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CHAPTER 3
Number Theory
1. Factors or not
According to Carl Friedrich Gauss (1777-1855) mathematics is the “queen of sciences”
and number theory is the “queen of mathematics”, where “queen” stands for elevated
and beautiful. Number theory is mainly the study of the system of integers Z =
{0, ±1, ±2, . . .} and the consequences of the fact that division is not always possible
within Z. E.g. 15/3 ∈ Z but 15/4 ∈
/ Z.
Let us start with the well-known “long division” that has rather surprising consequences.
Example 1.1.
300
090
006
396
223 | 88323
669
223 | 88323
66900 (= 300 · 223)
21423
2007
21423
20070 (= 90 · 223)
1353
1338
1353
1338 (= 6 · 223)
45
45
What is really being done and what has been achieved? On the right some zeros are
filled in that are not written on the left. We now see that
88323 − 300 · 223 − 90 · 223 − 6 · 223 = 45.
In effect, multiples of 223 were subtracted from 88323 until 223 could not be subtracted anymore without running into negative numbers. The mathematical content
of “long division” is the following theorem and “long division” is just an efficient way
of finding q and r in the “Division Theorem”.
1
2
3. NUMBER THEORY
Theorem 1.2. Division Theorem. Given integers a, b, b > 0, there exist unique
integers q and r such that
a = qb + r, 0 ≤ r < b
The quotient q and the remainder r can be found by repeated subtraction of b.
The division algorithm is just an efficient way for computing q and r by repeated
subtractions.
Example 1.3. Let a = 567923 and b = 735. Find non-negative integers q and r such
that 567923 = 735q + r and 0 ≤ r < 735.
Answer: By Long Division we find q = 772 and r = 503.
Example 1.4. Let a = 3257397 and b = 7593. Find non-negative integers q and r
such that 3257397 = 7593q + r and 0 ≤ r < 7593.
Answer: By Long Division we find q = 429 and r = 0.
Definition 1.5. An integer a is even if a = 2x for some integer x, i.e., r = 0 in
Theorem 1.2. An integer b is odd if b = 2x + 1 for some integer x in Z, i.e., r = 1
in Theorem 1.2.
Proposition 1.6. Every integer is either even or odd. The product of two odd integers is odd.
Proof. Let a and b be odd integers. According to the definition of “odd integer”
there exist integers x and y such that a = 2x + 1 and b = 2y + 1. Then
ab = (2x + 1)(2y + 1) = (2x + 1)(2y) + (2x + 1) · 1
= 2(y(2x + 1)) + 2x + 1
= 2(y(2x + 1) + x) + 1.
Here z = y(2x + 1) + x is an integer by the closure properties of Z, hence ab = 2z + 1
is an odd number.
√
Theorem 1.7. 2 is not rational.
√
Proof. (Aristotle) By way of contradiction assume that 2 = a/b where either
a or b is odd. Then 2b2 = a2 , hence a2 is even and therefore a is even. This means
that a = 2a′ and substituting 2b2 = 4a′ 2 . Thus b2 = 2a′ 2 . This says that b is even, a
contradiction.
Definition 1.8.
(1) Let a, f be integers. We say that f is a factor of a if
a = f · some integer
or
a = f · a′
for some integer a′ .
(2) Let a and b be given integers. An integer f is a common factor of a and
b if f is a factor of a and a factor of b.
1. FACTORS OR NOT
3
(3) The greatest common factor of two integers a and b is the largest among
the common factors of a and b. The greatest common factor of a and b is
denoted by gcf(a, b).
Remark 1.9. In the literature it is much more common to say f divides a (or f
divides a evenly) than to say that f is a factor of a. However, the first gets confused
with other uses of “divide” and therefore we will avoid its use.
Remark 1.10. Suppose that a and b are positive integers. Then b is a factor of a
if and only if in the Division Theorem a = qb + r the remainder r = 0. Therefore
instead of saying b is a factor of a it is also said that b divides a evenly.
Factoring Rules. Let a be a positive integer given in base 10 representation. Then
the following rules are true.
(1) 2 is a factor of a if its units digit is even.
(2) 3 is a factor of a if 3 is a factor of the sum of the digits of a.
