Section 4.4
93
vi.
No
x
exists. The values of
4x
2
+
3x
+
7 (mod 5) are 2, 3 and 4.
11. (a) [BB] Multiplying the second congruence by 2 gives
2x
==
0 (mod 6) and, hence,
x
==
0 or
x
==
3
(mod 6).
If
x
==
0, then the first congruence says
y
==
1 and it is easily checked that
x
=
0,
y
=
1
satisfies both congruences.
If
x
==
3, then the first congruence says
y
==
1, but these values for
x
and
y
do not satisfy the second congruence. There is just one solution mod 6; namely,
x
=
0,
y
=
1.
(b) [BB] Subtracting the first congruence from the second gives
3x
==
2
==
7 (mod 9). But the
values of
3x
mod 9 are just 0, 3 and 6. There is no solution.
(c) Subtracting the first congruence from the second gives
3x
==
2
==
6 (mod 8) and, hence,
x
==
2.
The first congruence then says 2
+
5y
==
3; that is,
5y
==
1 (mod 8). So
y
==
5 (mod 8). One
checks easily that
x
=
2,
y
=
5 is a solution to the given congruences.
(d) Multiply the first congruence by 2 to get
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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, Congruence

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