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A brief look at Number Theory
A problem that has occasionally been used in Math 300 as a Habits of Mind problem is:
Example 1: An Open and Shut Case: In a certain school there are 100 lockers lining a long
hallway. The lockers are numbered 1, 2, 3,..., 99, 100. All are closed. Suppose that 100
students walk down the hall in single file, one after another. Suppose the first student (who we
will call “Student #1” for obvious reasons) opens every locker. The second student (i.e.
Student #2) comes along and closes every 2nd locker beginning with locker #2. (i.e. lockers #2,
4, 6,…, 98, 100). Along comes Student #3 who changes the position of every third locker; if it
is open, this student closes it; if it is closed, this student opens it. (I.e., (s)he closes locker #3,
opens locker #6, closes locker #9, etc.) Student #4 changes the “open or shut” position of
every fourth locker, and so forth, until the 100th student changes the position of locker #100.
Which lockers are open at the end of this event?
To solve the problem, you had to think about the factors of each number between 1 and 100
and have the insight that the perfect squares were the only numbers that had an odd number
of factors.
Here are two other problems that are in the same vein as our problem about the lockers.
Example 2: A Rare Occurrence: In May of 1998, Missouri and many other parts of the U.S.
were inundated by two species of cicadas. One species had a 13-year life cycle and others had
a 17-year life cycle. To appreciate that this was a rare occurrence, we ask two questions:
a) When was the most recent time before 1998 that these 13-year cicadas were heard in
Missouri? What about the 17-year cicadas?
b) When was the most recent time before 1998 that both the 13-year and the 17-year cicadas
were heard in Missouri?
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Example 3: Riding the Ferris Wheel: You and your little sister go to a carnival that has both
a large and a small Ferris wheel. You get on the large Ferris wheel at the same time your sister
gets on the small Ferris wheel. The rides begin as soon as you are both buckled into your
seats. Determine the number of seconds that will pass before you and your sister are both at
the bottom again. Note : The Ferris Wheel problem is taken from a 6th grade unit in Connected
Mathematics.
a) Assume the large wheel makes one revolution in 60 seconds and the small wheel makes
one revolution in 20 seconds.
b) Assume the large wheel makes one revolution in 50 seconds and the small wheel makes
one revolution in 30 seconds.
These last two problems call for us to investigate multiples of a number.
Note: Our basic assumption is that in discussing these concepts, we are talking about integers.
Definition: In particular, if a, b are integers (and a  0), then we say:
a is a divisor of b (factor), and b is a multiple of a if there exists an integer c such that ac = b.
 One exception involves zero. Since 0 * 3 = 0, we would say 0 is a factor of 0, but we
would never say that 0 is a divisor of 0.
Our notation is a|b.
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Here are some key divisibility facts about integers a, b, c, d.
1) If a|b, then it is also true that a|bc.
2) Transitivity: If a|b and b|c, then it is also true that a|c.
3) The 2 out of 3 Idea: Given a nonzero integer d, it is possible to find examples of numbers
a, b, c such that
a + b = c, and d factors none of the three numbers, exactly one of the
three numbers, or all three of the numbers, but it is impossible to find an example where d
factors exactly two of the three numbers. That is,
a. If a + b = c, and d|a and d|b, then d|c;
b. If a + b = c, and d|a and d|c, then d|b; and
c. If a + b = c, and d|b and d|c, then d|a.
Given two integers, a, b, we might make a list of the divisors of each number and ask “what
divisors do they have in common?” Or, “does there exist a greatest common divisor?”
Similarly, we might look at the multiples of each number and ask “do they have a least
common multiple?”
Greatest common divisor: Let a, b be integers (not both 0). Then the gcd(a, b) = d if
1. d|a and d|b; and
2. If g|a and g|b, for some integer g, then g < d.
Practice Problems:
a) gcd(15,60)
b) gcd(17,21)
Least common multiple: Let a, b be nonzero integers. Then the lcm(a, b) = m if
1. m is positive;
2. a|m and b|m; and
3. If a|c and b|c, for some positive integer c, then m < c.
Practice Problems:
a) lcm(15,60)
b) lcm(17,21)
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One of our goals is to gain insight into how many divisors a number might have and into how
we might find the greatest common divisor or least common multiple of two integers. One
approach calls for us to have a good understanding of prime numbers.