(3) 9 is a factor of a if 9 is a factor of the sum of the digits of a.
Recall that gcf(a, b) is the largest among the common factors of a and b. Therefore
gcf(a, b) can be found in the following way which is instructive but not very efficient.
Remark 1.11. Finding greatest common factors Let a and b be given integers.
(1) List the positive factors of both a and b.
(2) List the common (positive) factors of a and b.
(3) Pick the largest in the list of common factors.
Example 1.12. Finding greatest common factors. Let a = 12 and b = 28.
(1) The positive factors of 12 are 1, 2, 3, 4, 6, 12.
(2) The positive factors of 28 are 1, 2, 4, 7, 14, 28.
(3) The common factors are 1, 2, 4.
(4) The greatest common factor is 4.
Example 1.13. Let a = 240 and b = 330. The positive factors 0f 240 are
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120, 240,
and the positive factors of 330 are
1, 2, 3, 5, 6, 10, 11, 15, 22, 30, 33, 55, 66, 110, 165, 330.
The positive common factors of a and b are
1, 2, 3, 5, 6, 10, 15, 30,
and the greatest common factor of a and b is 30, gcf(240, 330) = 30.
The following lemmas will be used over and over again and should be memorized.
Lemma 1.14. Let a, b, c be integers and suppose that c = a + b. If a number f is a
factor of two of the three integers a, b, c, then f is a factor of the third.
4
3. NUMBER THEORY
Lemma 1.15. Let a and b be integers. If f is a factor of a, then f is a factor of ab.
The Euclidean Algorithm is a very beautiful and efficient way of finding the greatest common factor of two integers. It is bases on the following fact.
Lemma 1.16. Let a and b be integers and a = bq + r for some integers q and r. Then
the common factors of a and b are exactly the same as those of b and r. In particular,
gcf(a, b) = gcf(b, r).
Example 1.17. Let a = 56371 and b = 3476. Then a = 16 · b + 755, hence
gcf(56371, 3476) = gcf(3476, 755).
Further, 3476 = 4 · 755 + 456, so
gcf(56371, 3476) = gcf(3476, 755) = gcf(755, 456).
Further, 755 = 1 · 456 + 299, 456 = 1 · 299 + 157, 299 = 1 · 157 + 122, 157 = 1 · 122 + 35,
hence
gcf(56371, 3476) = gcf(755, 456) = gcf(456, 299)
= gcf(299, 157) = gcf(157, 122) = gcf(122, 35).
It is clear that 35 has the positive factors 1, 5, 7, 35 and of these only 1 is a factor of
122. Hence gcf(56371, 3476) = 1.
Theorem 1.18. Let a and b be (positive) integers. Then there exist integers u and v
such that
gcf(a, b) = ua + vb.
Consequently, every common factor of a and b is a factor of gcf(a, b). The integers
u, v and gcf(a, b) can be found efficiently using the Euclidean Algorithm described
below.
Example 1.19. Let a = 569321 and b = 347. Then gcf(69321, 347) = gcf(a, b) =
ua + vb where gcf(a, b) = 1, u = 36 and v = −59065.
Proof.
(1) 569321 =
1 · 569321 +
(2)
347 =
0 · 569321 +
(3) = (1) − 1640 · (2)
241 =
1 · 569321 −
(4) = (2) − 1 · (3)
106 = −1 · 569321 +
(5) = (3) − 2 · (4)
29 =
3 · 569321 −
(6) = (4) − 3 · (5)
19 = −10 · 569321 +
(7) = (5) − (6)
10 =
13 · 569321 −
(8) = (6) − (7)
9 = −23 · 569321 +
(9) = (7) − (8)
1 =
36 · 569321 −
0
1
1640
1641
4922
16407
21329
37736
59065
·
·
·
·
·
·
·
·
·
347
347
347
347
347
347
347
347
347
1. FACTORS OR NOT
5
The long divisions used:
569321
347
241
106
29
19
10
= 1640 · 347 + 241
=
1 · 241 + 106
=
2 · 106 + 29
=
3 · 29 + 19
=
1 · 19 + 10
=
1 · 10 +
9
=
1 ·
9 +
1
The same process can be done in a short form that only lists the essential data.
action
(1)
(2)
(3) = (1) − 1640 · (2)
(4) = (2) − 1 · (3)
(5) = (3) − 2 · (4)
(6) = (4) − 3 · (5)
(7)=(5)-(6)
(8)=(6)-(7)
(9)=(7)-(8)
569321
347
241
106
29
19
10
9
1
a
1
0
1
-1
3
-10
13
-23
36
b
0
1
- 1640
+ 1641
- 4922
+ 16407
- 21329
+ 37736
- 59065
Example 1.20. Let a = 3675 and b = 791. Then gcf(a, b) = 7 = 48 · 3675 − 223 · 791.
