Solutions to Problem Set 6
1. Section 6.2 - Problem 2
Note 45 = 9 5 is composite.
We have 174 1 (mod 5) and 174 (1)4 1 (mod 9). Thus by the Chinese remainder
theorem, 174 1 (mod 45) and so 1745 = 17 17411 17 (mod 45). Thus 45 is a pseudoprime
to the base
Math 312 Homework Set 3 Solutions
May 24, 2013
1. Suppose that n lcm(a, b) and gcd(a, b)|n. Observe that ab = lcm(a, b) gcd(a, b),
so that
a
b
ab
lcm(a, b)
=
=
gcd(a, b) gcd(a, b)
gcd(a, b)2
gcd(a, b)
Now the integer
n
gcd(a,b)
satisfies
lcm(a, b)
a
b
n
=
Math 312 Homework Set 2 Solutions
May 22, 2013
1. Base case: 34 = 81 > 64 = 43 . Now suppose that 3n > n3 . We need to show
that 3n+1 > (n + 1)3 . In fact, since 3n+1 = 3 3n it suffices to show that
3
n+1
3
n
But the inequality simplifies to
n+1
3
3
n
an
Be sure that this examination has 7 pages, including this cover.
The University of British Columbia
Final Examinations December 2009
Mathematics 312
Instructor: V. Vatsal
Time: 2.5 hours
Name:
Student Number:
Signature:
Special instructions:
1. No calcula
Math 312-Final
Name:
1. T RUE OR FALSE
1. (7 points) Suppose n is an integer with exactly 3 positive divisors. Then
n = p2 for some prime p.
A. True,
B. False.
2. (7 points) Suppose that a b (mod m) and c d (mod m) with c|a
and d|b. Then
a
b
(mod m).
c
d
Mathematics 312, Section 101. Final Exam
December 17, 2010. Instructor: Z. Reichstein.
This is a closed book exam. You can use one 8.5 11 note sheet. No other materials or
calculators are allowed. In order to receive credit for a problem you need to show
Math 312, Section 101
Final Exam
December 5, 2008
Duration: 150 minutes
Please identify yourself in ink with name, student number and signature.
Student Number:
Name:
Signature:
Do not open this test until instructed to do so! This exam should have 10 pag
Math 312 Homework Set 5 Solutions
June 4, 2013
1. If n = ab, a 6= b, and 1 < a, b < n, then the product 1 2 . . . (n 1) contains
a and b as two factors, and so ab min(n 1)! as required. This also includes the
case gcd(a, b) = 1. In case n = pk , k > 1, n
Math 312 Homework Set 7 Solutions
June 11, 2013
1. We compute the first few values as 3, 10, 101, 10202, where 10202 1084
mod 4559. We have gcd(1081, 4559) = 47, which lets us factor 4559 = 47
97. Likewise, in the case of 2, 741, 054, 681, we compute the
Math 312 Homework Set 4 Solutions
May 28, 2013
1. We will apply Chinese remainder theorem to x y and 0 rather than to x
and y, since we are guaranteed to have gcd(a, b)|x y, 0. Specifically, recall that
a
b
gcd
,
=1
gcd(a, b) gcd(a, b)
and so by Chinese r
Solutions to Problem Set 5
1. Section 5.1 - Problem 2
a) The last 3 digits of 112250 are 250, which is divisible by 53 = 125 but the last 4 digits are
2250, which is not divisible by 54 = 625. Thus the largest power of 5 dividing 11250 is
53 = 125.
b) The
1. Section 4.1 - Problem 4
If a is even, we have a 0 (mod 4) or a 2 (mod 4). As 02 = 0 and 22 = 4 0 (mod 4) in
this case a2 0 (mod 4).
Suppose a is odd. If a 1 (mod 4), then a2 12 (mod 4). If a 3 (mod 4), then a2 9
1 (mod 4) as well.
