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MATH 3012 Homework Problems and Solutions
Chapter 3, Spring 2009, WTT and MTK
Group 1: Basic Problems
1.
Factor
m
= 2205 and
n
= 1925 into products of primes. Then use this factorization to ﬁnd the
greatest common divisor and the least common multiple of
m
and
n
.
Answer: 2205 = 3
2
·
5
·
7
2
and 1925 = 5
2
·
7
·
11, so gcd(2205
,
1925) = 5
·
7 = 35 and lcm(2205
,
1925) =
3
2
·
5
2
·
7
2
·
11.
2.
Use the algorithm developed in our class to ﬁnd the greatest common divisor
d
of
m
= 2205
and
n
= 1925. Also, ﬁnd integers
a
and
b
such that 2205
a
+ 1925
b
=
d
.
Answer: First, note by long division that:
2205 = 1
·
1925 + 280
1925 = 6
·
280 + 245
280 = 1
·
245 + 35
245 = 7
·
35 + 0.
From this it follows that gcd(2205
,
1925) = 35.
Also, we can rewrite the ﬁrst three calculations as:
280 = 1
·
2205

1
·
1925
245 = 1
·
1925

6
·
280
35 = 1
·
280

1
·
245.
Now we start with the last of these statements and backsubstitute as follows:
35 = 1
·
280

1
·
(1
·
1925

6
·
280) =

1
·
1925 + 7
·
280
35 =

1
·
1925 + 7(1
·
2205

1
·
1925) = 7
·
2205

8
·
1925.
That is, we may take
a
= 7 and
b
=

8. Note that 7
·
2205 = 15435 and 8
·
1925 = 15400.
3.
Can you ﬁnd integers
a
and
b
so that 2205
a
+ 1925
b
= 280?
The answer is yes if and only if the greatest common divisor of 2205 and 1925 is also a divisor of
280. Note that 280 = 8
·
35, so the answer is yes. Then from the preceding work, we can take
a
= 56 = 8
·
7 and
b
=

64 = 8(

8).
Can you ﬁnd integers
a
and
b
so that 2205
a
+ 19255
b
= 285?
Here the answer is no since 35 is not a divisor of 285.
4.
Let
r
n
denote the number of regions into which the plane is divided by
n
circles that intersect
in general position. Note that
r
1
= 2 and
r
2
= 4. Develop a recursive formula satisﬁed by
r
n
and
use it to ﬁnd
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 Spring '08
 COSTELLO
 Math

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