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CHAPTER 5
Number Theory
5.1. Primes, Composites, and Tests for Divisibility
Definition. A counting number with exactly two di↵erent factors is called
a prime number or a prime. A counting number with more than two factors is
called a composite number, or a composite.
The Sieve of Eratosthenes is a tool used to find primes.
Start by circling 2 and then crossing out every second number. Then circle
3 and cross out every third number. Do the same for 5, 7, etc. The circled
numbers are primes and the crossed out numbers are composites.
Note. The number 1 is neither prime nor composite.
Factor trees can be used to express a composite as a product of smaller numbers
and even as a product of primes.
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5. NUMBER THEORY
We see that each of the four cases above give the same prime factorizasion.
This is an example of the following theorem:
Theorem (Fundamental Theorem of Arithmetic). Each composite number can be expressed as the product of primes in exactly one way (except
for the order of the factors).
Example.
(1)
(2)
192 = 2 · 96 = 2 · 12 · 8 = 2 · 4 · 3 · 2 · 4 = 2 · 2 · 2 · 3 · 2 · 2 · 2 = 26 · 3
360 = 9 · 40 = 32 · 8 · 5 = 32 · 23 · 5
5.1. PRIMES, COMPOSITES, AND TESTS FOR DIVISIBILITY
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Definition. Let a and b be whole numbers with a 6= 0. We say that a
divides b, and write a | b, if and only if there is a whole number x such that
ax = b. The symbol a - b means tha a does not divide b.
In other words, a | b if and only if a is a factor of b. When a | b, we say that a
is a divisor of b, a is a factor of b, b is a multiple of a, and b is divisible by a.
Example.
(1) 21 | 210 since 21 · 10 = 210.
(2) 2 | (22 · 5 · 7) since 2 · (2 · 5 · 7) = (22 · 5 · 7).
Tests for Divisibility
Theorem (Tests for Divisibility by 2, 5, and 10).
A number is divisible by 2 if and only if its ones digit is 0, 2, 4, 6, or 8.
A number is divisible by 5 if and only if its ones digit is 0 or 5.
A number is divisible by 10 if and only if its ones digit is 0.
We will prove the first part of this below.
Example. 4 | 36 and 4 | 24. But we also have then that 4 | (36 + 24) and
4 | (36 24).
This suggests the following theorem.
Theorem. Let a, m, n, and k be whole numbers with a 6= 0.
(a) If a | m and a | n, then a | (m + n).
(b) If a | m and a | n, then a | (m
(c) If a | m, then a | km.
n) for m
n.
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5. NUMBER THEORY
Proof. (of (b))
a | m =) ax = m for some whole number x.
a | n =) ay = n for some whole number y.
m
n =) m
n is a whole number. Then
ax
ay = m
n =)
a(x
y) = m
a | (m
n and x
y is a whole number =)
⇤
n).
We now can look at the proof of the test for divisibility by 2 for an arbitrary
3-digit number. The same idea used in this proof works for any number of
digits.
Proof.
Let r = a · 102 + b · 10 + c be any 3-digit number (a, b, and c are the digits
from L to R).
Note a · 102 + b · 10 = 10(a · 10 + b).
Since 2 | 10,
2 | 10(a · 10 + b) =)
2 | (a · 102 + b · 10) for any digits a and b.
Thus if 2 | c, then 2 | [10(a · 10 + b) + c], so
2 | (a · 102 + b · 10 + c), or 2 | r.
[Thus 2 | c =) 2 | r.]
Now suppose 2 | r or 2 | (a · 102 + b · 10 + c).
Since 2 | (a · 102 + b · 10),
2 | [(a · 102 + b · 10 + c)
[Thus 2 | r =) 2 | c.]
(a · 102 + b · 10)] or 2 | c.
We have shown that 2 | r if and only if 2 | c.
⇤
5.1. PRIMES, COMPOSITES, AND TESTS FOR DIVISIBILITY
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Theorem (Tests for Divisibility by 4 and 8).
A number is divisible by 4 if and only if the number represented by the last
two digits is divisible by 4.
A number is divisible by 8 if and only if the number represented by the last
three digits is divisible by 8.
Example.
2536 is divisible by 4 since 36 is.
4 - 239 since 4 - 39.
8 | 53192 since 8 | 192.
8 - 2357 since 8 - 357.
Theorem (Tests for Divisibility by 3 and 9).
A number is divisible by 3 if and only if the the sum of its digits is divisible
by 3.
A number is divisible by 9 if and only if the the sum of its digits is divisible
by 9.
Example.
