n−1) = φ(3φ(k2)) = k2 (2) = 4.22m+n−1(1 + 1/2) (1)and (2) give the result. = 3.22m+n = 3.2nk2 Theorem 2.2. If k > 1,1 + 2 + 4 + 8 + ... + k = 2k − 1, where 2k − 1 is a prime and k(2k − 1) is a perfect number, Hence the theorem follows. φ n φ ψ 2 then (2 ) ( ( (k(2k − 1)))) = k , n ≥ 0. Theorem 2.6. For any even perfect number n, Proof: We prove the result by applying induction on n. (φ ◦ σ)(n) (φ ◦ ρ)(n)= By equation ( 2 ), the result is true for n = 0. Lets assume 2 the result is true for any positive integer n > 0. Then p−1 p 2φ n φ ψ k 2k 1 k2 Proof An even perfect number n is of the form 2 (2 − ( ) ( ( ( ( − )))) = p Now, 1), where 2 − 1 is a Mersenne prime. Now (2φ)n+1(φ(ψ(k(2k − 1)))) = 2φ((2φ)n(φ(ψ(k(2k − 1))))) = 2φ(k2) (By hypothesis) (φ ◦ ρ)(n)= φ(ρ(n)) = φ(n)= φ(2p−1(2p − 1)) 2 2 1 = 2k (1 − 1/2)= k = 2p−1(1 − )(2p − 2) Hence the theorem follows. 2 = 2p−1(2p−1 − 1) Theorem 2.3. If k > 1,1 + 2 + 4 + 8 + ... + k = 2k − 1, where 2k − 1 is a prime and k(2k − 1) is a perfect number, Again, then (φ ◦ 2ψ)n ◦ (φ ◦ ψ)(k(2k − 1)) = k2, n ≥ 0. Proof: We prove the result by applying induction on n. By (φ ◦ σ)(n)= φ(σ(n)) = φ(2n)= φ(2p(2p − 1)) equation (2), the result is true for n = 0. Lets assume the 1 result is true for n > 0. Then = 2p(1 − )(2p − 2) (φ ◦ 2ψ)n ◦ (φ ◦ ψ)(k(2k − 1)) = k2 2 p p−1 Now = 2 (2 − 1) φ ψ n+1 φ ψ ( ◦ 2 ) ◦ ( ◦ )(k(2k − 1)) (φ ◦ σ)(n) φ ψ φ ψ n φ ψ Thus (φ ◦ ρ)(n)= = ( ◦ 2 )(( ◦ 2 ) ◦ ( ◦ )(k(2k − 1))) 2 φ ψ 2 = (2 (k )) (By hypothesis) Theorem 2.7. If n = pk−1(pk − p + 1) where p and pk − = φ(3k2)= k2 p + 1 are primes, then n is (p − 1)-hyperperfect number. c 2014 NSP Natural Sciences Publishing Cor. Math. Sci. Lett. 3, No. 3, 249-253 (2014) / www.naturalspublishing.com/Journals.asp 251 Proof: From definition and basic results of the divisor (3) and(5) give, function σ, it follows that: 1 m + p − 1 k−1 k ψ(n) = (1 − )( )σ(n) (6) σ(n)= σ(p )σ(p − p + 1) p2 m + p − 2 pk − 1 (4) and(5) give, = (pk − p + 2) p − 1 (p − 1)(m − 1) pk(pk − p + 1)+ p − 2 φ(n)= ψ(n) (7) = (p + 1)(m + 1) p − 1 (6) and(7)give, p p − 2 = pk−1(pk − p + 1)+ p − 1 p − 1 (p − 1)(m − 1) 1 m + p − 1 p p − 2 φ(n)= (1 − )( )σ(n) = n + 2 p − 1 p − 1 (p + 1)(m + 1) p m + p − 2 This gives, Hence n is (p − 1)-hyperperfect number. p m + 1 m + p − 2 σ(n) = ( )2( )( )φ(n) (8) We generalize conjecture 2 introduced by Antal Bege and p − 1 m − 1 m + p − 1 Kinga Fogarasi in [1] as follows: Conjecture 1. All (p−1)-hyperperfectnumbers are of the For 2-hyperperfect number n, and m = 3k − 2, one can form n = pk−1(pk − p + 1), where p and pk − p + 1 are easily obtain the following result as a particular case of the theorem 2.8.
