Math 326
Exam 1 Solutions
Fall ’16
Show all of your work.
1. Use the Euclidean Algorithm to find (125, 56), and find a solution
to Bezout’s identity. Solution:
125 = 2 · 56 + 13
56∗ = 4 · 13 + 4
13 = 3 · 4 + 1
1 = 13 − 3 · 4 = 13 − 3(56 − 4 · 13)
= −3 · 56 + 13 · 13
= −3 · 56 + 13(125 − 2 · 56)
= 13 · 125 − 29 · 56.
2. True or False. If true say why. If false, give a counterexample.
(a) If a 6≡ 0 mod 24, and if ax ≡ ay mod 24, then x ≡ y mod
24
.
(24,x)
Solution: False. One counterexample is given by taking
a = 8, x = 1, y = 4.
(b) If a 6≡ 0 mod 15, and if ax ≡ b mod 15 has a solution, then
the solution is unique mod 15.
Solution: False. One counterexample is given by taking
a = 5 and b = 0. Then the solutions are given by
x ≡ 0, 3, 6, 9, 12 mod 15.
3. Find all solutions or say why none exist:
(a) 36x + 60y = 20 Solution: (36, 60) = 12 6 | 20. There are
no solutions.
(b) 36x + 20y = 60 Solution: (36, 20) = 4|60, so solutions
exist. We first solve Bezout’s identity for 36 and 20:
36 = 20 + 16
20 = 16 + 4
16 = 4 · 4 + 0
4 = 20 − 16 = 20 − (36 − 20)
= −36 + 2 · 20.
Multiplying through by 15, we get
60 = 15 · 4 = (−15)36 + 30 · 20,
so [ −15
30 ] is a solution. The general solution is now given by
20 −15
−15 + 5n 4
+n
n∈Z .
n∈Z =
30
30 − 9n − 36
4
2
Exam 1 Solutions
4. Suppose (a, b) = 10, [a, b] = 600. What are the possible values
for a and b (as a pair)?
Solution: (a, b) = 21 30 51 , [a, b] = 23 31 52 . So
a = 2r1 3r2 5r3 ,
b = 2s1 3s2 5s3 ,
where
min(r1 , s1 ) = 1
min(r2 , s2 ) = 0
min(r3 , s3 ) = 1
max(r1 , s1 ) = 3
max(r2 , s2 ) = 1
max(r3 , s3 ) = 2.
We may fix the exponents of 2 in a and b to avoid repetition. So
set r1 = 1, s1 = 3.
Case 1. r2 = 0, r3 = 1. So s2 = 1, s3 = 2. Then
a = 21 30 51 = 10,
b = 23 31 52 = 600.
Case 2. r2 = 0, r3 = 2. So s2 = 1, s3 = 1. Then
a = 21 30 52 = 50,
b = 23 31 51 = 120.
Case 3. r2 = 1, r3 = 1. So s2 = 0, s3 = 2. Then
a = 21 31 51 = 30,
b = 23 30 52 = 200.
Case 4. r2 = 1, r3 = 2. So s2 = 0, s3 = 1. Then
a = 21 31 52 = 150,
b = 23 30 51 = 40.
5. Find all nonnegative solutions less than the modulus or show
why no solutions exist:
(a) 18x ≡ 120 mod 108. Solution: (18, 108) = 18 6 |120.
There are no solutions.
(b) 18x ≡ 108 mod 120. Solution: (18, 120) = 6|18, so solutions exist. Dividing through by 6 we get
3x ≡ 18 mod 20.
Exam 1 Solutions
3
We can see a solution visibly here: x = 6. We could also
obtain it by finding an inverse for 3 mod 20. Visibly 7 · 3 ≡
1 mod 20, so
x ≡ 7 · 3x ≡ 7 · 18 = 126 ≡ 6 mod 20.
Thus,
x ≡ 6, 6 + 20, 6 + 40, 6 + 60, 6 + 80.6 + 100 mod 120
≡ 6, 26, 46, 66, 86, 106 mod 120.
6. (a) What is the inverse of 11 in Z×
75 ? Solution: We solve for
Bezout’s identity for 11 and 75:
75 = 6 · 11 + 9
11 = 9 + 2
9=4·2+1
1 = 9 − 4(2) = 9 − 4(11 − 9) = −4 · 11 + 5(9)
= −4 · 11 + 5(75 − 6 · 11)
= 5 · 75 − 34 · 11
Thus, 1 ≡ (−34) · 11 mod 75. So the inverse of 11 is −34 in
Z×
75 . Now, −34 ≡ −34 + 75 ≡ 41 mod 75, so −34 = 41 in
Z×
75
(b) What is the inverse of 11 in Z×
45 ? Solution: Here, 45 =
4 · 11 + 1, so
1 = 45 − 4 · 11 ≡ (−4) · 11 mod 45.
Thus, the inverse of 11 is −4 in Z×
45 . As above, −4 =
−4 + 45 = 41 in Z×
.
45