3 | 43812 and 9 | 43812 since 4 + 3 + 8 + 1 + 2 = 18, 3 | 18, and 9 | 18.
3 - 454 and 9 - 454 since 4 + 5 + 4 = 13, 3 - 13, and 9 - 13.
Theorem (Test for Divisibility by 11).
A number is divisible by 11 if and only if 11 divides the di↵erence of the
sum of the digits whose place values are odd powers of 10 and the sum of
the digits whose place values are even powers of 10, with the smaller sum
subtracted from the larger.
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5. NUMBER THEORY
Example.
11 | 89969 since 8 + 9 + 9 = 26, 9 + 6 = 15, 26
11 | 3421 since 4 + 1 = 5, 3 + 2 = 5, 5
11 - 8732 since 7 + 2 = 9, 8 + 3 = 11, 11
15 = 11, and 11 | 11.
5 = 0, and 11 | 0.
9 = 2, and 11 - 2.
Theorem (Test for Divisibility by 6).
A number is divisible by 6 if and only if it is divisible by both 2 and 3.
Theorem.
A number is divisible by the product ab of two nonzero whole numbers a
and b if and only if it is divisible by both a and b, and a and b have only
1 as a common whole number factor.
Example. Is 557 prime?
Theorem (Prime Factor Test).
To test for prime factors of a p
number n, one need only search for prime
2
factors p where p  n or p  n.
Example.
p
By calculator, 557 ⇡ 23.6, 232 = 529 and 242 = 576. Thus we need check
only whether 2, 3, 5, 7, 11, 13, 17, 19, or 23 divide 557.
Since none of these primes divide 557, 557 is prime.
5.1. PRIMES, COMPOSITES, AND TESTS FOR DIVISIBILITY
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Example. Factor 5899 into primes.
2 - 5899.
5 + 8 + 8 + 9 = 30, 3 | 30 but 9 - 30, so 5889 = 3 · 1963 .
5 - 1963, 7 - 1963,
11 - (12
7), so 11 - 1963.
13 | 1963, so 5889 = 3 · 13 · 151 .
p
12 < 151 < 13, so 151 is prime since we have already tried each prime factor
less than 12.
Problem (Page 189 # 37). Ink is spilled on a bill for 36 sweatshirts. If
only the first and last digits were covered, and the other three digits were, in
order, 839 as in ?83.9?, how much did each cost?
Solution.
Strategy 10 – Use properties of numbers.
36 | bill.
Since 9 · 4 = 36 and 9 and 4 have only 1 as a common factor, 4 | bill and 9 | bill.
Thus 4 | 9?, so
? is 2 or 6 since 4 | 92 and 4 | 96.
Suppose we have ?83.92.
Since 9 | sum of the digits and 8 + 3 + 9 + 2 = 22, the first digit is 5.
Thus the bill is $583.92 and the cost of each sweatshirt is 583.92/16= $16.22 .
Now suppose we have ?83.96.
Since 9 | sum of the digits and 8 + 3 + 9 + 6 = 26, the first digit is 1.
Thus the bill is $183.96 and the cost of each sweatshirt is 183.92/16 = $5.11 .
But this cost is unrealistic, so the cost per sweatshirt is $16.22.
⇤
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5. NUMBER THEORY
5.2. Counting Factors, Greatest Common Factor, and Least
Common Multiple
Counting Factors
Example. How many factors (or divisors) does 10,800 have?
First, 10, 800 = 24 · 33 · 52 as a product of primes.
Then a divisor of 10,800 can only have 2, 3, and 5 as prime factors, with no
more than 4 twos, no more than 3 threes, and no more than 2 fives.
There are five possibilities for the number of twos to include: 0, 1, 2, 3, 4.
There are four possibilities for the number of threes to include: 0, 1, 2, 3.
There are three possibilities for the number of fives to include: 0, 1, 2.
Thus there are 5 · 4 · 3 = 60 di↵erent factors of 10,800.
Theorem. Suppose that a counting number n is expressed as a product
of distinct primes with their respective exponents, say
n = (pn1 1 )(pn2 2 ) · · · (pnmm ).
Then the number of factors of n is the product
(n1 + 1)(n2 + 1) · · · (nm + 1).
Greatest Common Factor
Definition (Greatest Common Factor).
The greatest common factor (GCF) of two (or more) nonzero whole numbers
is the largest whole number that is a factor of both (all) of the numbers. The
GCF of a and b is written GCF(a, b).
5.2. COUNTING FACTORS, GREATEST COMMON FACTOR, AND LEAST COMMON MULTIPLE
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Set Intersection Method
Example. Find GCF(36, 48).
