Math 3 - Spring 2012
Solutions to Midterm
Problem 1
Note rst that 7 is its own inverse mod 12, since 7 7 = 49 1(mod 12). So the
rst equation is equivalent to X 5 7 11(mod 12). Similarly 7 is an invers
Math 3 - Spring 2012
Solutions to Problem Set 1
Problem 1
We give a proof by induction on n. First note that the case n = 1 holds trivially:
1 = 1(2)/2. Now suppose the claim holds for n 1, we show it
Math 3 - Spring 2012
Solutions to Problem Set 2
Problem 1
We give a proof by induction on n. Note that n = 1 holds trivially: gcd(a1 ) =
a1 = a1 1. Note also by Section 1.8. p.21 of the book, we know
Math 3 - Spring 2012
Solutions to Problem Set 3
Problem 1
(a) For m = pn1 pnk , we have the formula
1
k
pni 1 (pi 1)
i
(pni ) =
i
(m) =
1ik
1ik
So for (m) 10, the prime factorization of m can have at
Math 3 - Spring 2012
Solutions to Problem Set 4
Problem 1
One rst computes the Euler-phi function (1000) = 400. Since (7, 1000) = 1 we
have 7400 1(mod 1000). One can check that in fact 7 has order 20,
Math 3 - Spring 2012
Solutions to Problem Set 5
Problem 1
Let P = (p 1)/2. The numbers 5, 10, 15, . . . , 5P are all less than 5 p, so the
2
1
ones which are negative when reduced to lie between 1 p a
Math 3 - Spring 2012
Solutions to Problem Set 6
Problem 1:
(a) Set d = 21 c mod q (note that 2 is invertible since q is an odd prime). Let
u = x + y + d, v = y x d. Then
uv = y 2 x2 d2 2dx = y 2 x2 d2
Math 3 - Spring 2012
Solutions to Problem Set 6
Problem 1: We follow the method described on pg. 105-107. First we observe
that (67)2 + 1 = 10 449. Now we reduce 67 and 1 modulo 10 to obtain -3 and 1.
Math 3 - Spring 2012
Solutions to Problem Set 8
Problem 1: Suppose that such a triangle existed. Clearly we can assume that one
of the vertices is at the origin, let (x1 , y1 ) and (x2 , y2 ) be the o
MATH 3 2016 ASSIGNMENT 1 SOLUTIONS
Exercise 1.
The protocol consists of drawing 3 balls out of the urn (and recording their numbers). Here we do not
distinguish the order in which we draw the balls, s