1. Miniconference on the Mathematics of Computation
MTH 210
Test 2 Review
Dr. Anthony Bonato
Ryerson University

2. Notes on Test 2
Test 2 is in-class on Thursday, March 14 in VIC 205, starting at 10:30 am
90 minutes, 5 questions (multiple parts), 25 marks total
Material covers all material on number theory, up to and including material covered in the March 7 lecture.
Need to know: definitions, examples, exercises, assigned problems, quiz material, theorems, key facts
Two questions short answer, one question fill in the blank; two questions long answer
NOTE: scientific calculators are allowed. Phones or other internet connected devices cannot be used. You are responsible for bringing your own calculator.

3. Key facts an integer m is a divisor of n if n = pm for some integer p
integer n > 1 is prime if its only divisors are 1 and n
a real number is irrational if it is not rational
that is, it isn’t a fraction of integers
egs: 2 , 3 , 5 , …

4. General congruences m > 2 an integer a ≡ b (mod n) if n | (a - b)
congruences behave like =
[m] = {x | x ≡ m (mod n)}
called equivalence classes (mod n) or congruence classes

5. Greatest Common Divisors
m and n integers > 1
among all the divisors of both m and n, the largest is called the greatest common divisor
write gcd(m,n)

6. Key Facts gcd(n,0) = n If a = bq + r, then gcd(a,b) = (b,r)

7. Euclidean Algorithm Goal: find gcd(a,b) Step 1: Express a = bq + r
Step 2: Find gcd(b,r).
Step 3: Express b = sr + t.
Step 4: Find gcd(r,t). …
Keep going until remainder is zero.

8. x = x1 + tb/d, y = y1 – ta/d, where t is an integer.
How to solve ax + by = c?
Calculate gcd(a,b) by using the Euclidean Algorithm.
Does gcd(a,b) divide c?
If NO, then there is no solution to the linear Diophantine equation.
If YES then
Solve ax + by = gcd(a,b) = d by working backwards through your steps in the Euclidean Algorithm. This gives a particular solution (x1,y1)
The general solution is:
x = x1 + tb/d, y = y1 – ta/d, where t is an integer.