Definition: A positive integer p is a prime number if it has exactly two distinct positive
factors. A positive integer that has more than two distinct factors is said to be a composite
number. We treat 1 as a very special number that we say is a unit. In particular, 1 is not a
prime. Note that we have chosen our definitions so that 1 is not a prime. Note that 0 and 1 are
neither prime nor composite. Why do you think this is a good idea?
One of our first tasks is to investigate the prime numbers. How many primes are there under
100, 200, 1000? How are primes arranged among the positive integers?
Sieve of Eratosthenes
Eratosthenes, c. 275 – 195 BC, was a Greek scholar who made a map of the world, devised a
system of chronology, and accurately estimated the circumference of the Earth and the
distance to the moon.
One can create a list of the prime numbers by writing down the whole numbers 2,3,4,5,… and
successively checking each one. Of course, the even numbers larger than 2 are not prime, so
there is no need to check them. So, we can circle 2 and then cross out all larger multiples of 2.
Move to the next number that is not crossed out (it should be 3) and circle it. Now cross out all
multiples of 3. Move to the next number that is not crossed out (it should be 5) and circle it.
Now cross out all multiples of 5. Continue this procedure, at each step circling the first
number that is not circled or crossed out, and then crossing out all its multiples. The circled
numbers are the primes (less than the size you chose) and the numbers crossed out are
composites.
Question: Why is 2 the only even prime?
Example 4: Find the first 50 prime numbers using the Sieve of Eratosthenes. (See next page)
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5
11
21
31
41
51
61
71
81
91
101
111
121
131
141
151
161
171
181
191
201
211
221
231
241
2
12
22
32
42
52
62
72
82
92
102
112
122
132
142
152
162
172
182
192
202
212
222
232
242
3
13
23
33
43
53
63
73
83
93
103
113
123
133
143
153
163
173
183
193
203
213
223
233
243
4
14
24
34
44
54
64
74
84
94
104
114
124
134
144
154
164
174
184
194
204
214
224
234
244
5
15
25
35
45
55
65
75
85
95
105
115
125
135
145
155
165
175
185
195
205
215
225
235
245
6
16
26
36
46
56
66
76
86
96
106
116
126
136
146
156
166
176
186
196
206
216
226
236
246
7
17
27
37
47
57
67
77
87
97
107
117
127
137
147
157
167
177
187
197
207
217
227
237
247
8
18
28
38
48
58
68
78
88
98
108
118
128
138
148
158
168
178
188
198
208
218
228
238
248
9
19
29
39
49
59
69
79
89
99
109
119
129
139
149
159
169
179
189
199
209
219
229
239
249
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
The first 50 primes are:
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Having learned an algorithm for finding the primes less than a given number, we next turn our
attention to asking, given a number n, how can we tell whether n is prime or composite? Or
more generally, if a number is not prime, what process enables us to find factors of that
number other than 1 and the number itself?
If we have lots of patience, we could start checking numbers one at a time. Is 2 a 2 factor of
the number? Is 3 a factor of the number? What about 4, 5, 6, 7, 8, etc.? Actually, this process
would be horribly inefficient, because we are able to argue:
Prime number test: An integer n > 1 is either prime or it has a prime divisor d such that
1 < d < n . Thus to test whether n is prime one need only check divisibility by the primes
 n.
Example 5:
a) Is 221 prime?
b) Is 1067 prime?
Lemma: Every whole number greater than 1 is a multiple of a prime.
Encouraged by what we have already learned about prime numbers, we set our sights on the
Fundamental Theorem of Arithmetic.
Fundamental Theorem of Arithmetic: Every whole number greater than 1 can be written as
a finite product of primes, and this factorization is unique except for the order of the primes.
(In other words, every number has a unique prime factorization.)
Example 6: Find the factorizations for the following numbers:
a) 180 = 2*2*3*3*5
(Note that some primes occur more than once in this factorization.)
2 2
= 235
b) 1430
c) 80
d) 591600
e) 7 = 7
(It bothers some people to call this a factorization into a product
of one prime. They restate the fundamental theorem as below.)
