Problem 1 Let Be Any Prime Number. Then, by Definition

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Problem 1 Let Be Any Prime Number. Then, by Definition MAT 102 Solutions – Take-Home Exam 1 (Number Theory) Problem 1 Let 푝 be any prime number. Then, by definition, 푝 has only itself and 1 as factors. This implies that 1 is the only proper factor of 푝. So 푝 must be deficient since all prime numbers are greater than 1. As a result, all prime numbers are deficient. Problem 2 First, we check with 푝 = 7, the prime that follows 5: 209 7# = 7 × 5 × 3 × 2 = 210. So 7# ± 1 = { . 211 While 211 is prime, 209 is composite: 209 = 11 × 19. So we need to check with the next prime 푝 = 11: 2311 11# = 11 × (7#) = 11 × 210 = 2310. So 11# ± 1 = { . 2309 Since both of these naturals are prime (check this), we conclude that (2309, 2311) is the next pair of primordial twin primes after (29, 31). Problem 3 Lucky Prime List of Proper Sum of Proper Kind of Number Factorization Divisors Divisors Number 18 18 = 2 × 32 1, 2, 3, 6, 9 21 ABUNDANT 28 28 = 22 × 7 1, 2, 4, 7, 14 28 PERFECT 51 51 = 3 × 17 1, 3, 17 21 DEFICIENT 102 102 = 2 × 3 × 17 1, 2, 3, 6, 17, 34, 51 114 ABUNDANT 495 495 = 32 × 5 × 11 1, 3, 5, 9, 11, 15, 441 DEFICIENT 33, 45, 55, 99, 165 496 496 = 24 × 31 1, 2, 4, 8, 16, 31, 496 PERFECT 62, 124, 248 Since there are two perfect numbers (28, 496), two deficient numbers (51, 495), and two abundant numbers (18, 102), your friend is correct. Problem 4 Number Prime Factorization Proper Divisors Sum of Proper Divisors 2620 2620 = 22 × 5 × 131 1, 2, 4, 5, 10, 20, 131, 2924 262, 524, 655, 1310 2924 2924 = 22 × 17 × 43 1, 2, 4, 17, 34, 43, 68, 2620 86, 172, 731, 1462 From this table, we conclude that 2620 and 2924 are indeed amicable numbers. Problem 5 ퟏퟕ = ퟑퟐ + ퟐퟐ + ퟐퟐ + ퟎퟐ or 1ퟕ = ퟒퟐ + ퟏퟐ + ퟎퟐ + ퟎퟐ ퟏퟕퟕ = ퟏퟐퟐ + ퟓퟐ + ퟐퟐ + ퟐퟐ or ퟏퟕퟕ = ퟏퟑퟐ + ퟐퟐ + ퟐퟐ + ퟎퟐ or ퟏퟕퟕ = ퟏퟎퟐ + ퟖퟐ + ퟑퟐ + ퟐퟐ ퟕퟏퟕ = ퟐퟔퟐ + ퟔퟐ + ퟐퟐ + ퟏퟐ ퟕퟕퟏ = ퟐퟕퟐ + ퟓퟐ + ퟒퟐ + ퟏퟐ (Note that there are many other possible answers for some of these numbers.) Problem 6 a) 945 = 33 × 5 × 7 is abundant since the sum of its 15 proper factors is greater than itself, as computed below. 1 + 3 + 5 + 7 + 9 + 15 + 21 + 27 + 35 + 45 + 63 + 105 +135 + 189 + 315 = ퟗퟕퟓ > ퟗퟒퟓ b) Since there are multiple ways to get proper factors of 945 to add up to 30 (e.g. 3 + 27, 21 + 5 + 3 + 1), there are also multiple ways to get some of the 15 proper factors of 945 to add up to itself. For instance, 945 = 7 + 9 + 15 + 27 + 35 + 45 + 63 + 105 + 135 + 189 + 315 As a result, 945 is semiperfect. Problem 7 23 8 푭ퟑ = 2 + 1 = 2 + 1 = ퟐퟓퟕ is prime. 24 16 푭ퟒ = 2 + 1 = 2 + 1 = ퟔퟓ, ퟓퟑퟕ is prime. In 1732, Leonhard Euler famously refuted the conjecture that all Fermat numbers are prime by decomposing the next Fermat number: 퐹5 = 4,294,967,297 = 631 × 6,700,417. Remarkably, it is now believed that 퐹4 is the largest Fermat prime! All Fermat numbers 퐹5, 퐹6, 퐹7, … are thus believed to be composite. Problem 8 List of all the primes that could be used in this table: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 Note: Most of the numbers in this table can be decomposed as a sum of 3 primes in multiple ways. For that reason, at least two possible answers are given whenever possible. Number Sum of 3 Primes Number Sum of 3 Primes Number Sum of 3 Primes 20 11+7+2 21 17+2+2 22 13+7+2 or 13+5+2 or 11+7+3 or 17+3+2 23 17+3+3 24 19+2+3 25 19+3+3 or 13+5+5 or 17+2+5 or 17+3+5 26 19+5+2 27 13+11+3 28 23+2+3 or 13+11+2 or 23+2+2 or 19+2+7 29 19+7+3 30 23+5+2 31 11+17+3 or 13+13+3 or 11+17+2 or 23+5+3 32 23+7+2 33 29+2+2 34 29+2+3 or 19+11+2 or 23+7+3 or 19+2+13 Problem 9 [3 pts] The following counter-example shows why the Fundamental Theorem of Arithmetic fails for this system of even numbers: 60 = 2 × 30 and 60 = 6 × 10 Here the “composite” number 60 has two distinct factorizations with “primes” 2, 6, 10 and 30. Problem 10 푝 = 2 ⇒ 22 − 1 = 4 − 1 = ퟑ is prime. 푝 = 3 ⇒ 23 − 1 = 8 − 1 = ퟕ is prime. 푝 = 5 ⇒ 25 − 1 = 32 − 1 = ퟑퟏ is prime. 푝 = 7 ⇒ 27 − 1 = 128 − 1 = ퟏퟐퟕ is prime. 푝 = 11 ⇒ 211 − 1 = 2048 − 1 = ퟐퟎퟒퟕ = ퟐퟑ × ퟖퟗ is not prime. Therefore, ퟐퟏퟏ − ퟏ is the smallest Mersenne number that is not prime. Problem 11 a) Here is the largest prime desert amongst the first 100 naturals (length 7): 90, 91, 92, 93, 94, 95, 96 b) Below are listed all the distinct prime deserts larger than 6 between 1 and 200: Prime Deserts of length 7 90 – 96 114 – 120 115 – 121 116 – 122 117 – 123 118 – 124 119 – 125 120 – 126 140 – 146 141 – 147 142 – 148 182 – 188 183 – 189 184 – 190 Prime Deserts of length 8 114 – 121 115 – 122 116 – 123 117 – 124 118 – 125 119 – 126 140 – 147 141 – 148 182 – 189 183 – 190 Prime Deserts of length 9 114 – 122 115 – 123 116 – 124 117 – 125 118 – 126 140 – 148 182 – 190 Prime Deserts of length 10 114 – 123 115 – 124 116 – 125 117 – 126 Prime Deserts of length 11 114 – 124 115 – 125 116 – 126 Prime Deserts of length 12 114 – 125 115 – 126 Prime Deserts of length 13 114 – 126 c) Here is the largest prime desert between 100 and 300 (length 13): 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126 .
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