Proof.
action
(1)
(2)
(3) = (1) − 4 · (2)
(4) = (2) − 1 · (3)
(5) = (3) − 1 · (4)
(6) = (4) − 4 · (5)
(7) = (5) − 1 · (6)
(8) = (6) − 1 · (7)
(9) = (7) − 2 · (8)
3675
791
511
280
231
49
35
14
7
a
1
0
1
-1
2
-3
14
-17
48
b
0
1
-4
+5
-9
+ 14
-65
+ 79
-223
Definition 1.21. An integer m is a multiple of the integer a if a is a factor of m.
An integer m is a common multiple of the integers a and b if m is a multiple of
both a and b. The smallest positive common multiple of a and b is the least common
multiple of a and b denoted by lcm(a, b).
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3. NUMBER THEORY
Given two fractions a/b and c/d, the least common denominator of the fractions
is lcm(b, d).
Example 1.22. Let a = 35 and b = 21. Some multiples of a (there are infinitely
many of them) are
a = 35, 70, 105, 140, 175, . . .
and some multiples of b are
b = 21, 42, 63, 84, 105, 126, . . .
We can now see that lcm(35, 21) = 105. It is easy to see that gcf(35, 21) = 7, and we
note the curious fact that
gcf(a, b) lcm(a, b) = 7 · 105 = 735 = ab.
Theorem 1.23. For any postive integers a and b,
gcf(a, b) lcm(a, b) = ab.
Remark 1.24. To find the least common multiple of two integers a and b, we use the
Euclidean algorithm to find gcf(a, b) and then compute
lcm(a, b) =
a·b
.
gcf(a,b)
Example 1.25.
= 240 · 11 = 2640. (See Example 1.13).
(1) lcm(240, 330) = 240·330
30
56371·3476
= 56371 · 3476 = 195945596. (See Exam(2) lcm(56371, 3476) =
1
ple 1.17).
(3) lcm(569321, 347) = 569321 · 347 = 197554387. (See Example 1.19).
(4) lcm(3675, 791) = 3675·791
= 3675 · 113 = 415275. (See Example 1.20).
7
Example 1.26.
1
240
+
1
330
=
11
2640
+
6
2640
=
17
.
2640
2. The Fundamental Theorem of Arithmetic
Given a positive integer, say 113, it can always be factored as 113 = 1 · 113, in general
a = 1 · a = a · 1. This is an uninteresting “trivial” factorization.
Definition 2.1. An (positive) integer a is composite if it can be factored as a = b·c
where neither b nor c is equal to one. In other words, the (positive) integer a is
composite if it is the product of two positive integers that are both smaller than a.
A positive integer that is not composite - can only be factored trivially - is a prime
number or simply a prime.
Lemma 2.2. Let p be a prime. Then p has exactly two positive factors, namely 1
and p. Let a be any other integer. Then gcf(a, p) = 1 or gcf(a, p) = p depending on
whether p is a factor of a or not.
2. THE FUNDAMENTAL THEOREM OF ARITHMETIC
7
Example 2.3. The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
53, 59, 61, 67, 71, 73, 79, 83, 89, 91, 97 are all prime numbers. The number 123 is
composite because 123 = 3 · 41. The number 3127 is composite because 3127 = 53 · 59.
The number 1111 is composite because 1111 = 11 · 101.
Proposition 2.4. Let n be a positive integer and suppose that
n=a·b
√
√
where a and b are again positive integers. If b ≥ n, then a ≤ n.
√
√ √
√
Proof. If b ≥ n and also a > n then a · b > n · n = n while a · b = n by
hypothesis.
Corollary 2.5. If the positive integer p has no factors that are less than or equal
√
to p then p is a prime number.
