tomit4/discrete_mathematics_with_applications
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

:constructdion: Fin 4.4

Browse source
This commit is contained in:
tomit4 2026-06-08 06:30:46 -07:00
parent 5941828233
commit 65852f69e4
3 changed files with 740 additions and 4 deletions
Download patch file Download diff file Expand all files Collapse all files

1
appendix_a.txt Normal file
View file

@ -0,0 +1 @@
895

727
chapter_4/exercises.md
View file

@ -3021,32 +3021,62 @@ Page 220
Give a reason for your answer in each of 1-13. Assume that all variables
represent integers.
1. is $52$ divisible by $13$?
1. Is $52$ divisible by $13$?
Yes $52 = 13 \cdot 4$
2. Does $7 \mid 56$?
Yes $56 = 7 \cdot 8$
3. Does $5 \mid 0$?
Yes, $0 = 5 \cdot 0$
4. Does $3$ divide $(3k + 1)(3k + 2)(3k + 3)$?
Yes $3 | 3(3k + 1)(3k + 2)(k + 1)$
5. Is $6m(2m + 10)$ divisible by $4$?
Yes $6m(2m + 10) = 12m^2 + 60m = 4 \cdot (3m^2 + 15m)$
6. Is $29$ a multiple of $3$?
No, $\dfrac{29}{3} \approx 9.666666\dots$ which is not an integer.
7. Is $-3$ a factor of $66$?
Yes, $66 = -3(-22)$
8. Is $6a(a + b)$ a multiple of $3a$?
Yes, $6a(a + b) = 3a(2)(a + b)$
9. Is $4$ a factor of $2a \cdot 34b$?
Yes. $2a \cdot 34b = 68ab = 4(17ab) $
10. Does $7 \mid 34$?
No $\dfrac{34}{7} = 4 + \dfrac{6}{7}$ which is not an integer.
11. Does $13 \mid 73$?
No $\dfrac{73}{13} = 5 + \dfrac{8}{13}$ which is not an integer.
12. If $n = 4k + 1$, does $8$ divide $n^2 - 1$?
Yes.
$$ n^2 - 1 = (4k + 1)^2 - 1 = 16k^2 + 8k + 1 - 1 = 16k^2 + 8k = 8(2k^2 + k) $$
13. If $n = 4k + 3$, does $8$ divide $n^2 - 1$?
Yes.
$$ n^2 - 1 = (4k + 3)^2 - 1 = 16k^2 + 24k + 9 - 1 = 16k^2 + 24k + 8 = 8(2k^2 + 3k + 1) $$
14. Fill in the blanks in the following proof that for all integers $a$ and $b$,
if $a \mid b$ then $a \mid (-b)$.
@ -3062,64 +3092,354 @@ both $-1$ and $r$ are integers. Thus, by substitution, $-b = at$, where $t$ is
an integer, and by the definition of divisibility, __ (d) __, as was to be
shown.
a. $a \mid b$
b. $b = a \cdot r$
c. $-r$
d. $a | (-b)$
Prove statements 15 and 16 directly from the definition of divisibility.
15. For all integers $a$, $b$, and $c$, if $a \mid b$ and $a \mid c$ then
$a \mid (b + c)$.
**Proof:**
Suppose $a$, $b$, and $c$ are any integers such that $a \mid b$ and $a \mid c$
By definition of divisibility, $b = ar$ and $c = as$ for some integers $r$ and
$s$.
Then:
$$ b + c = ar + as = a(r + s) $$
Let $t = r + s$, where $t$ is an integer because the sum of integers are
integers. And thus $b + c = a \cdot t$. By the definition of divisibility then
$a \mid (b + c)$.
Q.E.D.
16. For all integers $a$, $b$, and $c$, if $a \mid b$ then $a \mid c$ then
$a \mid (b - c)$.
**Proof:**
Suppose $a$, $b$, and $c$ are any integers such that $a \mid b$ and $a \mid c$
By definition of divisibility, $b = ar$ and $c = as$ for some integers $r$ and
$s$.
Then:
$$ b - c = ar - as = a(r - s) $$
Let $t = r - s$, where $t$ is an integer because the difference of integers are
integers. And thus $b - c = a \cdot t$. By the definition of divisibility then
$a \mid (b - c)$.
