MATH 433
January 26, 2015
Quiz 1: Solutions
Problem 1. Find all integers 1 ≤ x ≤ 100 such that gcd(x, 24) = 12 and gcd(x, 35) = 5.
Solution: x = 60.
Since gcd(x, 24) = 12, the number x is divisible by 12 but not divisible by 24. Hence x = 12k, where
k is an odd integer. In the range from 1 to 100, there are four such numbers: 12, 36, 60 and 84. Since
gcd(x, 35) = 5, the number x is divisible by 5, which leaves the only possibility: x = 60. It remains to check
that gcd(60, 24) = 12 and gcd(60, 35) = 5.
Problem 2. Find an integer solution of the equation 488m + 28n = 4.
Solution: m = −2 + 7k, n = 35 − 122k, where k ∈ Z.
First we apply the Euclidean algorithm to find the greatest common divisor of 488 and 28:
488 = 17 · 28 + 12,
28 = 2 · 12 + 4,
12 = 3 · 4.
Then we use back substitution to represent gcd(488, 28) = 4 as an integral linear combination of 488 and
28:
4 = 28 − 2 · 12 = 28 − 2 · (488 − 17 · 28) = 28 − 2 · 488 + 34 · 28 = (−2) · 488 + 35 · 28.
Hence m = −2, n = 35 is a solution. Moreover, (488 · 28)/4 = 122 · 28 = 488 · 7 so that 7 · 488 − 122 · 28 = 0.
Therefore 488(−2 + 7k) + 28(35 − 122k) = 4 for all k ∈ Z.
Alternatively, we can find a solution via the matrix method:
1 −17 12
1 −17 12
7 −122 0
1 0 488
→
→
→
.
0 1 28
0
1
−2
35 4
−2
35 4
28
From the second row of the final matrix we read off that (−2) · 488 + 35 · 28 = 4. In addition, from the first
row we read off that 7 · 488 − 122 · 28 = 0, which implies that (−2 + 7k)488 + (35 − 122k)28 = 4 for every
k ∈ Z.