√
√
Proof. If p had a factor larger than n then it would have a factor ≤ n and
these have all been eliminated.
Corollary 2.6. If the positive integer p has no prime factors which are less than
√
or equal to p then p is a prime number.
√
Proof. If p has no prime factors ≤ n then there is no composite factor ≤ n
either, since such a factor would in turn contain a prime factor of n.
Theorem 2.7. There exist infinitely many primes.
Proof. This is an example of a proof by contradiction. We make an assumption, then derive an impossibility from it and conclude that the assumption was
false.
Assume that there are only finitely many primes. List them in increasing order:
(2.8)
2, 3, 5, 7, 11, 13, 17, . . . , P,
so P is the largest prime. Study the number n obtained by multiplying all the primes
and adding 1:
n = 2 · 3 · 5 · 7 · 11 · 13 · 17 · · · p · · · P + 1.
Now 2 is not a factor of n since 2 is not a factor of 1; 3 is not a factor of n since 3 is
not a factor of 1; 5 is not a factor of n since 5 is not a factor of 1; in general, p is not
a factor or n because p is not a factor of 1. So n is either a prime itself different from
any one in the list (2.8) or it is a product of primes, each of which is not contained in
the list (2.8). This means that the list (2.8) did not contain all the primes after all,
so the assumption that there are only finitely many primes is false. Therefore, there
are infinitely many primes.
Theorem 2.9. Fundamental Theorem of Arithmetic. Every integer a > 1 can
be factored uniquely as
a = pn1 1 pn2 2 · · · pnk k ,
8
3. NUMBER THEORY
ni ≥ 1,
p 1 < p2 < · · · < pk ,
pi primes.
We will indicate a proof later. For the moment we observe that every integer a > 1
is a product of primes for the following reason. If a is itself a prime it is considered
a “product of primes”. If a is composite, then a = a1 · a2 and both a1 and a2 are
smaller than a. Now look at a1 and a2 . Factor them if they are composite, and keep
going until you arrive at primes. The process have to come to a halt because integers
cannot become smaller and smaller forever.
Example 2.10. 155771 = 539 · 289 = (11 · 49) · (17 · 17) = 11 · (7 · 7) · 17 · 17 =
11 · 7 · 7 · 17 · 17 = 72 · 11 · 172 .
Before getting into the uniqueness proof, let us see what it can do for us. The
uniqueness says that however we arrive at a factorization into a product of primes,
the result is always the same.
Example 2.11. Let a = 23 ·52 ·7. Now suppose that b is a factor of a and a = b·c. Let
mk
1
b = pn1 1 · · · pnk k and c = pm
1 · · · pk , where p1 < p2 < · · · < pk , and 0 ≤ ni , mi , be the
prime factorizations of b and c as in Theorem 2.9. Now a = b · c = p1n1 +m1 · · · pknk +mk .
By the uniqueness of the prime factorization we must have
a = 23 · 52 · 7 = pn1 1 +m1 · · · pknk +mk .
This means that p1 = 2, n1 + m1 = 3, p2 = 5, n2 + m2 = 2, p3 = 7, and n3 + m3 = 1.
We conclude that k = 3, n1 ≤ 3, n2 ≤ 2 and n3 ≤ 1.
Using this we can list the positive factors of a = 23 ·52 ·7 = 1400 (there are 4×3×2 = 24
of them) as follows:
20 · 50 · 70
20 · 51 · 71
21 · 50 · 70
21 · 51 · 71
22 · 50 · 70
22 · 51 · 71
23 · 50 · 70
23 · 51 · 71
= 1, 20 · 50 · 71 = 7, 20 · 51 · 70 = 5,
= 35, 20 · 52 · 70 = 25, 20 · 52 · 71 = 175,
= 2, 21 · 50 · 71 = 14, 21 · 51 · 70 = 10,
= 70, 21 · 52 · 70 = 50, 21 · 52 · 71 = 700,
= 4, 22 · 50 · 71 = 28, 22 · 51 · 70 = 20,
= 140, 22 · 52 · 70 = 100, 22 · 52 · 71 = 700,
= 8, 23 · 50 · 71 = 56, 23 · 51 · 70 = 40,
= 280, 23 · 52 · 70 = 200, 23 · 52 · 71 = 1400.
