Exercise 2.13.
#8: Prove or disprove that if
a

bc
, where
a, b
, and
c
are positive
integers and
a
6
= 0, then
a

b
or
a

c
.
Proof.
This is false: for example, take
a
= 6,
b
= 2 and
c
= 3.
Then clearly
a
divides neither 2 nor 3, though it divides their product, since every number divides
itself. (Note, however, that if we assumed
a
was prime, then this statement would
be true!)
Exercise 2.14.
#10: What is the quotient and remainder when
(d)

1 is divided by 23
(e)

2002 is divided by 87.
Proof.
(d) The greatest integer multiple of 23 less than

1 is

23. The remainder
is 22.
(e) The greatest integer multiple of 87 less than

2002 is

2088. The remainder
is 86.
Exercise 2.15.
#12: What time does a 24 hour clock read
(c) 168 hours after it reads 19 : 00?
Proof.
(c) It reads 19 : 00  168 hours is exactly 24 hours
×
7.
Exercise 2.16.
#14: Suppose that
a
and
b
are integers,
a
≡
11
mod 19 and
b
≡
3
mod 19. Find the integer
c
with 0
≤
c
≤
18 such that
(d)
c
≡
7
a
+ 3
b
mod 19
Proof.
a
≡
11
mod 19 implies
7
a
≡
77
≡
1
mod 19
,
and, likewise,
b
≡
3
mod 19 implies 3
b
≡
9
mod 19. It follows that
7
a
+ 3
b
≡
10
mod 19
.
Exercise 2.17.
#18: Show that if
a
is an integer and
d
is an integer greater than
1, then the quotient and remainder obtained when
a
is divided by
d
are
b
a/d
c
and
a

d
b
a/d
c
, respectively.
Proof.
The quotient is the greatest integer multiple of
d
less than or equal to
a
,
divided by
d
, and
r
is
a

qd
. Certainly
b
a
d
c
d
is an integer multiple of
d
less than
or equal to
a
so
q
≥ b
a
d
c
.
Suppose towards contradiction that
q >
b
a
d
c
.
Then
q >
b
a
d
c
d
+ 1 so we have
a
≥
qd
≥ b
a
d
c
d
+
d,
14
MATH 55  HOMEWORK 4 SOLUTIONS
which gives the inequality
a

d
≥ b
a
d
d
. Dividing by
d
and taking floor functions,
we get
b
a

d
d
c
=
b
a
d

1
c ≥ b
a
d
c
,
a contradiction.
So
q
=
b
a
d
c
.
Then the remainder is clearly
r
=
a

d
b
a
d
c
, since
r
=
a

qd
.
Exercise 2.18.
Decide whether each of these integers is congruent to 3 modulo 7:
(a) 37
(d)

67
Proof.
(a) No: the difference between 3 and 37 is 34 which is not divisible by 7.
(d) Yes: the difference between

67 and 3 is 70 = 7
×
10.
Exercise 2.19.
#32: Find each of these values:
(c) (7
3
mod 23)
2
mod 31
Proof.
First, we calculate the number inside the square:
7
3
≡
7
×
49
≡
7
×
3
≡
21
mod 29
.
So the number inside the square is 21. Then we calculate using reduction modulo
31:
21
2
≡
7
×
3
×
21
≡
7
×
63
≡
7
×
1
≡
7
mod 31
.
So the answer is 7.
Exercise 2.20.
#36:
Show that if
a, b, c
and
m
are integers such that
m
≥
2,
c >
0, and
a
≡
b
mod
m
then
ac
≡
bc
mod
mc
.
Proof.
To show
ac
≡
bc
mod
mc
we have to show that
mc

(
ac

bc
). Since
a
≡
b
mod
m
, we know
m

(
a

b
) so there is some integer
d
so that
md
= (
a

b
).
It
follows then that
(
mc
)
d
= (
a

b
)
c
= (
ac

bc
)
.
So we’ve shown for some integer
d
, (
mc
)
d
= (
ac

bc
), i.e.
mc

(
ac

bc
) so
ac
≡
bc
mod
mc
.
Exercise 2.21.
#38: Show that if
n
is an integer, then
n
2
≡
0 or 1
mod 4.
Proof.
For any integer, we know
n
≡
0
,
1
,
2
,
or 3
mod 4.
It follows that
n
2
≡
0 or 1
mod 4 since 2
2
≡
0
mod 4 and 3
2
≡
1
mod 4.
Exercise 2.22.
#40:
Prove that if
n
is an odd positive integer, then
n
2
≡
1
mod 8.
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 Spring '08
 STRAIN
 Math, Natural number, Finite set, Countable set