Load more

Recommended publications
  • 9N1.5 Determine the Square Root of Positive Rational Numbers That Are Perfect Squares
    9N1.5 Determine the square root of positive rational numbers that are perfect squares. Square Roots and Perfect Squares The square root of a given number is a value that when multiplied by itself results in the given number. For ​ ​ example, the square root of 9 is 3 because 3 × 3 = 9 . Example Use a diagram to determine the square root of 121. Solution Step 1 Draw a square with an area of 121 units. Step 2 Count the number of units along one side of the square. √121 = 11 units √121 = 11 units The square root of 121 is 11. This can also be written as √121 = 11. A number that has a whole number as its square root is called a perfect square. Perfect squares have the ​ ​ unique characteristic of having an odd number of factors. Example Given the numbers 81, 24, 102, 144, identify the perfect squares by ordering their factors from smallest to largest. Solution The square root of each perfect square is bolded. Factors of 81: 1, 3, 9, 27, 81 ​ ​ Since there are an odd number of factors, 81 is a perfect square. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 Since there are an even number of factors, 24 is not a perfect square. Factors of 102: 1, 2, 3, 6, 17, 34, 51, 102 Since there are an even number of factors, 102 is not a perfect square. Factors of 144: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144 ​ ​ Since there are an odd number of factors, 144 is a perfect square.
    [Show full text]
  • On Fixed Points of Iterations Between the Order of Appearance and the Euler Totient Function
    mathematics Article On Fixed Points of Iterations Between the Order of Appearance and the Euler Totient Function ŠtˇepánHubálovský 1,* and Eva Trojovská 2 1 Department of Applied Cybernetics, Faculty of Science, University of Hradec Králové, 50003 Hradec Králové, Czech Republic 2 Department of Mathematics, Faculty of Science, University of Hradec Králové, 50003 Hradec Králové, Czech Republic; [email protected] * Correspondence: [email protected] or [email protected]; Tel.: +420-49-333-2704 Received: 3 October 2020; Accepted: 14 October 2020; Published: 16 October 2020 Abstract: Let Fn be the nth Fibonacci number. The order of appearance z(n) of a natural number n is defined as the smallest positive integer k such that Fk ≡ 0 (mod n). In this paper, we shall find all positive solutions of the Diophantine equation z(j(n)) = n, where j is the Euler totient function. Keywords: Fibonacci numbers; order of appearance; Euler totient function; fixed points; Diophantine equations MSC: 11B39; 11DXX 1. Introduction Let (Fn)n≥0 be the sequence of Fibonacci numbers which is defined by 2nd order recurrence Fn+2 = Fn+1 + Fn, with initial conditions Fi = i, for i 2 f0, 1g. These numbers (together with the sequence of prime numbers) form a very important sequence in mathematics (mainly because its unexpectedly and often appearance in many branches of mathematics as well as in another disciplines). We refer the reader to [1–3] and their very extensive bibliography. We recall that an arithmetic function is any function f : Z>0 ! C (i.e., a complex-valued function which is defined for all positive integer).
