Download MATH 1: Homework 4

Your name:
Arvind Borde
MATH 1: Homework 4
Solutions
A] Are the following true or false?
1. 12 ≡ 5 (mod 5)
2. 12 ≡ 5 (mod 7)
3. 12 ≡ 7 (mod 5)
4. −1 ≡ 5 (mod 6)
5. 397 ≡ 2 (mod 3)
6. 397 ≡ 2 (mod 5)
Answers
1.
2.
3.
4.
5.
6.
False. 12 6≡ 5 (mod 5), because 5 − 12 = −7 is not divisible by 5.
True. 12 ≡ 5 (mod 7), because 5 − 12 = −7 is divisible by 7.
True. 12 ≡ 7 (mod 5), because 7 − 12 = −5 is divisible by 5.
True. −1 ≡ 5 (mod 6), because 5 − (−1) = 6 is divisible by 6.
False. 397 6≡ 2 (mod 3), because 2 − 397 = −395 is not divisible by 3.
True. 397 ≡ 2 (mod 5), because 2 − 397 = −395 is divisible by 5.
Note: Understanding how these equivalence questions are settled is crucial to understanding the rest of the
homework. See pages 11 through 14 in the class notes.
B] Solve
1. 2x + 1 ≡ 3 (mod 5)
2. 2x + 1 ≡ 13 (mod 5)
3. 2x + 3 ≡ 1 (mod 5)
4. x − 1 ≡ 3 (mod 7)
5. 1 − 3x ≡ 16 (mod 3)
6. y + 3 ≡ 3 (mod 3)
7. 3x ≡ 1 (mod 5)
8. 3x ≡ 1 (mod 7)
9. 4x ≡ 1 (mod 5)
10. 4x ≡ 0 (mod 5)
11. 4x ≡ 0 (mod 6)
12. 6x ≡ 0 (mod 10)
Answers
1. 2x + 1 ≡ 3 (mod 5)
i) Solve 2x + 1 = 3 normally: x = 1.
ii) Is 0 ≤ x < 5 (the base)? Yes.
ii) Answer: x ≡ 1.
2. 2x + 1 ≡ 13 (mod 5)
i) Solve 2x + 1 = 13 normally: x = 6.
ii) Is 0 ≤ x < 5 (the base)? No.
iii) Since x ≥ 5, you repeatedly subtract the base:
6 − 5 = 1: in the correct range.
iv) Answer: x ≡ 1.
3. 2x + 3 ≡ 1 (mod 5)
i) Solve 2x + 3 = 1 normally: x = −1.
ii) Is 0 ≤ x < 5 (the base)? No.
iii) Since x < 0, you repeatedly add the base:
−1 + 5 = 4: in the correct range.
iv) Answer: x ≡ 4.
4. x − 1 ≡ 3 (mod 7)
i) Solve x − 1 = 3 normally: x = 4.
ii) Is 0 ≤ x < 7 (the base)? Yes.
ii) Answer: x ≡ 4.
5. 1 − 3x ≡ 16 (mod 3)
i) Solve 1 − 3x = 16 normally: x = −5.
ii) Is 0 ≤ x < 3 (the base)? No.
iii) Since x < 0, you repeatedly add the base:
−5 + 3 = −2: Still negative.
−2 + 3 = 1: in the correct range.
iv) Answer: x ≡ 1.
6. y + 3 ≡ 3 (mod 3)
i) Solve y + 3 = 3 normally: y = 0.
ii) Is 0 ≤ y < 3 (the base)? Yes.
ii) Answer: y ≡ 0.
7. 3x ≡ 1 (mod 5)
Solve 3x = 1 normally: x = 1/3. Fraction.
So, plug in values for x in 3x:
3(0) = 0. Is 0 ≡ 1 (mod 5)? No.
3(1) = 3. Is 3 ≡ 1 (mod 5)? No.
3(2) = 6. Is 6 ≡ 1 (mod 5)? Yes.
x ≡ 2 is your whole number answer.
9. 4x ≡ 1 (mod 5)
Solve 4x = 1 normally: x = 1/4. Fraction.
So, plug in values for x in 3x:
4(0) = 0. Is 0 ≡ 1 (mod 5)? No.
4(1) = 4. Is 4 ≡ 1 (mod 5)? No.
4(2) = 8. Is 8 ≡ 1 (mod 5)? No.
4(3) = 12. Is 12 ≡ 1 (mod 5)?No.
4(4) = 16. Is 16 ≡ 1 (mod 5)? Yes.
x ≡ 4 is your whole number answer.
8. 3x ≡ 1 (mod 7)
Solve 3x = 1 normally: x = 1/3. Fraction.
So, plug in values for x in 3x:
3(0) = 0. Is 0 ≡ 1 (mod 7)? No.
3(1) = 3. Is 3 ≡ 1 (mod 7)? No.
3(2) = 6. Is 6 ≡ 1 (mod 7)? No.
3(3) = 9. Is 9 ≡ 1 (mod 7)? No.
3(4) = 12. Is 12 ≡ 1 (mod 7)? No.
3(5) = 15. Is 15 ≡ 1 (mod 7)?Yes.
x ≡ 5 is your whole number answer.