2. Section 4.1 - Pro
1. Section 3.5 - Problem 8
Let n be a positive integer. By the fundamental theorem of arithmetic we can factor n as
n = pe1 pek where the pi are distinct primes and ei 0 for each i. For each ei , by the
1
k
division algorithm write ei = 2fi + ri with ri c
1. Section 1.5 - Problem 8 (Problem 16 in the 6th edition)
Yes there are. For example, 6 | 2 3 = 6, but 6 | 2 and 6 | 3.
2. Section 1.5 - Problem 18 (Problem 26 in the 6th edition)
First we prove existence. By the division algorithm there exist integers q
1. Section 1.3 - Problem 4
We see that
1
12
=
1
2
1
12
and
+
1
23
=
n
k=1
2
3
and
1
12
+
1
23
+
1
34
=
3
4
and conjecture that
1
n
=
k(k + 1)
n+1
This is true for the base case, where n = 1, as shown above.
For the inductive step we assume the conjecture
Math 312 Homework Set 8Due 2013/6/18
June 18, 2013
1. Since id is multiplicative, f is multiplicative. This reduces the problem to
finding f (pe ). Now f is defined by the relation
(
X
1 if e = 0
pe
f (d) =
d
0 if e > 0
d|pe
When e = 0 we have pe = 1 and
Math 312 Homework Set 1 Solutions
May 17, 2013
1. We check that 13|52 so gcd(13, 52) = 13. We have 143 = 11 13 so
gcd(11, 144) = gcd(143, 144) = 1. We have 21|42, 105 but then 42
21 = 2 and
105
=
5
and
then
gcd(2,
5)
=
1
so
gcd(42,
105)
=
21.
We
have
13|2
Math 312 Quiz: Notes on Section 9.2
Let p denote a prime, and let a denote a number with 1 a p 1. For each number d which
divides p 1, we want to count the size of the set Xd which consists of the numbers a satisfying
ordp (a) = d. We wish to show that fo
MATH 312
Assignment 3
Solutions
1. Suppose
c
a
+ Z.
b
d
Then
ad + bc
=n
bd
for some n Z, so ad + bc = bdn. Since b|bdn and b|bc, we have b|ad. Since (a, b) = 1, we must have
b|d. Therefore b d. Similarly, repeating the argument with d instead of b gives d
MATH 312
Assignment 2
Solutions
1. (a) Player 2 has two options: either remove 1 or 2 matches from a pile (WLOG we can assume it is
the rst pile.) If Player 2 removes 1 match, then Player 1 can respond by removing a match from
the other pile, so there are
MATH 312
Assignment 6
Solutions
1. From Wilsons Theorem, as 101 is prime,
100! 1 (mod 101) .
We also have
102! = 102 101 100!,
So
100! + 102! 1 100 (mod 101) .
2. Suppose p is an odd prime. From Wilsons Theorem, we have
2(p 3)! (1)(2)(p 3)! (p 1)(p 2)(p 3
Solutions to Practice Problems, Math 312
1 Find all prime numbers of the form 24k+2 + 1.
Solution Since 24k+2 +1 0 modulo 5, if k 1, it follows that 24k+2 +1
is composite and hence the only prime is with k = 0 (i.e. 5).
2 A child has six dollars and fty c
Solutions to Practice Problems, Math 312
1 Prove that the fourth power of an odd integer is expressible in the
form 16n + 1 for n Z.
Solution If m is odd, we can write m = 2k + 1 for some k Z and so
m4 = (2k + 1)4 = 16k 4 + 32k 3 + 24k 2 + 8k + 1
and so w
Final Practice Problems, Math 312
1 Use mathematical induction to show that 3n < n! for n 7.
Solution We check that this inequality is satised for n = 7. Suppose
that it is true for some n 7. Then
(n + 1)! = (n + 1) n! > (n + 1) 3n > 3 3n = 3n+1 .
2 Show
MATH 312
Assignment 7
Solutions
1. The factors of 1146115723 are 20507 and 55889, so
(n) = 1146038328
and d is 85525323. This gives the plaintext
00061400
07000304
11081800
00031108
03170400
13040905
12220819
07051012
which gives, using 00 = A, 01 = B, et