Corollary 2.12. Let
a = pn1 1 pn2 2 · · · pnk k ,
ni ≥ 1, p1 < p2 < · · · < pk , pi primes.
Then the positive factors of a are exactly the integers
f = pi11 pi22 · · · pikk ,
where 0 ≤ ij ≤ ni . There are (n1 + 1)(n2 + 1) · · · (nk + 1) such factors.
2. THE FUNDAMENTAL THEOREM OF ARITHMETIC
9
Example 2.13. How many positive factors does 120 posses? We factor 120 = 23 ·3·5.
There are 4 × 2 × 2 = 16 factors.
Example 2.14.
(1) The number
a = 52 74 132 176
has 3 · 5 · 3 · 7 = 315 different positive factors.
(2) The number
512 = 29
has 10 different positive factors.
(3) The number
30030 = 2 · 3 · 5 · 7 · 11 · 13
has 64 different positive factors.
We come to a crucial result.
Theorem 2.15. Let a, b, c be integers. If a is a factor of bc and gcf(a, b) = 1, then a
is a factor of c.
Proof. It is given that gcf(a, b) = 1. By Theorem 1.18 there are integers u, v
such that 1 = ua + vb. Multiplying the equation by c we obtain c = uac + vab.
Now a is a factor of uac trivially, and c is a factor of ab, so vab by hypothesis. By
Lemma 1.14 a is a factor of c.
Corollary 2.16. Let p, p1 , p2 primes. If p is a factor of p1 p2 , then p = p1 or p = p2 .
Corollary 2.17. Let p, p1 , . . . , pn be primes. If p is a factor of p1 · p2 · · · pn , then
p = p1 or p = p2 , or ... or p = pn .
Corollary 2.17 is the result that is needed to prove the uniqueness part of the Fundamental Theorem of Arithmetic. We illustrate the formal proof by an example.
Example 2.18. We find that
10500 = 2 · 2 · 3 · 5 · 5 · 5 · 7
Somebody else come up with a “mysterious” factorization
(2.19)
10500 = pn1 1 pn2 2 · · · pnk k ,
Therefore we have that
p 1 < p2 < · · · < pk .
2 · 2 · 3 · 5 · 5 · 5 · 7 = pn1 1 pn2 2 · · · pnk k
Take a factor pi of the right hand side which is then also a factor of the left hand
side. By Corollary 2.17 pi = 2 or pi = 3 or pi = 5 or pi = 7. Hence (2.19) must look
like
10500 = 2n1 3n2 5n3 7n4
and we have
2 · 2 · 3 · 5 · 5 · 5 · 7 = 2n1 3n2 5n3 7n4
10
3. NUMBER THEORY
Now 2 is a factor of the left hand side so it is a factor of the right hand side by
Corollary 2.17. This means that n1 ≥ 1 and we can cancel 2 to get
2 · 3 · 5 · 5 · 5 · 7 = 2n1 −1 3n2 5n3 7n4
The prime 2 is still a factor of the left hand side, so of the right hand side, and we
must have n1 − 1 ≥ 1. This means we can cancel another 2 and obtain
3 · 5 · 5 · 5 · 7 = 2n1 −2 3n2 5n3 7n4 .
Now there is no 2 on the left so there cannot be a 2 on the right and we conclude
that n1 − 2 = 0 or n1 = 2. We next look at the prime 3 that appears on the left. It
must appear on the right also, so n2 ≥ 1. Canceling the 3 we have
5 · 5 · 5 · 7 = 3n2 −1 5n3 7n4 .
We conclude that n2 = 1 because no 3 appears on the left and have now
5 · 5 · 5 · 7 = 5n3 7 n4 .
We have a five on the left, so n3 ≥ 1 enabling us to cancel a 5 and get
5 · 5 · 7 = 5n3 −1 7n4 .
We still have a five on the left, so n3 − 1 ≥ 1 enabling us to cancel another 5 to get
5 · 7 = 5n3 −2 7n4 .
We still have a five on the left, so n3 − 2 ≥ 1 enabling us to cancel another 5 to get
7 = 5n3 −3 7n4 .
