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
...
alO
is correct. Thus
a
=
at
+2a2+"
·+iai+··
·+1OalO
==
0
(mod 11).
If
digit
ai
is miscopied
as
x,
then the test computes
b
=
at
+2a2+
.
+ix+·
+1OalO.
The difference
a

b
=
iai

ix
=
i(ai

x),
so
b
==
i(x

ai)
(mod 11). Now
11
I
i(ai

x)
implies
11
I
i
or
11
I
(ai

x)
(because
11
is prime). Since 1
~
i
~
10, the first possibility cannot
occur. Since the numbers
ai
and
x
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 Summer '10
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 Graph Theory, ISBN, Prime number, different values

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