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Fundamental Theorem of Arithmetic (2). Every whole number greater than 1 is either
prime or it can be written as a finite product of primes that is unique except for the order of the
primes.
One approach that can be used to keep track of our efforts to factor a number into a product of
primes is called a factor tree. At the “top” of the factor tree is the number. If the product of
two numbers is that number, we write them under the number and “connect” the two numbers
to their product with “limbs” of the factor tree. We repeat this process until finally every
number at the bottom of any set of branches is a prime. (Think of these as the “low hanging
fruit.”)
Example 7: What is the prime factorization of 5040? 10584?
Most curricula include work on factorizations and primes sometime in Grades 5-7. This work
solidifies understanding of multiplication, division, and divisibility tests, and many students
find the topic interesting. The Fundamental Theorem of Arithmetic shows that each whole
number is built out of primes, much as molecules are built out of atoms (in both cases
everything is assembled from a collection of indivisible ‘particles’.) This same idea shows up
again in the context of factoring polynomials such as x 2  5 x  6 . Binomials such as ( x  3)
and ( x  12) are simple polynomials. The Fundamental Theorem of Algebra states that, if we
consistently use complex numbers, every polynomial p (x ) can be written as a constant times a
product of binomials. These binomials can be considered as ‘prime factors’ of p (x ) .
Question: Does the list of primes 2, 3, 5, 7, 11, 13, 17, 19, … end? Well, either this list ends
or it never ends.
Theorem: There are infinitely many primes. The list of primes never ends.
We will show the list of primes never ends by using an indirect argument or sometimes called
a proof by contradiction. Let’s suppose the list of primes does end. In fact, we could write
them in order: 2,3,5, 7,11, , P , where P is the largest prime number. We could then
multiply all the primes together, add 1, and call the resulting number N:
N  (2*3*5*7 *11* * P)  1 If we divide N by any prime on the list, the remainder is 1.
Thus, N is not a multiple any prime. But that contradicts that every whole number is a
multiple of a prime! (by the Lemma preceding the Fundamental Theorem of Arithmetic).
This contradiction means that our assumption that there is a largest prime P is NOT true.
Thus, the list of primes never ends.
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Now let us return to a fairly basic question. How many factors does a given number have?

One approach is to check numbers, one at a time, up to the square root of the number.
Note: if ab = c, then either a or b is smaller than the square root of c. Why?

Another approach is to find the prime factorization of the number and then use this to
determine the answer.
Example 8: How many factors does a given number have?
a) 12  22 *3
b) 63  32 *7
c) 180  22 *32 *5
d) a  25 *36 *57
e) b  pi q j r k , p, q, r are prime numbers and i, j, k are whole numbers.
Example 9: Find a number T with exactly 20 factors such that
a) T has exactly 2 distinct prime factors
b) T has exactly 3 distinct prime factors
Example 10: Explain your answer to each of the following:
a) Is 340 a factor of 8130 ?
b) Is 340 a factor of 6330 ?
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Example 11: What is the smallest positive integer with exactly 9 proper divisors?
Note: In 1982 Jim’s oldest son was an 8th grader at Pound. He (and his teacher) encountered
this problem as a practice problem when they were preparing for the Math Counts
competition.
Divisibility Tests: If someone tells you that there are 63 shoes in the room, then you would
know something is peculiar since every even number ends with a 0, 2, 4, 6, or 8. That familiar
fact is one example of a divisibility test. We will develop divisibility tests for 2, 3, 4, 5, 6, 8,
9, and 10.
When you look at the number a = 412763895, you immediately know that a is divisible by 5.
Similarly, you know that b = 37475064 is divisible by 2 and that a is not divisible by 2. Why?
Because you know a few “divisibility tests” for whether a number is divisible by 5 (the last
digit is divisible by 5) or by 2 (the last divisible by 2). But do you know whether a or b is
divisible by 9? Or, can you tell whether b is divisible by 8?
Because we often want a quick answer to whether a number is divisible one or more small
positive integers, there are divisibility tests (some call them divisibility tricks) that tell us
whether a number is divisible by 2, 3, 4, 5, 6, 8, 9, 10, or 11. There are divisibility tests for 7,
but are more complicated and we will not discuss divisibility tests for 7 at this time.
