Download Solutions for Assignment 3

Math 160(A1). Assignment 3 Solutions.
Pages 36 and 37.
1. Which of the following statements are true?
(a) 3 | 6,
(b) 1 | 5,
(c) 6 | 3,
(d) 2 | 0,
(e) 0 | 2.
Solution. (a), (b), (d) are true, (c) and (e) are false.
6. Classify each of the following as true or false, and if the statement is false, give an example showing it
is false or explain why it is false.
(a) If positive integer is divisible by a and by b, then it must be divisible by a · b.
(b) If a number is divisible by 8, then it is also divisible by 4.
(c) If a 6 | b and a 6 | c, then a 6 | (b + c).
Solution.
(a) False. 4 and 6 both divide 12, but 24 does not.
(b) True.
(c) False. 56 | 6 and 56 | 9 but 5 |(6 + 9).
7. List all divisors of x and y, circle the common divisors, and find the gcd(x, y).
(a) x = 108, y = 144.
(b) x = 132, y = 0.
(c) x = 17, y = 38.
Solution.
(a) 108: 1 , 2 , 3 , 4 , 6 , 9 , 12, 18 , 27, 36 , 54, 108.
144: 1 , 2 , 3 , 4 , 6 , 8, 9 , 12 , 16, 18 , 24, 36 , 72, 144.
So gcd 108, 144 = 36.
(b) 132: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.
0: 1, 2, 3, 4, 5, 6, . . . .
Every divisor of 132 is a divisor of 0, so gcd(0, 132) = 132.
(c) 17: 1, 17
38: 1, 2, 19, 38
So gcd(17, 38) = 1.
13. (a) Make up and prove a test for divisibility by 15.
(b) For what pairs of digits x and y is the number 411xy divisible by 15?
Solution.
(a) A number is divisible by 15 iff it is divisible by 3 and 5. If it is divisible by 15, certainly it is
divisible by both 3 and 5. If x is divisible by 3, then x = 3m for some m. If x is also divisible
by 5 then 5 | 3m. By the Relatively Prime Divisor Theorem, it follows that 5 | m, so m = 5n for
some n. Then x = 15n showing that x is divisible by 15.
(b) Since 15 = 3 · 5, and since (3, 5) = 1, then 411xy is divisible by 15 iff it is divisible by 3 and by 5.
In order that 411xy be divisible by 5, y must be either 0 or 5.
If y = 0, then to be divisible by 3, the sum of the digits, 4 + 1 + 1 + x + 0, or 6 + x, would have
to be divisible by 3. In this case, x would have to be 0, 3, 6, or 9.
If y = 5, then to be divisible by 3, the sum of the digits, 4 + 1 + 1 + x + 5, or 11 + x, would have
to be divisible by 3. In this case, x would have to be 1, 4, or 7.
To summarize, the pairs (x, y) of digits for which 411xy is divisible by 15 are (0, 0), (3, 0), (6, 0),
(9, 0), (1, 5), (4, 5), (7, 5).
14. Which of the following numbers are divisible by 77? By 91?
(a) 3,400,927,788,104
(b) 999,876,543,209,544
(c) 40,454,565,676,707
Solution. You can’t just try these on a calculator (at least on most when the question was posed)
because there are too many digits. So let us try to use the divisibility tests.
We note that 77 = 7 × 11 and 91 = 7 × 13. So we test for divisibility by 7, 11 and 13.
(a) 3−400+927−788+104 = −154. This is divisible by 7 and 11, but not by 13, so 3,400,927,788,104
is divisible by 77 but not by 91.
(b) 999 − 876 + 543 − 209 + 544 = 1001. This is divisible by 7, 11, and 13, so 999,876,543,209,544 is
divisible by both 77 and 91.
(c) 40−454+565−676+707 = 182. This is divisible by 7 and 13, but not by 11, so 40,454,565,676,707
is divisible by 91 but not by 77.
Page 50.
1. Find the prime factorizations of the following numbers.
(a)
(d)
5040
499
(b)
(e)
299
123123
(c)
(f)
33000
8437
Solution.
(a) 5040 = 24 32 5 · 7
(d) 499 (prime)
(b)
(e)
299 = 13 · 23
(c) 33000 = 23 3 · 53 11
123123 = 7 · 11 · 13 · 123 (f) 8437 = 11 · 13 · 59
2. Use the sieve of Eratosthenes to find all the primes up to 200. (It serves somewhat as a visual check
if you put the integers in rows that contain a multiple of 6 elements.)
Solution. Here are the primes you should end up with:
2
31
73
127
179
3
37
79
131
181
5
41
83
137
191
7
43
89
139
193
11
47
97
149
197
13
53
101
151
199
17
59
103
157
19
61
107
163
23
67
109
167
29
71
113
173
4. Gerry is trying to factor the number 704791. By testing all primes in succession, she has found that
the smallest prime that divides 704791 is 89. The square root of 704791 is approximately 839.5, so at
first she thinks that she still has a way to go. Then she realized that she didn’t need to go any farther,
and she was correct. Why?
Solution. Dividing by the factor 89, she gets 704791 = 89 × 7919. Checking to see if 89 | 7919 she
finds it does not. She knows that no prime less than 89 divides 7919 (if there was one, it would have
divided 704791). Since 892 = 7921 > 7919 she concludes that 7919 has no prime factors less than 7919.
So 7919 is a prime number and she is finished.