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MATH 3360 HOMEWORK SOLUTION 3
(Some computational problems are omitted.)
p72, E2
1729 = 7
×
13
×
19. Hint: 3
6
≡
1(
mod
7), 3
12
≡
1(
mod
13) and 3
18
≡
1(
mod
19).
p72, E3
0
,
9
,
18
,
27
p72, E7
We know
r
(
a

b
) =
km
, then divide both sides by
r
, we get:
a

b
=
km
r
. If you factor out the gcd of
m
and
r
, it becomes:
a

b
=
k
r/
(
r,m
)
m
(
r,m
)
k
r/
(
r,m
)
must be an integer because
r
(
r,m
)
is relatively prime to
m
(
r,m
)
.
p73, E2
All are possible except 24.
p74, E8
Find
as
+
mt
= 1, then multiply both sides by
r

b
.
p86, E8
(a) Use page 72 E7.
(b) Suppose (
b,m
) =
r >
1. Then we claim that it cannot happen that
ba
i
≡
1(
modm
), as
r
divides
both
ba
i
and
m
but doesn’t divide 1. So
{
ba
1
,ba
2
,...,ba
m
}
is not a complete list.
p89, E5
[3][7] = [1], [9][9] = [1], [11][11] = [1], [13][17] = [1], [19][19] = [1].
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Unformatted text preview: p90, E8(i) Empty set. p90, E12 (i) If [ c ] and [ d ] are inverses of [ a ] and [ b ], respectively, then [ cd ][ ab ] = [ c ][ a ][ d ][ b ] = [1]. (ii) If [ ba ] = [ ba ], then by multiplying both sides by the inverse of [ b ] one can get [ a ] = [ a ], contradiction. (iii) First of all, each [ ba i ] is a unit by (i), and they are diﬀerent by (ii). The number of units is a ﬁxed number for each m , so [ ba 1 ] ,..., [ ba r ] is a complete list of units. 1...
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This note was uploaded on 09/05/2011 for the course MATH 3360 taught by Professor Billera during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 BILLERA
 Math, Algebra

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