We finally get n3 = 3 and n4 = 1 showing that the “mysterious” factorization is
identical with ours.
Proposition 2.20. Common Factors. Suppose
a = pn1 1 pn2 2 pn3 3 . . . pnk k
is the factorization of the integer a into a product of primes, and
mk
1 m2 m3
b = pm
1 p2 p3 . . . pk
is the factorization of the integer b into a product of primes, then the common factors
of a and b are all integers
Ek
1 E2 E3
f = pE
1 p2 p3 . . . pk
where Ei is less than or equal to both ni and mi . Therefore the greatest common
factor of a and b is
gcf(a, b) = pe11 pe22 pe33 . . . pekk
where ei is the lesser of ni and mi .
Proposition 2.21. Multiples.
(1) The positive multiples of an integer a are the integers a · b where b is any
positive integer.
2. THE FUNDAMENTAL THEOREM OF ARITHMETIC
11
(2) If
a = pn1 1 pn2 2 pn3 3 . . . pnk k
is the factorization of a into a product of primes, then making the exponents
ni larger produces multiples, also bringing in additional primes produces multiples.
(3) The multiples of a are all integers of the form
mk s1 s2 s3
1 m2 m3
m = pm
1 p2 p3 . . . pk q1 q2 q3 · · ·
where mi ≥ ni and si ≥ 0.
Proposition 2.22. Common Multiples. Suppose
a = pn1 1 pn2 2 pn3 3 . . . pnk k
is the factorization of the integer a into a product of primes, and
mk
1 m2 m3
b = pm
1 p2 p3 . . . pk
is the factorization of the integer b into a product of primes, then the common
multiples of a and b are all integers
Mk s 1 s 2 s 3
1 M2 M3
m = pM
1 p2 p3 . . . pk q1 q2 q3 · · ·
where Mi is greater or equal to both ni and mi and si ≥ 0. Therefore the least
common multiple of a and b is
Mk
1 M2 M3
lcm(a, b) = pM
1 p2 p3 . . . pk
where Mi is the greater of ni and mi .
Theorem 2.23. Let a and b be positive integers. Then
lcm(a, b) · gcf(a, b) = a · b.
Example 2.24. Let
a = 34 52 136 173 232 292
and let
b = 53 132 172 292
Then
lcm(a, b) = 34 53 136 173 232 292
and
gcf(a, b) = 52 132 172 292
and hence
lcm(a, b) · gcf(a, b) = (34 53 136 173 232 292 )(52 132 172 292 ) = ab.
12
3. NUMBER THEORY
3. Exercises
Exercise 3.1. For each of the following values of a and b find the unique integers q
and r such that
a = qb + r, 0 ≤ r < b.
(1) a = 723, b = 23.
(2) a = 1582, b = 231.
(3) a = 123456789, b = 12345.
(4) a = 12345, b = 123456789.
(5) a = 0, b = 13.
(6) a = 365, 904, b = 2376.
Exercise 3.2. Decide whether or not the following numbers have factors of 2, 3, 4,
5 and 9.
12345, 182734, 293045, 29480, 298754, 123456789.
Exercise 3.3. For the following integers a and b list the positive factors and find the
greatest common factor gcf(a, b).
(1) a = 115, b = 225.
(2) a = 1111, b = 333.
(3) a = 237, b = 1659.
(4) a = 17 · 23, b = 17 · 19.
Exercise 3.4. Find the greatest common factor of the following integers a and b
using the method of Euclid.
(1) a = 543, b = 113.
(2) a = 4563, b = 981.
(3) a = 451, b = 85.
(4) a = 1111, b = 11111.
(5) a = 12345, b = 1234.
Exercise 3.5. Find the greatest common factor of the following integers a and b and
find integers u, v such that gcf(a, b) = ua + vb.
(1) a = 543, b = 113.
(2) a = 4563, b = 981.
(3) a = 451, b = 85.
(4) a = 1111, b = 11111.
(5) a = 12345, b = 1234.
Exercise 3.6. Find the least common multiple of the following integers a and b by
listing common multiples and looking for the least one among these.
(1) a = 543, b = 113.
(2) a = 56, b = 98.
(3) a = 45, b = 85.
3. EXERCISES
13
(4) a = 111, b = 111.
