104
Solutions to Review Exercises
3. Inbaseb, 31
=
1+3b,
156
=
6+5b+b
2
and 220
=
2b+2b
2
.
The equation
(1+3b)+(6+5b+b
2
)
=
2b
+
2b
2
implies 7 +
8b
+
b
2
=
2b
+
2b
2
,
so
b
2

6b

7
=
0
=
(b

7)(b
+ 1). Buddy is correct in
base
b
=
7.
4. The integer
n
has prime divisors 2, 3 and 7, so
n
=
2(3)(7)k
for some integer
k.
Since 2(3)(7) has
eight factorsI, 2, 3, 7, 6, 14, 21,
42k
=
1 and
n
=
42.
5. The given integer is
1 +
2b
+
3b
2
+
4b
3
+
5b
4
+
6b
5
+
7b
6
+
8b
7
+
9b
B
=
(1 +
2b
+
4b
3
+
6b
5
+
8b
7
)
+
(3b
2
+
5b
4
+
7b
6
+
9b
B
),
the sum of an odd and an even integer. The given integer is never even.
6. (a) Since 8215
=
5 + 1(10) + 2(100) + 8(1000), the stated "observation"popularly called "casting
out 9's"follows from Proposition 4.4.7 and the fact that any power of 10 is
==
1 (mod 9). This
holds for any integer, because
(an
la
n
2
...
alaoho
is just
ao+
lOal +
10
2
a2+·
.
·+10
n

l
an_l.
7.
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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