MATH 135
Fall 2008
Assignment #4 Solutions
HandIn Problems
1. In each part, we must ensure that the remainder
r
satisﬁes 0
≤
r <

b

= 12.
(a) Since 261 = 21(12) + 9, then the quotient is 21 and the remainder is 9.
(b) Since

261 =

22(12) + 3, then the quotient is

22 and the remainder is 3.
(c) Since 261 = (

21)(

12) + 9, then the quotient is

21 and the remainder is 9.
(d) Since

261 = 22(

12) + 3, then the quotient is 22 and the remainder is 3.
2. We prove the result by contradiction.
Assume that
a, b, c
are consecutive terms in a geometric sequence.
Thus,
b
=
ar
and
c
=
ar
2
for some real number
r
.
(Since none of the three terms equals 0, then
a
6
= 0 and
r
6
= 0.)
So our quadratic equation becomes
ax
2
+
arx
+
ar
2
= 0.
Since
a
6
= 0, we can divide through by
a
to obtain
x
2
+
rx
+
r
2
= 0.
The discriminant is Δ =
r
2

4(1)(
r
2
) =

3
r
2
which is strictly less than 0 since
r
6
= 0.
Since the discriminant is negative, the quadratic equation cannot have real roots.
This is a contradiction, as we are told that the quadratic equation does have real roots.
Therefore,
a, b, c
cannot be consecutive terms in a geometric sequence.
3. (a) Using the Euclidean Algorithm,
1989 = 7(251) + 232
251 = 1(232) + 19
232 = 12(19) + 4
19 = 4(4) + 3
4 = 1(3) + 1
3 = 3(1) + 0
so gcd(1989
,
251) = 1, the second last remainder.
(b) Reversing the Euclidean Algorithm above,
gcd(1989
,
251) = 1
= 4

3
= 4

(19

4(4)) = 5(4)

1(19)
= 5(232

12(19))

1(19) = 5(232)

61(19)
= 5(232)

61(1(251)

1(232)) = 66(232)

61(251)
= 66(1989

7(251))

61(251) = 66(1989)

523(251)
so one integer solution to 1989
x
+ 251
y
= gcd(1989
,
251) is (
x, y
) = (66
,

523).