    [Show full text]
  • Grade 7/8 Math Circles the Scale of Numbers Introduction
    Faculty of Mathematics Centre for Education in Waterloo, Ontario N2L 3G1 Mathematics and Computing Grade 7/8 Math Circles November 21/22/23, 2017 The Scale of Numbers Introduction Last week we quickly took a look at scientific notation, which is one way we can write down really big numbers. We can also use scientific notation to write very small numbers. 1 × 103 = 1; 000 1 × 102 = 100 1 × 101 = 10 1 × 100 = 1 1 × 10−1 = 0:1 1 × 10−2 = 0:01 1 × 10−3 = 0:001 As you can see above, every time the value of the exponent decreases, the number gets smaller by a factor of 10. This pattern continues even into negative exponent values! Another way of picturing negative exponents is as a division by a positive exponent. 1 10−6 = = 0:000001 106 In this lesson we will be looking at some famous, interesting, or important small numbers, and begin slowly working our way up to the biggest numbers ever used in mathematics! Obviously we can come up with any arbitrary number that is either extremely small or extremely large, but the purpose of this lesson is to only look at numbers with some kind of mathematical or scientific significance. 1 Extremely Small Numbers 1. Zero • Zero or `0' is the number that represents nothingness. It is the number with the smallest magnitude. • Zero only began being used as a number around the year 500. Before this, ancient mathematicians struggled with the concept of `nothing' being `something'. 2. Planck's Constant This is the smallest number that we will be looking at today other than zero.
    [Show full text]
  • Arxiv:Math/0201082V1
    2000]11A25, 13J05 THE RING OF ARITHMETICAL FUNCTIONS WITH UNITARY CONVOLUTION: DIVISORIAL AND TOPOLOGICAL PROPERTIES. JAN SNELLMAN Abstract. We study (A, +, ⊕), the ring of arithmetical functions with unitary convolution, giving an isomorphism between (A, +, ⊕) and a generalized power series ring on infinitely many variables, similar to the isomorphism of Cashwell- Everett[4] between the ring (A, +, ·) of arithmetical functions with Dirichlet convolution and the power series ring C[[x1,x2,x3,... ]] on countably many variables. We topologize it with respect to a natural norm, and shove that all ideals are quasi-finite. Some elementary results on factorization into atoms are obtained. We prove the existence of an abundance of non-associate regular non-units. 1. Introduction The ring of arithmetical functions with Dirichlet convolution, which we’ll denote by (A, +, ·), is the set of all functions N+ → C, where N+ denotes the positive integers. It is given the structure of a commutative C-algebra by component-wise addition and multiplication by scalars, and by the Dirichlet convolution f · g(k)= f(r)g(k/r). (1) Xr|k Then, the multiplicative unit is the function e1 with e1(1) = 1 and e1(k) = 0 for k> 1, and the additive unit is the zero function 0. Cashwell-Everett [4] showed that (A, +, ·) is a UFD using the isomorphism (A, +, ·) ≃ C[[x1, x2, x3,... ]], (2) where each xi corresponds to the function which is 1 on the i’th prime number, and 0 otherwise. Schwab and Silberberg [9] topologised (A, +, ·) by means of the norm 1 arXiv:math/0201082v1 [math.AC] 10 Jan 2002 |f| = (3) min { k f(k) 6=0 } They noted that this norm is an ultra-metric, and that ((A, +, ·), |·|) is a valued ring, i.e.
    [Show full text]
  • Generalized Perfect Numbers
    Acta Univ. Sapientiae, Mathematica, 1, 1 (2009) 73–82 Generalized perfect numbers Antal Bege Kinga Fogarasi Sapientia–Hungarian University of Sapientia–Hungarian University of Transilvania Transilvania Department of Mathematics and Department of Mathematics and Informatics, Informatics, Tˆargu Mure¸s, Romania Tˆargu Mure¸s, Romania email: [email protected] email: [email protected] Abstract. Let σ(n) denote the sum of positive divisors of the natural number n. A natural number is perfect if σ(n) = 2n. This concept was already generalized in form of superperfect numbers σ2(n)= σ(σ(n)) = k+1 k−1 2n and hyperperfect numbers σ(n)= k n + k . In this paper some new ways of generalizing perfect numbers are inves- tigated, numerical results are presented and some conjectures are estab- lished. 1 Introduction For the natural number n we denote the sum of positive divisors by arXiv:1008.0155v1 [math.NT] 1 Aug 2010 σ(n)= X d. d|n Definition 1 A positive integer n is called perfect number if it is equal to the sum of its proper divisors. Equivalently: σ(n)= 2n, where AMS 2000 subject classifications: 11A25, 11Y70 Key words and phrases: perfect number, superperfect number, k-hyperperfect number 73 74 A. Bege, K. Fogarasi Example 1 The first few perfect numbers are: 6, 28, 496, 8128, . (Sloane’s A000396 [15]), since 6 = 1 + 2 + 3 28 = 1 + 2 + 4 + 7 + 14 496 = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 Euclid discovered that the first four perfect numbers are generated by the for- mula 2n−1(2n − 1).