A = {x|x is a factor of 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}
B = {x|x is a factor of 48} = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
A \ B = {x|x is a factor of 36 and 48} = {1, 2, 3, 4, 6, 12}
Thus GCF(36, 48) = 12, the largest element in A \ B.
Prime Factorization Method
Example. Find GCF(36, 60).
(1) Factor each number into primes.
36 = 22 · 32
60 = 22 · 3 · 5
(2) For the GCF, take each factor the fewest number of times it is used in any
one number.
GCF(36, 60) = 22 · 3 = 12.
Euclidean Algorithm Method
Suppose a and b are whole numbers with a
If c | a and c | b, c | a
Also, if c | b and c | a
of a and b.
b.
b, so c is a common factor of b and a
b, c | [(a
b.
b) + b)], so c | a and c is a common factor
We have proven the following theorem:
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5. NUMBER THEORY
Theorem. If a and b are whole numbers with a
GCF(a, b) = GCF(a
b, then
b, b).
Example.
GCF(98, 28) = GCF(70, 28)
= GCF(42, 28)
= GCF(14, 28)
= GCF(28, 14)
= GCF(14, 14) = 14
Recalling that division can be viewed as repeated subtraction, and that
we can shorten the above to
GCF(98, 28) = GCF(14, 28) = 14
since 14 | 28.
In general,
5.2. COUNTING FACTORS, GREATEST COMMON FACTOR, AND LEAST COMMON MULTIPLE
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Theorem (Euclidean Algoritm Method).
If a and b are whole numbers with a
b and a = bq + r where r < b, then
GCF(a, b) = GCF(r, b).
Example. Find GCF(399, 102).
The final divisor that leads to a remainder of 0 is the GCF. This is the
Euclidean Algorithm method.
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5. NUMBER THEORY
Least Common Multiple
Definition (Least Common Multiple).
The least common multiple (LCM) of two (or more) nonzero whole numbers
is the smallest nonzero whole number that is a multiple of each (all) of the
numbers. The LCM of a and b is written LCM(a, b).
Set Intersection Method
Example. Find LCM(36, 48).
A = {x|x is a multiple of 36}
= {36, 72, 108, 144, 180, 216, 252, 288, 324, 360, . . . }.
B = {x|x is a multiple of 48}
= {48, 96, 144, 192, 240, 288, 336, 384, . . . }.
A \ B = {x|x is a multiple of 36 and 48}
= {144, 288, . . . }.
LCM(36, 48) = 144, the smallest number in A \ B.
Prime Factorization Method
Example. Find GCF(36, 60).
(1) Factor each number into primes.
36 = 22 · 32
60 = 22 · 3 · 5
(2) For the GCF, take each factor the most number of times it is used in any
one number.
GCF(36, 60) = 22 · 32 · 5 = 4 · 9 · 5 = 180.
5.2. COUNTING FACTORS, GREATEST COMMON FACTOR, AND LEAST COMMON MULTIPLE
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Build-Up Method
Example. Find LCM(36, 60).
(1) Factor each number into primes.
36 = 22 · 32
60 = 22 · 3 · 5
(2) Start with one number’s prime factorization, say 36, and compare its prime
factorization to the prime factorization of the other number, 60 in this case.
We begin with
22 · 32.
Comparing with the prime factors of 60, there is no need to increase the exponents of 2 or 3. We simply need to add a factor of 5.
LCM(36, 60) = 22 ·2 ·5 = 180.
Example. Find GCF(90, 36, 54) and LCM(90, 36, 54).
90 = 2 · 32 · 5
36 = 22 · 32
54 = 2 · 33
GCF(90, 36, 54) = 2 · 32 = 18
Note.
LCM(90, 36, 54) = 22 · 33 · 5 = 4 · 27 · 5 = 540
GCF(36, 60) = 12 and LCM(36, 60) = 180.
12 · 180 = 2160 and 36 · 60 = 2160.
This is an instance of
Theorem. Let a and b be any two whole numbers. Then
GCF(a, b) ⇥ LCM(a, b) = ab.
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5. NUMBER THEORY
How many prime numbers are there?
Theorem. There is an infinite number of primes.
Proof. We us indirect reasoning.
Suppose there is a finite number of primes, say
2, 3, 5, 7, 11, . . . , p,
where p is the greatest prime. Let
N = 2 · 3 · 5 · 7 · 11 · · · p + 1.
N > 1 and N is greater than any prime.
But when N is divided by any prime, the remainder is always 1.
So N is not 1, is not prime, and is not a composite, which is impossible.
Thus there must be infinitely many primes.
⇤