This preview shows page 1. Sign up to view the full content.
Section 4.5
97
(c) We must have
1(9) + 2(1) + 3(2) + 4(3) +
5(A)
+ 6(4) + 7(5) + 8(5) + 9(1) + 10(10)
==
0
(mod 11);
hence,
5A
+ 237
==
0 (mod 11),
5A
+ 6
==
0 (mod 11),
5A
==
5 (mod 11) and
A
= 1.
(d) [BB] We must have
1(2) + 2(7) + 3(2) + 4(9) + 5(1) + 6(8) + 7(8) +
8(A)
+ 9(6) + 10(2)
==
0
(mod 11);
hence,
8A
+ 241
==
0 (mod 11),
8A
+ 10
==
0 (mod 11),
8A
==
1 (mod 11) and
A
=
7.
(e) We must have
1(8) + 2(9) +
3(A)
+ 4(5) + 5(3) + 6(9) + 7(8) + 8(2) + 9(8) + 10(4)
==
0
(mod 11);
hence,
3A
+ 299
==
0 (mod 11),
3A
+ 2
==
0 (mod 11),
3A
==
2
==
9 (mod 11) and
A
=
3.
3. The first digit
A
would have to satisfy
I(A)
+ 2(3) + 3(1) + 4(5) + 5(2) + 6(6) + 7(6) + 8(7) + 9(8) + 10(2)
==
0
(mod 11);
hence,
A
+ 265
==
0 (mod 11),
A
+ 1
==
0 (mod 11), and there is no
A
in the range 09 with this
property, so no such ISBN number exists.
4. [BB] at +
2a2
+
... + 9ag +
10alO
==
0 (mod 11)
~
at +
2a2
+
... + 9ag
==
10alO
~
at +
2a2
+ .
..
+ 9ag
==
alO
since 1OalO
==
alO (mod 11).
5. (a) Suppose the ISBN numberata2
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details