    [Show full text]
  • Fixed Points of Certain Arithmetic Functions
    FIXED POINTS OF CERTAIN ARITHMETIC FUNCTIONS WALTER E. BECK and RUDOLPH M. WAJAR University of Wisconsin, Whitewater, Wisconsin 53190 IWTRODUCTIOW Perfect, amicable and sociable numbers are fixed points of the arithemetic function L and its iterates, L (n) = a(n) - n, where o is the sum of divisor's function. Recently there have been investigations into functions differing from L by 1; i.e., functions L+, Z.,, defined by L + (n)= L(n)± 1. Jerrard and Temperley [1] studied the existence of fixed points of L+ and /._. Lai and Forbes [2] conducted a computer search for fixed points of (LJ . For earlier references to /._, see the bibliography in [2]. We consider the analogous situation using o*, the sum of unitary divisors function. Let Z.J, Lt, be arithmetic functions defined by L*+(n) = o*(n)-n±1. In § 1, we prove, using parity arguments, that L* has no fixed points. Fixed points of iterates of L* arise in sets where the number of elements in the set is equal to the power of/.* in question. In each such set there is at least one natural number n such that L*(n) > n. In § 2, we consider conditions n mustsatisfy to enjoy the inequality and how the inequality acts under multiplication. In particular if n is even, it is divisible by at least three primes; if odd, by five. If /? enjoys the inequality, any multiply by a relatively prime factor does so. There is a bound on the highest power of n that satisfies the inequality. Further if n does not enjoy the inequality, there are bounds on the prime powers multiplying n which will yield the inequality.
    [Show full text]
  • Representing Square Numbers Copyright © 2009 Nelson Education Ltd
    Representing Square 1.1 Numbers Student book page 4 You will need Use materials to represent square numbers. • counters A. Calculate the number of counters in this square array. ϭ • a calculator number of number of number of counters counters in counters in in a row a column the array 25 is called a square number because you can arrange 25 counters into a 5-by-5 square. B. Use counters and the grid below to make square arrays. Complete the table. Number of counters in: Each row or column Square array 525 4 9 4 1 Is the number of counters in each square array a square number? How do you know? 8 Lesson 1.1: Representing Square Numbers Copyright © 2009 Nelson Education Ltd. NNEL-MATANSWER-08-0702-001-L01.inddEL-MATANSWER-08-0702-001-L01.indd 8 99/15/08/15/08 55:06:27:06:27 PPMM C. What is the area of the shaded square on the grid? Area ϭ s ϫ s s units ϫ units square units s When you multiply a whole number by itself, the result is a square number. Is 6 a whole number? So, is 36 a square number? D. Determine whether 49 is a square number. Sketch a square with a side length of 7 units. Area ϭ units ϫ units square units Is 49 the product of a whole number multiplied by itself? So, is 49 a square number? The “square” of a number is that number times itself. For example, the square of 8 is 8 ϫ 8 ϭ .