Questions 10–12 in this set all have x ≡ 0 as an obvious answer. But two of them also have an additional,
non-zero answer:
10. 4x ≡ 0 (mod 5)
Plug in nonzero values for x in 4x:
4(1) = 4. Is 4 ≡ 0 (mod 5)? No.
4(2) = 8. Is 8 ≡ 0 (mod 5)? No.
4(3) = 12. Is 12 ≡ 0 (mod 5)? No.
4(4) = 16. Is 16 ≡ 0 (mod 5)? No.
There is no alternative nonzero answer.
12. 6x ≡ 0 (mod 10)
Plug in nonzero values for x in 6x:
6(1) = 6. Is 6 ≡ 0 (mod 5)? No.
6(2) = 12. Is 12 ≡ 0 (mod 5)? No.
6(3) = 18. Is 18 ≡ 0 (mod 5)? No.
6(4) = 24. Is 24 ≡ 0 (mod 5)? No.
6(5) = 30. Is 30 ≡ 0 (mod 5)? Yes.
So x ≡ 5 is an alternative nonzero answer.
C ] What are the additive and multiplicative inverses of
1. 5 (mod 7)?
2. 2 (mod 7)?
Answers
1a. Additive inverse of 5 (mod 7):
Set up the equation 5 + x ≡ 0 (mod 7).
i) Solve 5 + x = 0 normally: x = −5.
ii) Is 0 ≤ x < 7 (the base)? No.
iii) Since x < 0, you repeatedly add the base:
−5 + 7 = 2: in the correct range.
So 2 is the additive inverse of 5 (mod 7).
11. 4x ≡ 0 (mod 6)
Plug in nonzero values for x in 4x:
4(1) = 4. Is 4 ≡ 0 (mod 6)? No.
4(2) = 8. Is 8 ≡ 0 (mod 6)? No.
4(3) = 12. Is 12 ≡ 0 (mod 6)? Yes.
So x ≡ 3 is an alternative nonzero answer.
NOTE: How the coefficient of x in these equations
relates to the base tells you if a nonzero answer will
exist. Can you guess at the connection?
3. 2 (mod 3)?
1b. Multiplicative inverse of 5 (mod 7):
Set up the equation 5x ≡ 1 (mod 7).
Plug in values for x in 5x:
5(0) = 0. Is 0 ≡ 1 (mod 7)? No.
5(1) = 5. Is 5 ≡ 1 (mod 7)? No.
5(2) = 10. Is 10 ≡ 1 (mod 7)? No.
5(3) = 15. Is 15 ≡ 1 (mod 7)? Yes.
So 3 is the multiplicative inverse of 5 (mod 7).
2a. Additive inverse of 2 (mod 7):
Set up the equation 2 + x ≡ 0 (mod 7).
i) Solve 2 + x = 0 normally: x = −2.
ii) Is 0 ≤ x < 7 (the base)? No.
ii) Since x < 0, you repeatedly add the base:
−2 + 7 = 5: in the correct range.
So 5 is the additive inverse of 2 (mod 7).
2b. Multiplicative inverse of 2 (mod 7):
Set up the equation 2x ≡ 1 (mod 7).
Plug in values for x in 5x:
2(0) = 0. Is 0 ≡ 1 (mod 7)? No.
2(1) = 2. Is 2 ≡ 1 (mod 7)? No.
2(2) = 4. Is 4 ≡ 1 (mod 7)? No.
2(3) = 6. Is 6 ≡ 1 (mod 7)? No.
2(4) = 8. Is 8 ≡ 1 (mod 7)? Yes.
So 4 is the multiplicative inverse of 2 (mod 7).
3a. Additive inverse of 2 (mod 3):
Set up the equation 2 + x ≡ 0 (mod 3).
i) Solve 2 + x = 0 normally: x = −2.
ii) Is 0 ≤ x < 3 (the base)? No.
ii) Since x < 0, you repeatedly add the base:
−2 + 3 = 1: in the correct range.
So 1 is the additive inverse of 2 (mod 3).
3b. Multiplicative inverse of 2 (mod 3):
Set up the equation 2x ≡ 1 (mod 3).
Plug in values for x in 5x:
2(0) = 0. Is 0 ≡ 1 (mod 3)? No.
2(1) = 2. Is 2 ≡ 1 (mod 3)? No.
2(2) = 4. Is 4 ≡ 1 (mod 3)? Yes.
So 2 is the multiplicative inverse of 2 (mod 3).
D] Write down the addition and multiplication tables for integers mod 5.
Answer
Set up normal addition and multiplication tables:
+
0
1
2
3
4
×
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
0
1
2
3
4
0
0
0
0
0
0
1
2
3
4
0 0
2 3
4 6
6 9
8 12
0
4
8
12
16
For each number that’s ≥ 5 in these tables, find a number that’s equivalent to it and < 5.
+
0
1
2
3
4
×
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
1
2
3
4
0
2
3
4
0
1
3
4
0
1
2
4
0
1
2
3
0
1
2
3
4
0
0
0
0
0
0
1
2
3
4
0
2
4
1
3
0
3
1
4
2
0
4
3
2
1