(5) a = 123, b = 234.
Exercise 3.7. Find the least common multiple of the following integers a and b using
Theorem 1.23.
(1) a = 543, b = 112.
(2) a = 4563, b = 981.
(3) a = 451, b = 85.
(4) a = 1111, b = 11111.
(5) a = 12345, b = 1234.
Exercise 3.8. Compute the following sums and differences using least common denominators.
28
25
(1)
+
.
543 213
23
12
(2)
+
.
323 437
61
51
(3)
−
.
245 300
23
14
+
.
(4)
289 221
1
1
(5)
−
.
2599 12769
Exercise 3.9. Which ones of the following numbers are prime and which ones are
composite?
211, 373, 453, 565, 463, 371, 637, 343, 111111, 3487 + 539.
Exercise 3.10. How many positive factors do the following integers possess?
24,
6
5,
7
67,
69,
p where p is a prime,
2445,
n
7 ,
1111,
999,
n
p where p is a prime.
Exercise 3.11. Using your calculator find integers q and r such that
1238459 = q4593 + r,
0 ≤ r < 4593.
Exercise 3.12. Let x and y be (unknown) integers that are related by the equality
Why are the
(1) 5 is
(2) 3 is
(3) If y
x − 130 = 75y.
following statements true?
a factor of x.
not a factor of x.
has a factor of 13, then x has a factor of 13.
14
3. NUMBER THEORY
(4) If y is even, then x is also even.
(5) If y is odd, then x is also odd.
(6) If gcf(13, y) = 1, then 13 cannot be a factor of x.
Exercise 3.13. Verify the following statements.
(1) odd × odd = odd.
(2) even × even = even.
(3) even × odd = even.
(4) odd + odd = even.
(5) even + even = even.
Exercise 3.14. Let a = 25 × 42 × 19 × b where b is some unknown positive integer
such that gcf(2 × 5 × 19, b) = 1. Which of the following numbers are factors of a and
which ones are not?
16, 80, 95, 361, 100, 76.
Exercise 3.15. The integer 378 has 16 different positive factors and a partial list of
factors is
1, 2, 3, 6, 7, 9, 14, 18, 21, 63, 378.
Complete the list.
4. Solution to exercises
3.1 Double check your answers by computing qb + r. It has to come to be a.
3.2 12345 has factors 3 and 5,
182734 has factor 2,
293045 has factor 5,
29480 has factors 4 and 5,
298754 has a factor 2,
123456789 has a factor of 9.
3.3 gcf(115, 225) = 5, 115 = 5 × 23, 225 = 32 52 .
gcf(1111, 333) = 1, 1111 = 11 × 101, 333 = 32 37.
gcf(237, 1659) = 237, 237 = 3 × 79, 1659 = 3 × 7 × 79.
gcf(17 · 23, 17 · 19) = 17.
3.4 gcf(543, 113) = 1; gcf(4563, 981) = 9; gcf(451, 85) = 1; gcf(1111, 11111) =
1; gcf(12345, 1234) = 1.
3.5 In the following the triples (∗, ∗, ∗) are the rows of the tables in the text.
(543, 1, 0) − 4(113, 0, 1) = (91, 1, −4)
(113, 0, 1) − (91, 1, −4) = (22, −1, 5)
(91, 1, −4) − 4 (22, −1, 5) = (3, 5, −24)
(22, −1, 5) − 7 (3, 5, −24) = (1, −36, 173).
Check: −36 × 543 + 173 × 113 = 1.
4. SOLUTION TO EXERCISES
15
(4563, 1, 0) − 4(981, 0, 1) = (639, 1, −4)
(981, 0, 1) − (639, 1, −4) = (342, −1, 5)
(639, 1, −4) − (342, −1, 5) = (297, 2, −9)
(342, −1, 5) − (297, 2, −9) = (45, −3, 14)
(297, 2, −9) − 6 (45, −3, 14) = (27, 20, −93)
(45, −3, 14) − (27, 20, −93) = (18, −23, 107)
(27, 20, −93) − (18, −23, 107) = (9, 43, −200)
Check: 43 × 4563 − 200 × 981 = 9.
