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Math 1271 3071 Assignment 1 Solutions

Math 1271 3071 Assignment 4 Solutions
Course

Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course
Academic year: 2023/2024
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Sidharth
Mohawk College
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Math 1271 / 3071: Assignment 1 Solutions

  1. (4 points) Let S = { 1 , 3 , − 1 }. Find the sum

∑

x∈S

x 2.

Solution

∑

x∈S x

2 = (1) 2 + (3) 2 + (−1) 2 = 1 + 9 + 1 = 11.

  1. (4 points) Find the sum ∑n

k=

(4k − 2).

Solution: n ∑

k=

(4k − 2) = 4

∑ n

k=

k −

∑ n

k=

2 = 4 ·

n(n + 1) 2

− 2 n = 2n 2.

  1. (12 points) Convert the following numbers to base 10

(a) (1271) 8. Solution: (1271) 8 = 1 · 83 + 2 · 82 + 7 · 8 + 1 = 697. (b) (CBC) 16. Solution: (CBC) 16 = 12 · 162 + 11 · 16 + 12 = 3260. (c) (12) 5. Solution: (12) 5 = 1 · 5 + 2 + 3/5 + 4/ 52 = 5 + 2 + 0 + 0 = 7.

  1. (4 points) Express 210 in base 2.

Solution: 2 | 210 2 | 105 · · · 0 2 | 52 · · · 1 2 | 26 · · · 0 2 | 13 · · · 0 2 | 6 · · · 1 2 | 3 · · · 0 2 | 1 · · · 1 | 0 · · · 1. Thus 120 = (11010010) 2.

  1. (4 points) Express 321 in base 3.

Solution: 3 | 321 3 | 107 · · · 0 3 | 35 · · · 2 3 | 11 · · · 2 3 | 3 · · · 2 3 | 1 · · · 0 | 0 · · · 1. Thus 321 = (102220) 3.

1

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Math 1271 3071 Assignment 1 Solutions

Course: Discrete Mathematics  (Mathematics 1271)

23 Documents
Students shared 23 documents in this course

University: Lakehead University

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Math 1271 / 3071: Assignment 1 Solutions
1. (4 points) Let S={1,3,−1}. Find the sum
X
x∈S
x2.
Solution Px∈Sx2= (1)2+ (3)2+ (−1)2= 1 + 9 + 1 = 11.
2. (4 points) Find the sum
n
X
k=1
(4k−2).
Solution: n
X
k=1
(4k−2) = 4
n
X
k=1
k−
n
X
k=1
2=4·n(n+ 1)
2−2n= 2n2.
3. (12 points) Convert the following numbers to base 10
(a) (1271)8.
Solution: (1271)8= 1 ·83+ 2 ·82+ 7 ·8 + 1 = 697.
(b) (CBC)16.
Solution: (CBC)16 = 12 ·162+ 11 ·16 + 12 = 3260.
(c) (12.34)5.
Solution: (12.34)5= 1 ·5+2+3/5+4/52= 5 + 2 + 0.6+0.16 = 7.76.
4. (4 points) Express 210 in base 2.
Solution:
2|210
2|105 · · · 0
2|52 · · · 1
2|26 · · · 0
2|13 · · · 0
2|6· · · 1
2|3· · · 0
2|1· · · 1
|0· · · 1.
Thus 120 = (11010010)2.
5. (4 points) Express 321.1 in base 3.
Solution:
3|321
3|107 · · · 0
3|35 · · · 2
3|11 · · · 2
3|3· · · 2
3|1· · · 0
|0· · · 1.
Thus 321 = (102220)3.
1

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