    [Show full text]
  • On Some New Arithmetic Functions Involving Prime Divisors and Perfect Powers
    On some new arithmetic functions involving prime divisors and perfect powers. Item Type Article Authors Bagdasar, Ovidiu; Tatt, Ralph-Joseph Citation Bagdasar, O., and Tatt, R. (2018) ‘On some new arithmetic functions involving prime divisors and perfect powers’, Electronic Notes in Discrete Mathematics, 70, pp.9-15. doi: 10.1016/ j.endm.2018.11.002 DOI 10.1016/j.endm.2018.11.002 Publisher Elsevier Journal Electronic Notes in Discrete Mathematics Download date 30/09/2021 07:19:18 Item License http://creativecommons.org/licenses/by/4.0/ Link to Item http://hdl.handle.net/10545/623232 On some new arithmetic functions involving prime divisors and perfect powers Ovidiu Bagdasar and Ralph Tatt 1,2 Department of Electronics, Computing and Mathematics University of Derby Kedleston Road, Derby, DE22 1GB, United Kingdom Abstract Integer division and perfect powers play a central role in numerous mathematical results, especially in number theory. Classical examples involve perfect squares like in Pythagora’s theorem, or higher perfect powers as the conjectures of Fermat (solved in 1994 by A. Wiles [8]) or Catalan (solved in 2002 by P. Mih˘ailescu [4]). The purpose of this paper is two-fold. First, we present some new integer sequences a(n), counting the positive integers smaller than n, having a maximal prime factor. We introduce an arithmetic function counting the number of perfect powers ij obtained for 1 ≤ i, j ≤ n. Along with some properties of this function, we present the sequence A303748, which was recently added to the Online Encyclopedia of Integer Sequences (OEIS) [5]. Finally, we discuss some other novel integer sequences.
    [Show full text]
  • Figurate Numbers
    Figurate Numbers by George Jelliss June 2008 with additions November 2008 Visualisation of Numbers The visual representation of the number of elements in a set by an array of small counters or other standard tally marks is still seen in the symbols on dominoes or playing cards, and in Roman numerals. The word "calculus" originally meant a small pebble used to calculate. Bear with me while we begin with a few elementary observations. Any number, n greater than 1, can be represented by a linear arrangement of n counters. The cases of 1 or 0 counters can be regarded as trivial or degenerate linear arrangements. The counters that make up a number m can alternatively be grouped in pairs instead of ones, and we find there are two cases, m = 2.n or 2.n + 1 (where the dot denotes multiplication). Numbers of these two forms are of course known as even and odd respectively. An even number is the sum of two equal numbers, n+n = 2.n. An odd number is the sum of two successive numbers 2.n + 1 = n + (n+1). The even and odd numbers alternate. Figure 1. Representation of numbers by rows of counters, and of even and odd numbers by various, mainly symmetric, formations. The right-angled (L-shaped) formation of the odd numbers is known as a gnomon. These do not of course exhaust the possibilities. 1 2 3 4 5 6 7 8 9 n 2 4 6 8 10 12 14 2.n 1 3 5 7 9 11 13 15 2.n + 1 Triples, Quadruples and Other Forms Generalising the divison into even and odd numbers, the counters making up a number can of course also be grouped in threes or fours or indeed any nonzero number k.
    [Show full text]
  • Multiplication Fact Strategies Assessment Directions and Analysis
    Multiplication Fact Strategies Wichita Public Schools Curriculum and Instructional Design Mathematics Revised 2014 KCCRS version Table of Contents Introduction Page Research Connections (Strategies) 3 Making Meaning for Operations 7 Assessment 9 Tools 13 Doubles 23 Fives 31 Zeroes and Ones 35 Strategy Focus Review 41 Tens 45 Nines 48 Squared Numbers 54 Strategy Focus Review 59 Double and Double Again 64 Double and One More Set 69 Half and Then Double 74 Strategy Focus Review 80 Related Equations (fact families) 82 Practice and Review 92 Wichita Public Schools 2014 2 Research Connections Where Do Fact Strategies Fit In? Adapted from Randall Charles Fact strategies are considered a crucial second phase in a three-phase program for teaching students basic math facts. The first phase is concept learning. Here, the goal is for students to understand the meanings of multiplication and division. In this phase, students focus on actions (i.e. “groups of”, “equal parts”, “building arrays”) that relate to multiplication and division concepts. An important instructional bridge that is often neglected between concept learning and memorization is the second phase, fact strategies. There are two goals in this phase. First, students need to recognize there are clusters of multiplication and division facts that relate in certain ways. Second, students need to understand those relationships. These lessons are designed to assist with the second phase of this process. If you have students that are not ready, you will need to address the first phase of concept learning. The third phase is memorization of the basic facts. Here the goal is for students to master products and quotients so they can recall them efficiently and accurately, and retain them over time.