Q.E.D.
17. For all integers $a$, $b$, $c$, and $d$, if $a \mid c$ and $b \mid d$ then
$ab \mid cd$.
**Proof:**
Suppose $a$, $b$, $c$, and $d$ are any integers such that $a \mid c$ and
$b \mid d$.
By definition of divisibility, $c = ar$ and $d = bs$ for some integers $r$ and
$s$.
Then:
$$ cd = (ar)(bs) = ab(rs) $$
Let $t = rs$, where $t$ is an integer because the product of integers are
integers. Then $cd = ab \cdot t$. By the definition of divisibility then
$ab \mid cd$.
Q.E.D.
18. Consider the following statement: The negative of any multiple of $3$ is a
multiple of $3$.
a. Write the statement formally using a quantifier and a variable.
$$ \forall x \in \mathbb{Z}((3 \mid x) \to (3 \mid -x)) $$
b. Determine whether the statement is true or false and justify your answer.
**Proof:**
Suppose $x$ is any integer such that $3 \mid x$.
By definition of a multiple, $x = 3k$ for some integer $k$.
Then:
$$ -x = -(3k) = 3(-k) $$
Then $-k$ is an integer because the product of integers are integers. Therefore,
by the definition of divisibility, $3 \mid -x$.
Q.E.D.
19. Show that the following statement is false: For all integers $a$ and $b$, if
$3 \mid (a + b)$ then $3 \mid (a - b)$.
**Proof by Counterexample:**
Let $a = 1$ and $b = 2$.
Then:
$$ a + b = 1 + 2 = 3 $$
So $3 \mid 3$ is true. But, then:
$$ a - b = 1 - 2 = -1 $$
$$ 3 \nmid -1 $$
Thus, for the given $a$ and $b$, $3 \mid (a + b)$, but $3 \nmid (a - b)$.
Therefore the statement is false.
Q.E.D.
For each statement in 20-32, determine whether the statement is true or false.
Prove the statement directly from the definitions if it is true, and give a
counterexample if it is false.
20. The sum of any three consecutive integers is divisible by $3$.
**Proof:**
Suppose $n$ is any integer.
Then:
$$ n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1) $$
Let $t = n + 1$, where $t$ is an integer because the sum of integers are
integers. Then $n + (n + 1) + (n + 2) = 3t$. Therefore, by the definition of
divisibility, $3 \mid n + (n + 1) + (n + 2)$.
Q.E.D.
21. The product of any two even integers is a multiple of $4$.
**Proof:**
Suppose $x$ and $y$ are any two even integers.
Since $x$ and $y$ are even integers, $x = 2k$ and $y = 2m$ where $k$ and $m$ are
some integers.
Then:
$$ xy = (2k)(2m) = 4km $$
Let $t = km$, where $t$ is an integer because the product of integers are
integers. Then $xy = 4t$. Therefore, by the definition of divsibility,
$4 \mid xy$.
Q.E.D.
22. A necessary condition for an integer to be divisible by $6$ is that it be
divisible by $2$.
_Quick Note:_
Recall that "$R$ ... necessary condition for $S$" means $S \to R$. Thus this is
saying:
If an integer is divisible by $6$, then it is divisible by $2$.
**Proof:**
Suppose $x$ is any integer and $6 \mid x$.
Then $x = 6k$ where $k$ is some integer.
Then:
$$ x = 6k = 2(3k) $$
Let $t = 3k$ where $t$ is an integer because the product of integers is
integers. Then $x = 2t$. Therefore by the definition of divisibility,
$2 \mid x$.
Q.E.D.
23. A sufficient condition for an integer to be divisible by $8$ is that it be
divisible by $16$.
_Quick Note:_
"$R$ is a sufficient condition for $S$" means "if $R$ then $S$."
This reads then as:
If an integer is divisible by $16$, then it is divisible by $8$.
**Proof:**
Let $x$ be any integer and $16 \mid x$.
Since $16 \mid x$, $x = 16k$ where $k$ is some integer.
Then:
$$ x = 16k = 8(2k) $$
Let $t = 2k$ where $t$ is an integer by the product of integers. Then $x = 8t$,
and thus $8 \mid x$ by the definition of divisibility.
Q.E.D.
