Download Solutions 7

Math 55 Worksheet
Adapted from worksheets by Rob Bayer, Summer 2009.
Warmup
1. Solve the congruence 5x ≡ 16( mod 21)x = −4
Linear Congruences and the CRT


x ≡ 3 (mod 4)
1. Solve the system of congruences x ≡ 2 (mod 5) Solution: x ≡ 47 mod 140.


x ≡ 5 (mod 7)


2x ≡ 1 (mod 3)
2. Solve the system of congruences 3x ≡ 2 (mod 5)
Solution: To start this one, notice what happens if


4x ≡ 1 (mod 7)
we multiply both sides of the first equation by 2! Since 2 · 2 ≡ 1 mod 3, the first equation is the same
as x ≡ 2 mod 3. Similarly, we take care of the other equations. Here’s a website which talks about the
CRT and has a calculator to help you solve problems like this: (And plenty of other good resources too!
http://www.math.mtu.edu/mathlab/COURSES/holt/dnt/chinese3.html)
3. F Suppose we want to find all solutions to x2 ≡ 4 (mod 35)
(
(a) Show that such solution must also be a solution to the system
x2 ≡ 4 (mod 5)
x2 ≡ 4 (mod 7)
and vice-versa.
(b) Show that, based on the conditions in (a), x ≡ 2 or 3 (mod 5) and that x ≡ 2 or 5 (mod 7).
(c) Use the CRT on each of the possibilities from (b) to find all solutions to x2 ≡ 4 (mod 35).
4. Compute the following: Try out this other cool calculator: (http://ptrow.com/perl/calculator.pl)
(a) 7130 mod 11,
(b) 416 mod 10, and
16
(c) 104
mod 11.
5. F Let p be a prime number and let A = {1, 2, · · · , p − 1}. Because all elements of A are relatively prime to p,
they all have inverses (mod p). Let f : A → A send each element of A to its inverse.
(a) Show that f is injective.
(b) Show that f is its own inverse function, meaning that f (f (x)) = x for every x ∈ A.
(c) Show that, if x ∈ A and x is not equal to 1 or p − 1, then f (x) is not 1 or −1. Therefore, when searching
by trial and error for the inverse of, let’s say, 3 (mod 11), one need only check the numbers form 2 to 9.
(d) (Super bonus) Show that there is some number m (it will be different for each p) Such that m, m2 , m3 , . . .
take on all the numbers in A. In other words, every element of A is of the form mi . Try lots of examples
first. The proofs I know of this is rather tricky, but maybe you can find a new one!
6. Determine whether each of the following congruences (separately) have solutions.
(a) 2x ≡ 3 (mod 4). No, the lefthand side is even and
(c) 12x ≡ 18 (mod 30). Yes, x = 4
the righthand side odd
(d) 12x ≡ 17 (mod 30). No, since the lefthand side is
(b) 4x ≡ 1 (mod 5). Yes, x = 4
even and the right is odd
7. Prove that if ac ≡ bc (mod mc), then a ≡ b (mod m). (Notice the modulus is changing)
8. Prove that if ac ≡ bc (mod m), and if c is coprime to m, then a ≡ b (mod m).
Induction
1. Find the flaw in each of the following “proofs.”
(a) Claim: In any group of n horses, all n are the same color.
Proof. We’ll go by induction on n. If n = 1, then clearly that goat is the same color as itself. Now, suppose
the claim holds for groups of k horses. We’ll show it holds also for groups of k + 1. Suppose we have some
group of k + 1 horses, numbered 1 through k + 1. By the IH, horses 1 through k are all the same color and
2 through k + 1 all are too. Since there is some horse in the overlap of thses subsets, all k + 1 must share
the same color. Thus, by induction, the result holds for all n.
The flaw is that the inductive step doesn’t work for arbitrary k. For the ’overlap’ between the two sets to
be nonempty, you need that k ≥ 2, and but then the induction is doomed to fail, because we’ll never be
able to prove that P (1) implies P (2). There’s an example in the book that has a similar problem.
(b) Claim: All integers are perfect squares.
Proof. Clearly 1 is a perfect square. Suppose the claim works for integers up to and including k. Then if we
write k + 1 = ab, the IH tells us that a = m2 and b = l2 for some integers m, l. Thus, k + 1 = m2 l2 = (ml)2
and k + 1 is a perfect square. Thus, by induction we conclude that all integers are perfect squares.
The flaw is that the inductive step doesn’t work for arbitrary k. For the ’overlap’ between the two sets to
be nonempty, you need that k ≥ 2, and but then the induction is doomed to fail, because we’ll never be
able to prove that P (1) implies P (2). There’s an example in the book that has a similar problem. Similar
to the part a)
2. (An induction-based logic puzzle!) An Induction-based Puzzle The 55 students of math 55 have found a large
bag of gold containing 555 gold pieces. In order to divide it up, they decide to use the following system: at the
first stage, the fiercest student proposes a distribution of the gold pieces and all students (including the fiercest)
vote on it. If at least half (that is, 55/2) vote in favor, they use that distribution. Otherwise, the student
is permanently banned from Evans Hall and the next fiercest student proposes a new distribution which now
requires at least half of the remaining students’ (ie, 54/2) votes to pass, etc, etc. You may assume students all
prefer more gold to less, prefer not to be banished from Evans, and all else being equal like to kick other people
out. How much gold does the ercest student get? Hint: It’s a lot.