    [Show full text]
  • The Math Problems Notebook
    Valentin Boju Louis Funar The Math Problems Notebook Birkhauser¨ Boston • Basel • Berlin Valentin Boju Louis Funar MontrealTech Institut Fourier BP 74 Institut de Technologie de Montreal URF Mathematiques P.O. Box 78575, Station Wilderton Universite´ de Grenoble I Montreal, Quebec, H3S 2W9 Canada 38402 Saint Martin d’Heres cedex [email protected] France [email protected] Cover design by Alex Gerasev. Mathematics Subject Classification (2000): 00A07 (Primary); 05-01, 11-01, 26-01, 51-01, 52-01 Library of Congress Control Number: 2007929628 ISBN-13: 978-0-8176-4546-5 e-ISBN-13: 978-0-8176-4547-2 Printed on acid-free paper. c 2007 Birkhauser¨ Boston All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkhauser¨ Boston, c/o Springer Science+Business Media LLC, 233 Spring Street, New York, NY 10013, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbid- den. The use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. 987654321 www.birkhauser.com (TXQ/SB) The Authors Valentin Boju was professor of mathematics at the University of Craiova, Romania, until his retirement in 2000. His research work was primarily in the field of geome- try.
    [Show full text]
  • Problem Solving and Recreational Mathematics
    Problem Solving and Recreational Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Summer 2012 Chapters 1–44 August 1 Monday 6/25 7/2 7/9 7/16 7/23 7/30 Wednesday 6/27 *** 7/11 7/18 7/25 8/1 Friday 6/29 7/6 7/13 7/20 7/27 8/3 ii Contents 1 Digit problems 101 1.1 When can you cancel illegitimately and yet get the cor- rectanswer? .......................101 1.2 Repdigits.........................103 1.3 Sortednumberswithsortedsquares . 105 1.4 Sumsofsquaresofdigits . 108 2 Transferrable numbers 111 2.1 Right-transferrablenumbers . 111 2.2 Left-transferrableintegers . 113 3 Arithmetic problems 117 3.1 AnumbergameofLewisCarroll . 117 3.2 Reconstruction of multiplicationsand divisions . 120 3.2.1 Amultiplicationproblem. 120 3.2.2 Adivisionproblem . 121 4 Fibonacci numbers 201 4.1 TheFibonaccisequence . 201 4.2 SomerelationsofFibonaccinumbers . 204 4.3 Fibonaccinumbersandbinomialcoefficients . 205 5 Counting with Fibonacci numbers 207 5.1 Squaresanddominos . 207 5.2 Fatsubsetsof [n] .....................208 5.3 Anarrangementofpennies . 209 6 Fibonacci numbers 3 211 6.1 FactorizationofFibonaccinumbers . 211 iv CONTENTS 6.2 TheLucasnumbers . 214 6.3 Countingcircularpermutations . 215 7 Subtraction games 301 7.1 TheBachetgame ....................301 7.2 TheSprague-Grundysequence . 302 7.3 Subtraction of powers of 2 ................303 7.4 Subtractionofsquarenumbers . 304 7.5 Moredifficultgames. 305 8 The games of Euclid and Wythoff 307 8.1 ThegameofEuclid . 307 8.2 Wythoff’sgame .....................309 8.3 Beatty’sTheorem . 311 9 Extrapolation problems 313 9.1 Whatis f(n + 1) if f(k)=2k for k =0, 1, 2 ...,n? . 313 1 9.2 Whatis f(n + 1) if f(k)= k+1 for k =0, 1, 2 ...,n? .
    [Show full text]