24. For all integers $a$, $b$, and $c$, if $a \mid b$ and $a \mid c$ then
$a \mid (2b - 3c)$.
**Proof:**
Suppose $a$, $b$, and $c$ are any integers where $a \mid b$ and $a \mid c$.
Since $a \mid b$ and $a \mid c$, $b = ak$ and $c = am$ where $k$ and $m$ are
some integers.
Then:
$$ 2b - 3c = 2(ak) - 3(am) $$
$$ \quad = a(2k - 3m) $$
Let $t = 2k - 3m$, where $t$ is an integer by the difference and product of
integers. Then $2b - 3c = at$. Thus $a \mid 2b - 3c$ by the definition of
divisibility.
Q.E.D.
25. For all integers $a$, $b$, and $c$, if $a$ is a factor of $c$ and $b$ is a
factor of $c$ then $ab$ is a factor of $c$.
**Proof by Counterexample:**
Let $a = 4$, $b = 8$, and $c = 8$.
Then $a \mid c$ is $4 \mid 8$ and $b \mid c$ is $8 \mid 8$, but $ab \mid c$ is
$32 \mid 8$, which is false since $32 \nmid 8$. Therefore this statement is
false.
Q.E.D.
26. For all integers $a$, $b$, and $c$, if $ab \mid c$ then $a \mid c$ and
$b \mid c$.
**Proof:**
Suppose $a$, $b$, and $c$ are integers and $ab \mid c$.
Since $ab \mid c$, $c = abk$ where $k$ is some integer.
Then:
$$ c = abk = a(bk) $$
And:
$$ c = abk = b(ak) $$
Let $t = bk$ and $u = ak$ where $t$ and $u$ are integers by the product of
integers. Then $c = at$ and $c = bu$. Therefore, by the definition of
divisibility, $a \mid c$ and $b \mid c$.
Q.E.D.
27. For all integers $a$, $b$, and $c$, if $a \mid (b + c)$ then $a \mid b$ or
$a \mid c$.
**Proof by Counterexample:**
Let $a = 3$, $b = 4$, and $c = 5$.
Then $a \mid (b + c)$ is $3 \mid 9$, which is true. Then, however, $a \mid b$ is
$3 \mid 4$, which is false since $3 \nmid 4$ and $a \mid c$ becomes $3 \mid 5$,
which is also false since $3 \nmid 5$.
Thus for the given $a$, $b$, and $c$, $a \mid (b + c)$, but $a \nmid b$ and
$a \nmid c$. Therefore the statement is false.
Q.E.D.
28. For all integers $a$, $b$, and $c$, if $a \mid bc$ then $a \mid b$ or
$a \mid c$.
**Proof by Counterexample:**
Let $a = 6$, $b = 2$, $c = 3$.
Then $a \mid bc$ is $6 \mid 6$, which is true. Then, however, $a \mid b$ is
$6 \mid 2$, which is false since $6 \nmid 2$ and $a \mid c$ is $6 \mid 3$ which
also false since $6 \nmid 3$.
Thus for the given $a$, $b$, and $c$, $a \mid bc$ is true, but then $a \nmid b$
and $a \nmid c$. Therefore the statement is false.
Q.E.D.
29. For all integers $a$ and $b$, if $a \mid b$ then $a^2 \mid b^2$.
**Proof:**
Suppose $a$ and $b$ are any integers where $a \mid b$.
Since $a \mid b$, then $b = ak$ for some integer $k$.
Then:
$$ b^2 = (ak)^2 $$
$$ b^2 = a^2k^2 $$
Let $t = k^2$ where $t$ is an integer by the product of integers. Then
$b^2 = a^2t$. Therefore, by the definition of divisibility, $a^2 \mid b^2$.
Q.E.D.
30. For all integers $a$ and $n$, if $a \mid n^2$ and $a \leq n$ then
$a \mid n$.
**Proof by Counterexample:**
Let $a = -36$ and $n = 6$.
Then $a \mid n^2$ is $-36 \mid 36$, which is true and $a \leq n$ is
$-36 \leq 6$, which is also true. Then, however, $a \mid n$ is $-36 \mid 6$,
which is false since $-36 \nmid 6$.
Thus for the given $a$ and $n$, $a \mid n^2$ and $a \leq n$, but $a \nmid n$.
Therefore this statement is false.
Q.E.D.
