Discussion 3B

Bijections

Consider two sets and , and a function that maps elements from to the elements of .

Chinese Remainder Theorem

Goal

Given a list of congruences , , , , find an that satisfies all congruences.

Theorem

Let . If all the are relatively prime () then there is exactly one value of in the set that satisfies all the congruences.

Finding

For and , we want to find a number that is a multiple of and , and similarly a number that is a multiple of and . Then is a solution because and .