Download MAT 102 SOLUTIONS – TAKE-HOME EXAM 1 (NUMBER THEORY

MAT 102
SOLUTIONS – TAKE-HOME EXAM 1 (NUMBER THEORY)
Problem 1
First we check with p  7 (the next prime after 5):
 211
7 #  7  5  3  2  7   5#   7  30  210 . So 7 # 1  
.
209
211 is prime, but since 209 is not prime ( 209  1119 ), we must check with p  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 2
Number
Prime
Factorization
List of Proper
Divisors
Sum of Proper
Divisors
Deficient?
Abundant?
Perfect?
18
18  2  3  3
1, 2, 3, 6, 9
21
ABUNDANT
28
28  2  2  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
315
315  3  3  5  7
1, 3, 5, 7, 9, 15, 21, 35, 309
45, 63, 105
DEFICIENT
414
414  2  3  3  23
1, 2, 3, 6, 9, 18, 23, 46, 522
69, 138, 207
ABUNDANT
So your friend is wrong!
Problem 3
Number
Prime
Factorization
Proper Divisors
Sum of Proper
Divisors
2620
2620  2  2  5 131
1, 2, 4, 5, 10, 20, 131, 262, 524, 2924
655, 1310
2924
2924  2  2 17  43
1, 2, 4, 17, 34, 43, 68, 86, 172, 2620
731, 1462
From this table, we conclude that 2620 and 2924 are indeed amicable numbers.
Problem 4
Let p be any prime number. Then, by definition, p has only itself and 1 as factors/divisors.
This implies that 1 is the only proper factor/divisor of p . So p must be deficient since all prime
numbers are greater than 1. As a result, we conclude that all prime numbers are
necessarily deficient.
Problem 5
List of primes that could be used for 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 or 13+5+2
21
17+2+2 or 11+7+3
22
13+7+2 or 17+3+2
23
17+3+3 or 13+5+5
24
19+2+3 or 17+2+5
25
19+3+3 or 17+3+5
26
19+5+2 or 13+11+2
27
13+11+3 or
23+2+2
28
23+2+3 or 19+2+7
29
19+7+3 or 13+13+3
30
23+5+2 or
11+17+2
31
11+17+3 or
23+5+3
32
23+7+2 or 19+11+2
33
29+2+2 or 23+7+3
34
29+2+3 or 19+2+13
35
11+11+13 or
31+2+2
36
31+2+3 or 29+2+5
37
29+3+5 or
11+13+13
38
31+5+2 or 23+13+2
39
31+5+3 or 29+5+5
Problem 6
7  12  12  12  22
17  32  22  22  02
177  122  52  22  22
(Note that these answers are not unique.)
1770  402  112  7 2  02
Problem 7
22  1  4  1  3 is prime.
23  1  8  1  7 is prime.
25  1  32  1  31 is prime.
27  1  128  1  127 is prime (check this!)
211  1  2,048  1  2,047  23  89 is not prime.
Therefore, 211  1 is the smallest Mersenne number that is not prime.
Problem 8
Prime desert of length 2016:
2017! 2
2017! 3
2017! 2016
2017! 2017
Problem 9
a)
The largest prime desert amongst the first 100 naturals is given below. It has length 7.
90, 91, 92, 93, 94, 95, 96
b)
Prime deserts larger than 4 between 1 and 100:
Deserts of length 5
24 – 28
32 – 36
74 – 78
84 – 88
48 – 52
90 – 94
54 – 58
91 – 95
Deserts of length 6
90 – 95
91 – 96
Desert of length 7
90 – 96
c)
Largest prime desert between 100 and 300:
114, 115, 116, … , 125, 126 (length = 13)
62 – 66
92 – 96