31. For all integers $a$ and $b$, if $a \mid 10b$ then $a \mid 10$ or
$a \mid b$.
**Proof:**
Let $a = 4$, $b = 2$.
Then $a \mid 10b$ is $4 \mid 20$ is true, but then $a \mid 10$ is $4 \mid 10$ is
false and $a \mid b$ is $4 \mid 2$ is false.
Thus for the given $a$ and $b$, $a \mid 10b$, but then $a \nmid 10$ and
$a \nmid b$. Therefore the statement is false.
Q.E.D.
32. A fast-food chain has a contest in which a card with numbers on it is given
to each customer who makes a purchase. If some of the numbers on the card
add up to $100$, then the customer wins $100. A certain customer receives a
@ -3129,19 +3449,50 @@ counterexample if it is false.
Will the customer win $100? Why or why not?
No, each of the given numbers is divisible by $3$, but $3 \nmid 100$, therefore
the sum of any of the given numbers can never equal $100.
33. Is it possible to have a combination of nickels, dimes, and quarters that
add up to $4.72? Explain.
No, nickels, dimes and quarters represent 5¢, 10¢, and 25¢ respectively. They
have a common divisor of 5, but $4.72 or 472¢, is not divisible by 5, and so it
is not possible to have a combination of nickels, dimes and quarters that will
add up to $4.72.
34. Consider a string consisting of _a_'s, _b_'s, and _c_'s where the number of
_b_'s is three times the number of _a_'s and the number of _c_'s is five
times the number of _a_'s. Prove that the length of the string is divisible
by $3$.
**Proof:**
Suppose $n$ is any integer that represents the number of _a_ characters in a
string.
Suppose also that $3n$ is the number of _b_ characters in the string and $5n$
represents the number of _c_ characters in the string.
Let $L$ be the length of the string.
Then the length of the string is:
$$ L = n + 3n + 5n = 9n = 3(3n) $$
Let $t = 3n$ where $t$ is an integer by the product of integers. Then $L = 3t$.
Thus, by the definition of divisibility, $3 \mid L$.
Q.E.D.
35. Two athletes run a circular track at a steady pace so that the first
completes one round in 8 minutes and the second in 10 minutes. If they both
start from the same spot at 4 P.M., when will be the first time they return
to the start?
We are looking at the LCM (least common multiple) of both $8$ and $10$ in this
case, which is $40$. Then the two athletes will return to the start for the
first time at 4:40 P.M.
36. It can be shown (see exercises 44-48) that an integer is divisible by 3 if,
and only if, the sum of its digits is divisible by 3; an integer is
divisible by 9 if, and only if, the sum of its digits is divisible by 9; an
@ -3152,66 +3503,310 @@ Will the customer win $100? Why or why not?
a. 637,425,403,705,125
Divisible by 3?
$$ 3 \stackrel{?}{\mid} (6 + 3 + 7 + 4 + 2 + 5 + 4 + 0 + 3 + 7 + 0 + 5 + 1 + 2 + 5) $$
$$ 3 \stackrel{?}{\mid} 54 $$
Yes, $\dfrac{54}{3} = 18$
Divisible by 4?
$$ 4 \stackrel{?}{\mid} 25 $$
No, $\dfrac{25}{4} = 6 + \dfrac{1}{4}$
Divisible by 5?
Last digit a $5$ or $0$?
Yes, last digit is $5$.
Divisible by 9?
$$ 9 \stackrel{?}{\mid} (6 + 3 + 7 + 4 + 2 + 5 + 4 + 0 + 3 + 7 + 0 + 5 + 1 + 2 + 5) $$
$$ 9 \stackrel{?}{\mid} 54 $$
Yes, $\dfrac{54}{9} = 6$
b. 12,858,306,120,312
Divisible by 3?
$$ 3 \stackrel{?}{\mid} (1 + 2 + 8 + 5 + 8 + 3 + 0 + 6 + 1 + 2 + 0 + 3 + 1 + 2) $$
$$ 3 \stackrel{?}{\mid} 42 $$
Yes $\dfrac{42}{3} = 14$
Divisible by 4?
$$ 4 \stackrel{?}{\mid} 12 $$
Yes $\dfrac{12}{4} = 3$
Divisible by 5?
Last digit a $5$ or $0$?
