Comp201: Discrete Computational Structures
Fall Term, 2010
Homework 3: Solutions
1 Answers for Section 3.7
3.7.10
We will use proof by contradiction
and assume that
x
is an inverse of
a
modulo
m
, i.e.,
ax
≡
1 (mod
m
). Then by deﬁnition of modular congruence,
ax

1 =
tm
for some integer
t
. If
a
and
m
in this equation both have a common divisor greater than 1, then 1 must also have this
same common divisor since 1 =
ax

tm
. This is not possible because the only positive divisor
of 1 is 1. Therefore, this is a contradiction and no such
x
exists.
3.7.12
By only one iteration of Euclidean algorithm, 17 = 8
·
2 + 1, we ﬁnd gcd(17,2) which
is equal to 1. Thus inverse of 2 in modulo 17 exists. In order to ﬁnd the inverse, reversing the
steps of Euclidean algorithm, we have to write 1 as a linear combination of 2 and 17. Easily we
get 1 =

8
·
2 + 1
·
17. Thus 8 is an inverse and so is 9 in modulo 17.
To solve the congruence, we use this inverse. Multiplying both sides of the equation by 9, we
will get
x
≡
9
·
7 (mod 17). 9
·
7
≡
63
≡
12 (mod 17), so the solutions to this congruence are all
integers congruent to 12 (mod 17), which are
{
...,

22
,

5
,
12
,
29
,...
}
.
3.7.28
a) By Fermat’s Little Theorem we know that 3
4
≡
1 (mod 5); therefore 3
300
= (3
4
)
75
≡
1
75
≡
1 (mod 5), and so 3
302
= 3
2
·
3
300
≡
9
·
1 = 9 (mod 5), so 3
302
mod
5 = 4. Similarly,
3
6
≡
1 (mod 7); therefore, 3
300
= (3
6
)
50
≡
1 (mod 5), and so 3
302
= 3
2
·
3
300
≡
9 (mod 7), so
3
302
mod
7 = 2. Finally, 3
10
≡
1 (mod 11); therefore, 3
300
= (3
10
)
30
≡
1 (mod 11), and so
3
302
= 3
2
·
3
300
≡
9 (mod 11), so 3
302
mod
11 = 2.
b) Since 3
302
is congruent to 9 modulo 5, 7, and 11, it is also congruent to 9 modulo 385 by
Chinese Remainder Theorem.
3.7.46
Translating the letters into numbers we have 0019 1900 0210. Thus we need to compute
C
=
M
13
mod(2537) for
M
1
= 19,
M
2
= 1900 and
M
3
= 210. For the calculations, the fast
modular exponentiation algorithm and Chinese Remainder Theorem will be used.
Fast Modular Exponentiation Algorithm:
We have four variables:
x
is the base,
e
is the exponent and
n
is the modular base. The
variable
y
is initially set to 1. For example for
C
1
= 19
13
mod(43);
x
= 19,
e
= 13 and
n
= 43.
The algorithm is as follows: