hp40g+.book Page 13 Friday, December 9, 2005 1:03 AM
Exercise 7
Step-by-Step Examples
×
(
b
=
999
c
–
so,
3
3
×
×
– (
b
1000
+
c
999
3
3
The calculator is not needed for finding the general
solution to equation [1].
x ⋅
b
+
c
We started with
3
and have established that
So, by subtraction we have:
⋅
(
)
⋅
(
b
x 1000
–
+
c
y
+
3
3
⋅
(
)
⋅
b
x 1000
–
=
–
c
or
3
3
According to Gauss's Theorem,
(
c
x 1000
–
is a divisor of
3
∈
k
Z
Hence there exists
(
)
×
x 1000
–
=
k c
3
and
(
)
×
–
y
+
999
=
k b
3
Solving for x and y, we get:
×
x
=
1000
+
k c
3
and
×
y
=
–
999
–
k b
3
∈
k
Z
.
for
This gives us:
⋅
⋅
×
b
x
+
c
y
=
b
1000
3
3
3
The general solution for all
×
x
=
1000
+
k c
3
×
y
=
–
999
–
k b
3
Let m be a point on the circle C of center O and radius 1.
Consider the image M of m defined on their affixes by the
1
-- - z
F : z >
–
transformation
2
)
b
+
1
, or
3
)
=
1
y ⋅
=
1
3
×
×
– (
)
b
1000
+
c
999
3
3
)
999
=
0
(
)
y
+
999
c
is prime with
3
)
.
such that:
×
– (
)
+
c
999
=
1
3
∈
k
Z
is therefore:
2
⋅
–
Z
. When m moves on
=
1
.
b
, so
3
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