MAT 102 Solutions – Take-Home Exam 1 ( Theory)

Problem 1

Let 푝 be any . 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 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 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 .

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, famously refuted the conjecture that all Fermat numbers are prime by decomposing the next : 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