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 1
1.
In each case find the greatest common divisor of
a
and
b
, and integers
r, s
such
that gcd(
a, b
) =
ra
+
sb
.
(
i
)
a
= 14,
b
= 35;
(
ii
)
a
= 168,
b
= 132;
(
iii
)
a
= 847,
b
= 510.
Solution.
(
i
)
A
B
35
14
7
0
Q
2
2
L
K
0
1
2
5
N
M
1
0
1
2
.
This table is computed as follows. Firstly, the first two columns of numbers
are filled in, using the given values of
a
and
b
(namely 14 and 35) as the
initial values of
A
and
B
, and 0, 1, 1, 0 as the initial values of
L
,
K
,
N
and
M
(respectively). Choose
A
to be the larger of the two given numbers and
B
the smaller – that is, put
A
= 35 and
B
= 14 – rather than vice versa.
(Putting
A
= 14 and
B
= 35 works too, it just takes one step longer.) Now
divide
B
into
A
, and put the remainder
R
into the top row (the new value
of
B
) and the quotient
Q
underneath it. Underneath
Q
put
QK
+
L
(the
new
K
) and under that put
QM
+
N
(the new
M
). Repeat until
B
= 0.
The second last entry in the top row will be the gcd (7 in this case) and
the integers
r
and
s
that we seek are either the final values of
N
and

L
(if completion occurs after an even number of steps) or the final values of

N
and
L
(for an odd number of steps). In this example we finished in
two steps; so
r
is 1 (the 2nd last number in the last row) and
s
is

2 (the
negative of the 2nd last number in the second last row). We readily check
that 1
×
35 + (

2)
×
14 = 7.
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 One '08
 Howlett
 Number Theory, Cryptography, Prime number, positive integers

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