(451, 1, 0) − 5(85, 0, 1) = (26, 1, −5)
(85, 0, 1) − 3 (26, 1, −5) = (7, −3, 16)
(26, 1, −5) − 3 (7, −3, 16) = (5, 10, −53)
(7, −3, 16) − (5, 10, −53) = (2, −13, 69)
(5, 10, −53) − 2 (2, −13, 69) = (1, 36, −191)
(11111, 1, 0) − 10(1111, 0, 1) = (1, 1, −10)
(12345, 1, 0) − 10(1234, 0, 1) = (5, 1, −10)
(1234, 0, 1) − 246 (5, 1, −10) = (4, −246, 2461)
(5, 1, −10) − (4, −246, 2461) = (1, 247, −2471)
Check: 247 × 12345 − 2471 × 1234 = 1
3.6 These problems are very time consuming and unpleasant. The lesson is that
with more mathematics life gets much easier. The answers given are obtained in the
“advanced fashion”.
3.7 gcf(543, 112) = 1, lcm(543, 112) = 543 × 112/ gcf(543, 112) = 60 816,
gcf(4563, 981) = 9; lcm(4563, 981) = 4563 × 981/ gcf(4563, 981) = 497 367
gcf(451, 85) = 1; 451 × 85 = 38 335
gcf(1111, 11111) = 1; lcm(1111, 11111) = 1111 × 11111 = 12 344 321
gcf(12345, 1234) = 1; lcm(12345, 1234) = 12345 × 1234 = 15 233 730
3.8
28
25
+ 213
= 122171
543
851
12
733
23
+
=
323
437
7429
61
51
387
−
=
245
300
4900
14
23
573
+ 221 = 3757
289
1
1
− 12769
= 29390687
2599
3.9
211 = 211 is prime; 373 = 373 is prime; 453 = 3 × 151 is composite; 565 = 5 × 113 is
composite; 463 = 463 is prime; 371 = 7 × 53 is composite 637 = 72 13 is composite;
343 = 73 is composite; 111111 = 3 × 7 × 11 × 13 × 37 is comp[osite; 3487 + 539 =
4026 = 2 × 3 × 11 × 61 is composite.
3.11 1238459 ÷ 4593 = 269.6 . . .. So q = 269 and r = 1238459 − 269 × 4593 = 2942.
3.12
(1) 5 is a factor of x because 5 is a factor of 75y and 130.
16
3. NUMBER THEORY
(2) 3 is not a factor of x because if it were, then 3 would be a factor of x and
75, hence of 75y and x, hence of 130 which is not true.
(3) If y has a factor of 13, then x has a factor of 13. True.
(4) If y is even, then x is also even. True.
(5) If y is odd, then x is also odd. True.
(6) If gcf(13, y) = 1, then 13 cannot be a factor of x. True. Assume to the
contrary that 13 is a factor of x. Then 13 is a factor of x and 130 hence of
75y. Since gcf(13, y) = 1, we know that 13 is not a factor of y so it would
have to be a factor of 75, which is false.
3.13
(1) odd × odd = odd. This was done in class.
(2) even × even = even. Take two even numbers x and y. Being even they are
of the form x = 2x′ , y = 2y ′ . Hence xy = 2x′ · 2y ′ = 4x′ y ′, so xy even has a
factor 4 which is more than having a factor 2.
(3) even × odd = even. Let x be even and y be odd. Then x = 2x′ and
y = 2y ′ + 1. Hence xy = 2x′ y which is even.
(4) odd + odd = even. True.
(5) even + even = even. True.
3.14 Let a = 25 × 42 × 19 × b where b is some unknown positive integer such that
Since gcf(2 × 5 × 19, b) = 1 any integer having only factors 2, 5, and 19 must be a
factor of 25 × 42 × 19 = 52 × 24 × 19. So
16 : Yes, 80 Yes, 95 Yes, 361 = 192 No, 100 Yes, 76 = 4 × 19 Yes.
3.15 Look for the complementary factors 378/1 = 378, 378/2 = 189, 378/3 = 126,
378/6 = 63, 378/7 = 54, 378/9 = 42, 378/14 = 27, 378/18 = 21. Hence the complete
list is
1, 2, 3, 6, 7, 9, 14, 18, 21, 27, 42, 54, 63, 126, 189, 378.