No.
Divisible by 9?
$$ 9 \stackrel{?}{\mid} (1 + 2 + 8 + 5 + 8 + 3 + 0 + 6 + 1 + 2 + 0 + 3 + 1 + 2) $$
$$ 9 \stackrel{?}{\mid} 42 $$
No, $\dfrac{42}{9} = 4 + \dfrac{2}{3}$.
c. 517,924,440,926,512
Divisible by 3?
$$ 3 \stackrel{?}{\mid} (5 + 1 + 7 + 9 + 2 + 4 + 4 + 4 + 0 + 9 + 2 + 6 + 5 + 1 + 2) $$
$$ 3 \stackrel{?}{\mid} 61 $$
No, $\dfrac{61}{3} = 20 + \dfrac{1}{3}$
Divisible by 4?
$$ 4 \stackrel{?}{\mid} 12 $$
Yes $\dfrac{12}{4} = 3$
Divisible by 5?
Last digit a $5$ or $0$?
No.
Divisible by 9?
$$ 9 \stackrel{?}{\mid} (5 + 1 + 7 + 9 + 2 + 4 + 4 + 4 + 0 + 9 + 2 + 6 + 5 + 1 + 2) $$
$$ 9 \stackrel{?}{\mid} 61 $$
No, $\dfrac{61}{9} = 6 + \dfrac{7}{9}$
d. 14,328,083,360,232
Divisible by 3?
$$ 3 \stackrel{?}{\mid} (1 + 4 + 3 + 2 + 8 + 0 + 8 + 3 + 3 + 6 + 0 + 2 + 3 + 2) $$
$$ 3 \stackrel{?}{\mid} 45 $$
Yes $\dfrac{45}{3} = 15$
Divisible by 4?
$$ 4 \stackrel{?}{\mid} 32 $$
Yes $\dfrac{32}{4} = 8$
Divisible by 5?
Last digit a $5$ or $0$?
No.
Divisible by 9?
$$ 9 \stackrel{?}{\mid} (1 + 4 + 3 + 2 + 8 + 0 + 8 + 3 + 3 + 6 + 0 + 2 + 3 + 2) $$
$$ 9 \stackrel{?}{\mid} 45 $$
Yes, $\dfrac{45}{9} = 5$
37. Use the unique factorization theorem to write the following integers in
standard factored form.
a. 1,176
$$ 1176 = 8 \cdot 147 = 8 \cdot 7 \cdot 21 = 2 \cdot 2 \cdot 2 \cdot 7 \cdot 7 \cdot 3 $$
$$ \quad = 2^3 \cdot 7^2 \cdot 3 $$
b. 5,733
$$ 5733 = 49 \cdot 117 = 7 \cdot 7 \cdot 9 \cdot 13 = 7^2 \cdot 3^2 \cdot 13 $$
$$ \quad = 7^2 \cdot 3^2 \cdot 13 $$
c. 3,675
$$ 3675 = 25 \cdot 147 = 5 \cdot 5 \cdot 7 \cdot 21 = 5^2 \cdot 7 \cdot 7 \cdot 3 $$
$$ \quad = 5^2 \cdot 7^2 \cdot 3 $$
38. Let $n = 8,424$.
a. Write the prime factorization for $n$.
$$ 8,424 = 24 \cdot 351 = 6 \cdot 4 \cdot 3 \cdot 117 $$
$$ \quad = 3 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 39 $$
$$ \quad = 3^3 \cdot 2^3 \cdot 39 $$
$$ \quad = 3^3 \cdot 2^3 \cdot 3 \cdot 13 $$
$$ \quad = 3^4 \cdot 2^3 \cdot 13 $$
b. Write the prime factorization for $n^5$.
$$ n = 3^4 \cdot 2^3 \cdot 13 $$
$$ n^5 = (3^4 \cdot 2^3 \cdot 13)^5 $$
0$ n^5 = 3^{20} \cdot 2^{15} \cdot 13^5 $$
c. Is $n^5$ divisible by 20? Explain.
$$ 20 = 2^2 \cdot 5 $$
So in order for $20 \mod n^5$, then $2^2 \mid n^5$ and $5 \mid n^5$,
$2^2 \mid n^5$ is possible since one of the prime factorizations of $n^5$ is
$2^{15}$ but none of the prime factorizations of $n^5$ is $5$. Therefore
$20 \nmid n^5$.
d. What is the least positive integer $m$ so that $8,424 \cdot m$ is a perfect
square?
