The University of Sydney
MATH2068
Number Theory and Cryptography
(http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/)
Semester 2, 2008
Lecturer: R. Howlett
Tutorial 5
1.
Let
m
be a positive integer. In our terminology, a
residue
mod
m
is a natural
number less than
m
, and
the residue of
k
mod
m
is the remainder when
k
is
divided by
m
. If
a
and
b
are residues mod
m
then we say that
a
is the
inverse
of
b
(mod
m
) if
ab
≡
1 (mod
m
). (Then, equally,
b
is the inverse of
a
(mod
m
).)
(
i
)
Let
m
= 21. Find, if possible, the inverses of 8, 2, 5, 17, 15, 7.
(
ii
)
Let
m
= 84. Find, if possible, the inverses of 17, 83, 33, 23.
Solution.
(
i
)
8
2
= 64
≡
1 (mod 21); so 8 is the inverse of 8. Obviously 11 is the inverse
of 2, since 22
≡
1 (mod 21). Since 5
×
4
≡ 
1 (mod 21) the inverse of 5 is the
residue of

4, namely 17. The inverse of 17 is therefore 5. Since gcd(15
,
21) = 1
there is no
k
such that 15
k
≡
1 (mod 21); so 15 has no inverse.
Similarly,
gcd(7
,
21) = 1, and so 7 has no inverse mod 21.
(
ii
)
We can immediately spot that 17
×
5
≡
1 (mod 84); so the inverse of 17
is 5. Since 83
≡ 
1 (mod 84) and (

1)
2
= 1, it follows that 83 must be its own
inverse. Since gcd(84
,
33) = 3 = 1, the residue 33 has no inverse mod 84. Finally
to find the inverse of 23 we probably need to use the Euclidean Algorithm. We
find 84 = 3
×
23 + 15, then 23 = 15 + 8, then 15 = 8 + 7, then 8 = 7 + 1 and
finally 7 = 7
×
1. We construct the Extended Euclidean Algorithm table
84
23
15
8
7
1
0
3
1
1
1
7
0
1
3
4
7
11
84
1
0
1
1
2
3
23
and deduce that 11
×
23
≡ ±
1 (mod 84). In fact 11
×
23 = 253 = 3
×
84 + 1. So
the inverse of 23 is 11.
2.
Solve the following systems of simultaneous congruences.
(
i
)
23
x
≡
5 (mod 84)
4
x
≡
15 (mod 37)
(
ii
)
3
x
≡
1 (mod 5)
2
x
≡
10 (mod 12)
7
x
≡
2 (mod 17)
2
Solution.
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 One '08
 Howlett
 Number Theory, Cryptography, Congruence, Prime number, Fermat

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