MATH 433
February 13, 2015
Quiz 4: Solutions
Problem 1. Solve a linear congruence 4x ≡ 18 mod 11.
Solution: x ≡ 10 mod 11.
To solve this linear congruence, we need to find the inverse of 4 modulo 11. For this, we need to represent
1 as an integral linear combination of 4 and 11. This can be done either by inspection or by the matrix
method:
1 0 4
3 −1 1
1 0 4
→
→
.
0 1 11
−2 1 3
−2
1 3
From the first row we read off that 3 · 4 − 11 = 1, which implies that 3 is the inverse of 4 modulo 11. Thus
4x ≡ 18 mod 11 ⇐⇒ x ≡ 3 · 18 mod 11 ⇐⇒ x ≡ 54 mod 11 ⇐⇒ x ≡ 10 mod 11.
Alternative solution: 4x ≡ 18 mod 11 ⇐⇒ 4x ≡ −4 mod 11 ⇐⇒ x ≡ −1 mod 11.
Problem 2. Solve a system of congruences
y ≡ 4 mod 13,
y ≡ 2 mod 7.
Solution: y ≡ 30 mod 91.
The moduli 13 and 7 are coprime. We need to represent 1 as an integral linear combination of these
moduli. It is easy to see that (−1) · 13 + 2 · 7 = 1. Then one of the solutions is y = 2(−1) · 13 + 4 · 2 · 7 = 30.
The solution set is the congruence class of 30 modulo 13 · 7 = 91.
Alternative solution: From the first congruence we get that y = 4 + 13k, where k ∈ Z. Substituting this
into the second congruence, we obtain 4+ 13k ≡ 2 mod 7 ⇐⇒ 13k ≡ −2 mod 7 ⇐⇒ −k ≡ −2 mod 7 ⇐⇒
k ≡ 2 mod 7. Thus k = 2 + 7l, where l ∈ Z. Then y = 4 + 13k = 4 + 13(2 + 7l) = 30 + 91l.