Section
4.2
81
(c) Repeat the preceding arguments, but instead, writing the second term as
9(al
+
11a2
+
... +
~an).
ntimes
33. (a) Let 91
=
gcd(a, b)
and 9
=
gcd(9l,
c).
We have to show that 9
=
gcd(a, b, c).
Since 9
I
91 and
91
I
a
we see that 9
I
a.
Similarly, 9
I
b.
Since also 9
I
c, we see that 9 is a common divisor of
a,
b,
c. Next suppose that
d
is any common divisor of
c. Since in particular,
d
is a common
divisor of
a
and
b, d
I
91 (by Corollary 4.2.11). Since also
d
I
c, we see that
d
is a common divisor
of 91 and c, so
d
::; 9. This proves that 9 is the largest of the common divisors of
a, b
and c.
(b) There exist integers m and
n
such that
gcd(a, b)
=
ma
+
nb
and integers
r
B
such that
gcd(gcd(a,
b),
c)
=
r gcd(a, b)+BC.
Since
gcd(a, b,
c)
=
gcd(gcd(a, b), c),
we get
gcd(a, b,
c)
=
r(ma
+
nb)
+
BC
=
(rm)a
+
(rn)b
+
which expresses
gcd(a, b,
c) as an integral linear com
bination of
a, b
and c.
(c) gcd(231, 165, 105)
=
gcd(gcd(231, 165), 105)
=
gcd(33, 105)
=
3.
Since 33
=
2(231) + 3(165) and 3
=
5(105) + 16(33), we have
3
= 5(105)
+ 16[2(231) + 3(165))

32(231) + 48(165).
(d) gcd(8580,6279)
=
39
=
30(8580)+41(6279) andgcd(2873, 39)
=
13
=
1(2873)+74(39).
So 13
=
gcd(6279, 8580, 2873)
~
1(2873) + 74[30(8580) + 41(6279))
=
1(2873)  2220(8580) + 3034(6279).
(e) gcd(5577, 18837, 25740)
=
gcd(gcd(25740, 18837), 5577)
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 Summer '10
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 Graph Theory, Transitive relation, gcd, yl

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