To make $8424 \cdot m$ square, all prime exponents must be even. Let's examine
our prime factorization form of $n$:
$$ n = 3^4 \cdot 2^3 \cdot 13 $$
Only our last two terms need an additional factor each to make their exponents
even, so to make $8424m$ a perfect square, $m = 2 \cdot 13 = 26$, this will
make:
$$ 8424m = 3^4 \cdot 2^4 \cdot 13^2 $$
39. Suppose that in standard factored form
$a = p_1^{e_1}p_2^{e_2} \dots p_k^{e^k}$, where $k$ is a positive integer;
$a = p_1^{e_1}p_2^{e_2} \dots p_k^{e_k}$, where $k$ is a positive integer;
$p_1, p_2, \dots, p_k$ are prime numbers; and $e_1, e_2, \dots, e_k$ are
positive integers.
a. What is the standard factored form for $a^3$?
$$ a^3 = p_1^{3e_1}p_2^{3e_2} \dots p_k^{3e_k} $$
b. Find the least positive integer $k$ such that
$2^4 \cdot 3^5 \cdot 7 11^2 \cdot k$ is a perfect cube (that is, it equals an
integer to the third power). Write the resulting product as a perfect cube.
$2^4 \cdot 3^5 \cdot 7 \cdot 11^2 \cdot k$ is a perfect cube (that is, it equals
an integer to the third power). Write the resulting product as a perfect cube.
$$ k = 2^2 \cdot 3 \cdot 7^2 \cdot 11 $$
40.
a. If $a$ and $b$ are integers and $12a = 25b$, does $12 \mid b$? does
$25 \mid a$? Explain.
Because $12a = 25b$, the unique factorization theorem guarantees that the
standard factored forms of $12a$ and $25b$ must be the same. Thus $25b$ contains
the factors $2^2 \cdot 3 (= 12)$. But since neither $2$ nor 3$ divides $25$, the
factors $2^2 \cdot 3$ must all occur in $b$, and hence $12 \mid b$. Similarly,
$12a$ contains the factors $5^2 = 25$, and since $5$ is not a factor of $124,
the factors $5^2$ must occur in $a$. So $25 \mid a$.
b. If $x$ and $y$ are integers and $10x = 9y$, does $10 \mid y$? does
$9 \mid x$? Explain.
Because $10x = 9y$, the unique factorization theorem guarantees that the
standard factored forms of $10x$ and $9y$ must be the same. Thus $10x$ contains
the factors $3^2$. But since $3$ does not divide $10$, the factor $3^2$ must all
occur in $x$, and hence $9 \mid x$. Similary $10x$ contains factors $2 \cdot 5$,
and since neither $5$ nor $2$ are factors of $9$, the factors $2 \cdot 5$ must
occur in $y$. So $10 \mid y$.
41. How many zeros are at the end of $45^8 \cdot 88^5$? Explain how you can
answer this question without actually computing the number. (_Hint:_
$10 = 2 \cdot 5$.)
If we find the standard factored form of $45^8 \cdot 88^5$ and then find the
total amount of $10$'s, this should tell us how many zeros are at the end of
$45^8 \cdot 88^5$.
$$ 45^8 \cdot 88^5 = (9 \cdot 5)^8 \cdot (4 \cdot 22)^5 $$
$$ \quad = (3^2 \cdot 5)^8 \cdot (2^2 \cdot 2 \cdot 11)^5 $$
$$ \quad = (3^2 \cdot 5)^8 \cdot (2^3 \cdot 11)^5 $$
$$ \quad = 3^{16} \cdot 5^8 \cdot 2^{15} \cdot 11^5 $$
The total amount of $10$'s we can find is the maximum amount of $5 \cdot 2$'s we
can factor from this expression, which is $8$ (while we have 15 $2$'s, we only
have 8 $5$'s).
Therefore there are $8$ zeros in $45^8 \cdot 88^5$.