Divisibility Tests: A number is divisible
 by 2 if and only if its last digit is 0, 2, 4, 6, or 8
 by 3 if and only if the sum of the digits is divisible by 3.
 by 4 if and only if its last two digits are a number divisible by 4
 by 5 if and only if its last digit is 0 or 5.
 by 6 if and only if the number is divisible by 2 and 3.
 by 8 if and only if its last three digits are a number divisible by 8.
 by 9 if and only if the sum of the digits is divisible by 9.
 by 10 if and only if its last digit is 0
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Example of the divisibility test for 2:
a) 144 = 10(14) + 4
Reason why the divisibility test for 2 is true: Using expanded form, any whole number N can
be written as N  10a  b where b is the last digit. But 10a  2(5a) is divisible by 2, so by the
2 out of 3 rule, N is divisible by 2 if and only if b is divisible by 2, i.e., b is 0,2,4,6, or 8.
Example of the divisibility test for 3:
a) 345 = 100(3) + 10(4) + 5 = 99(3) + 9(4) + (3 + 4 + 5)
b) 3822 = 1000(3) + 100(8) + 10(2) + 2 = 999(3) + 99(8) + 9(2) + (3 + 8 + 2 + 2)
Reason why the divisibility test for 3 is true: Consider a three digit number N which digits
abc. In expanded form N  100a  10b  c . Note that 100a  99a  a and 10b  9b  b . So,
N  100a  10b  c  (99a  9b)  (a  b  c)  9(11a  b)  (a  b  c) . Now, 9(11a  b) is
divisible by 3. Thus, N is divisible by 3 if and only if (a  b  c) is divisible by 3. But,
(a  b  c) is the sum of the digits. [Note: This reason is for a 3 digit number, however, a
similar argument shows why a number with more than 3 digits is divisible by 3.]
Example of the divisibility test for 4:
a) 144 = 100(1) + 44
b) 36 = 100(0) + 36
Reason why the divisibility test for 4 is true: Using expanded form, any whole number N can
be written as N  100a  b , where b is the last 2 digits. But 100a  4(25a) is divisible by 4.
So by the 2 out of 3 rule, N is divisible by 4 if and only if b (the last two digits) is divisible by
4.
Example of the divisibility test for 5:
a) 345 = 10(34) + 5
b) 670 = 10(67) + 0
Reason why the divisibility test for 5 is true: Adapt the reason for the divisibility test for 2.
Example of the divisibility test for 6:
a) 818
b) 2178
Reason why the divisibility test for 6 is true: Since gcd(2,3)=1 and 2 divides N and 3 divides
N, then 2*3=6 divides N.
Example of the divisibility test for 8:
a) 1248 = 1000(1) + 248
b) 3537208 = 1000(3537) + 208
Reason why the divisibility test for 8 is true: Adapt the reason for the divisibility test for 4.
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Example of the divisibility test for 9:
a) 342 = 100(3) + 10(4) + 2 = 99(3) + 9(4) + (3 + 4 + 2)
b) 3822 = 1000(3) + 100(8) + 10(2) + 2 = 999(3) + 99(8) + 9(2) + (3 + 8 + 2 + 2)
Reason why the divisibility test for 9 is true: Consider a three digit number N which digits
abc. In expanded form N  100a  10b  c . Note that 100a  99a  a and 10b  9b  b . So,
N  100a  10b  c  (99a  9b)  (a  b  c)  9(11a  b)  (a  b  c) . Now, 9(11a  b) is
divisible by 9. Thus, N is divisible by 9 if and only if (a  b  c) is divisible by 9. But,
(a  b  c) is the sum of the digits. [Note: This reason is for a 3 digit number, however, a
similar argument shows why a number with more than 3 digits is divisible by 3.]
Casting out nines. Here is a simple trick that makes checking divisibility by 3 and 9 very fast
and easy: When checking divisibility by 9, it is not actually necessary to find the sum of the
digits; we need only determine whether the sum is a multiple of 9. Thus, we can ignore any
digits that are 9, and pairs of digits that sum to 9.
Question: Is 429761 divisible by 9?
Question: Is 429761 divisible by 3?
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