Homework Chp 5 Update – problem #20

Hi everyone,

Problem #20 in Chapter 5 uses an idea that we have *not* yet covered in class.  It is no longer a required problem – you don’t have to do it – but I will give you extra credit if you turn in a solution.  This is excellent practice in reading and applying a definition (just as we have been doing for the definitions of odd, even, divides, and so on).  The problem relies on a new definition, that of congruence mod n – it appears in the book as Definition 5.1 on page 105, but I will also give it here:

Definition.  Given integers a and b, and $n \in \mathbb{N}$, we say “a is congruent to b mod n”, or $a \equiv b \pmod n$, if
$n | (b-a)$.

For example, if are told that $x \equiv 7 \pmod 3$, then we can conclude:
$3|(7-x)$  (by the definition of congruence), and
$7-x = 3k$ for some integer $k$ (by the definition of divides)

Hope this helps!  Please write back and let me know if you have any questions.

Best of luck,
Prof. Reitz