42. If $n$ is an integer and $n > 1$, then $n!$ is the product of $n$ and every
other positive integer that is less than $n$. For example,
$5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$.
a. Write $6!$ in standard factored form.
$$ 6! = 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 $$
$$ 6! = 3 \cdot 2 \cdot 5 \cdot 2 \cdot 2 \cdot 3 \cdot 2 $$
$$ 6! = 2^4 \cdot 3^2 \cdot 5 $$
b. Write $20!$ in standard factored form.
Writing this out would be exhaustive, instead let's use
[Legendre's formula](https://en.wikipedia.org/wiki/Legendre%27s_formula).
$$ 20! = 2^a \cdot 3^b \cdot 5^c \cdot 7^d \cdot 11^e \cdot 13^f \cdot 17^g \cdot 19^h $$
Take the floor of every possible prime factor, and those are the exponents of
each.
$$ \left\lfloor \frac{20}{2} \right\rfloor + \left\lfloor \frac{20}{4} \right\rfloor + \left\lfloor \frac{20}{8} \right\rfloor + \left\lfloor \frac{20}{16} \right\rfloor $$
$$ \quad = 10 + 5 + 2 + 1 = 18 $$
$$ 2^{18} $$
$$ \left\lfloor \frac{20}{3} \right\rfloor + \left\lfloor \frac{20}{9} \right\rfloor $$
$$ 6 + 2 = 8 $$
$$ 3^8 $$
$$ \left\lfloor \frac{20}{5} \right\rfloor $$
$$ 5^4 $$
$$ \left\lfloor \frac{20}{7} \right\rfloor $$
$$ 7^2 $$
$$ \boxed{20! = 2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19} $$
c. Without computing the value of $(20!)^2$ determine how many zeros are at the
end of this number when it is written in decimal form. Justify your answer.
We can square the answer for $20!$ from part b:
$$ (20!)^2 = (2^{18} \cdot 3^8 \cdot 5^4 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19)^2 $$
$$ \quad = 2^{36} \cdot 3^{16} \cdot 5^8 \cdot 7^4 \cdot 11^2 \cdot 13^2 \cdot 17^2 \cdot 19^2 $$
Then take all combinations of 2 and 5, and take the minimum exponent between
them as this will tell us how many zeros are at the end of this number when it
is written in decimal form.
$$ \min(36, 8) = 8 $$
So there are 8 zeros at the end of $(20!)^2$.
43. At a certain university 2/3 of the mathematics students and 3/5 of the
computer science students have taken a discrete mathematics course. The
number of mathematics students who have taken the course equals the number
@ -3220,6 +3815,49 @@ end of this number when it is written in decimal form. Justify your answer.
possible number of mathematics students and the least possible number of
computer science students at the university?
Let $M$ be the number of mathematics students and $C$ be the number of computer
science students.
Given:
$$ \frac{2}{3}M = \frac{3}{5}C $$
Set them equal:
$$ \frac{2}{3}M = \frac{3}{5}C = x $$
So:
$$ M = \frac{3}{2}x $$
$$ C = \frac{5}{3}x $$
To make both integers, $x$ must be a multiple of $6$:
$$ x = 6k $$
Then:
$$ M = 9k $$
$$ C = 10k $$
Now use the given condition:
$$ M \geq 100 $$
$$ 9k \geq 100 \Rightarrow k \geq 12 $$
Smallest $k = 12$:
$$ M = 9 \cdot 12 = 108 $$
$$ C = 10 \cdot 12 = 120 $$
Math Students: 108
Computer Science Students: 120
**Definition:** Given any nonnegative integer $n$, the **decimal
representation** of $n$ is an expression of the form
@ -3238,14 +3876,87 @@ $$ n = d_k \cdot 10^k + d_{k + 1} \cdot 10^{k + 1} + \dots + d_2 \cdot 10^2 + d_
nonnegative integer $n$ ends in $d_0$, then $n = 10m + d_0$ for some integer
$m$.)
**Proof:**
Suppose $n$ is any nonnegative integer whose decimal representation ends in $0$.
Because $n$ is a nonnegative integer, it can be shown in decimal representation
as:
$$ n = 10m + d_0 $$
Where $m$ is some integer.
Since we know that $n$'s decimal representation ends in $0$, that means that
$d_0 = 0$. So $n$ is:
$$ n = 10m $$
Then:
$$ n = 5(2m) $$
Then $2m$ is an integer by the product of integers. By the definition of
divisibility, $5 \mid n$.
Q.E.D.
45. Prove that if $n$ is any nonnegative integer whose decimal representation
ends in $5$, then $5 \mid n$.
**Proof:**
Suppose $n$ is any nonnegative integer whose decimal representation ends in $5$.
Because $n$ is a nonnegative integer, it can be shown in decimal representation
as:
$$ n = 10m + d_0 $$
Where $m$ is some integer.
Since we know that $n$'s decimal representation ends in $5$, that means that
$d_0 = 5$. So $n$ is:
$$ n = 10m + 5 $$
Then:
$$ n = 5(2m + 1) $$
Then $2m + 1$ is an integer by the product and sum of integers. By the
definition of divisibility, $5 \mid n$.
Q.E.D.
46. Prove that if the decimal representation of a nonnegative integer $n$ ends
in $d_1d_0$ and if $4 \mid (10d_1 + d_0)$, then $4 \mid n$. (_Hint:_ If the
decimal representation of a nonnegative integer $n$ ends in $d_1d_0$, then
there is an integer $s$ such that $n = 100s + 10d_1 + d_0$.)
**Proof:**
Suppose $n$ is any nonnegative integer whose decimal representation ends in
$d_1d_0$ and $4 \mid (10d_1 + d_0)$.
Since $4 \mid (10d_1 + d_0)$, then $10d_1 + d_0 = 4k$ for some integer $k$.
And since the decimal representation of $n$ ends in $d_1d_0$, then there is an
integer $s$ such that $n = 100s + 10d_1 + d_0$.
$$ n = 100s + 10d_1 + d_0 $$
Then, by substitution:
$$ n = 100s + 4k $$
$$ n = 4(25s + k) $$
Where $25s + k$ is an integer by the product and sum of integers. Thus
$4 \mid n$.
Q.E.D.
47. Observe that
$$
@ -3266,9 +3977,13 @@ Generalize the argument given in this example to any nonnegative integer $n$. In
other words, prove that for any nonnegative integer $n$, if the sum of the
digits of $n$ is divisible by $9r, then $n$ is divisible by $9$.
Omitted.
48. Prove that for any nonnegative integer $n$, if the sum of the digits of $n$
is divisible by $3$, then $n$ is divisible by $3$.
Omitted.
49. Given a positive integer $n$ written in decimal form, the alternating sum of
the digits of $n$ is obtained by starting with the right-most digit,
subtracting the digit immediately to its left, adding the next digit to the
@ -3277,6 +3992,10 @@ digits of $n$ is divisible by $9r, then $n$ is divisible by $9$.
fact that for any nonnegative integer $n$, if the alternating sum of the
digits of $n$ is divisible by 11, then $n$ is divisible by 11.
Omitted.
50. The integer 123,123 has the form _abc,abc_, where _a_, _b_, and _c_ are
integers from $0$ through $9$. Consider all six-digit integers of this form.
Which prime numbers divide every one of these integers? Prove your answer.
Omitted.

16
chapter_4/test_yourself.md
View file

@ -112,23 +112,39 @@ Page 220
1. To show that a nonzero integer $d$ divides an integer $n$, we must show that
______.
$n$ equals $d$ divided by some integer and $d \neq 0$.
2. To say that $d$ divides $n$ means the same as saying that ______ is divisible
by ______.
$n$; $d$
3. If $a$ and $b$ are positive integers and $a \mid b$, then ______ is less than
or equal to ______.
$a$; $b$
4. For all integers $n$ and $d$, $d \nmid n$ if, and only if, ______.
$\dfrac{n}{d}$ is not an integer.
5. If $a$ and $b$ are integers, the notation $a \mid b$ denotes ______ and the
notation $a/b$ denotes ______.
the sentence "$a$ divides $b$"; the number obtained when $a$ is divided by $b$
6. The transitivity of divisibility theorem says that for all integers $a$, $b$,
and $c$, if ______ then ______.
$a \mid b$ and $b \mid c$; $a \mid c$
7. The divisibility by a prime theorem says that every integer greater than $1$
is ______.
divisible by some prime number.
8. The unique factorization of integers theorem says that any integer greater
than $1$ is either ______ or can be written as ______ in a way that is unique
except possibly for the ______ in which the numbers are written.
prime; a product of prime numbers; order