Download (b) 4x ≡ 6 (mod 18)

SOLUTIONS: PROBLEM SET 8 (SECTION 5)
2.
(a)
2x ≡ 1 ≡ 20 (mod 19);
x ≡ 10 (mod 19).
(b) 4x ≡ 6 (mod 18) is equivalent to 2x ≡ 3 ≡ 12 (mod 9), so that
the solution is x ≡ 6 (mod 9), or x ≡ 6, 15 (mod 18).
(c)
3x ≡ 1 ≡ 39 (mod 19);
x ≡ 13 (mod 19).
(d) 20x ≡ 984 (mod 1984) is equivalent to 5x ≡ 246 ≡ −250 (mod 496),
so that the general solution is x ≡ −50 ≡ 446 (mod 496), or
x ≡ 446, 942, 1438, 1934
(mod 1984).
4.
(a)
x = 1 + 2t ≡ 2
t≡2
(mod 3);
2t ≡ 1 ≡ 4
x≡5
(mod 3);
(mod 3);
(mod 6).
(b)
x = 2 + 5t;
4 + 10t ≡ 4 + 3t ≡ 3 (mod 7);
3t ≡ −1 ≡ 6 (mod 7);
t = 2 + 7u;
x = 12 + 35u;
3x = 36 + 105u ≡ 3 + 6u ≡ 4 (mod 11);
6u ≡ 1 ≡ 12 (mod 11);
u ≡ 2 (mod 1)1;
x = 12+35u ≡ 82 (mod 385).
(c)
x = 31 + 41t ≡ 6 + 15t ≡ 59 ≡ 7 (mod 26);
15t ≡ 1 ≡ 27 (mod 26);
5t ≡ 9 ≡ 35 (mod 26);
t ≡ 7 (mod 26);
x ≡ 318 (mod 1066).
10.
(a) Subtracting the first congruence from twice the second one, we
obtain
5x ≡ −1 ≡ 35
(mod 9);
x≡7
(mod 9).
Substituting for x in the second equation, we get
21 + y ≡ 3 + y ≡ 2 (mod 9);
1
y ≡ −1 ≡ 8
(mod 9).
2
SOLUTIONS: PROBLEM SET 8 (SECTION 5)
(b) Since multiplying by 2 may lose information (mod 10), this time
we subtract the second equation from three times the first one to
obtain 5y ≡ 7 (mod 10), which has no solution, since 5 divides
10, but not 7. Hence this system of congruences has no solution.
12. Anything congruent to 1, (mod 4) and (mod 3), is also congruent
to 1 (mod 2) and (mod 6); hence it suffices to solve the system
7x ≡ 1
(mod 3),
7x ≡ 1
(mod 4),
7x ≡ 1 (mod 5),
x≡3
(mod 4),
x ≡ 3 (mod 5).
or equivalently,
x ≡ 1 (mod 3),
Since anything congruent to 3 both (mod 4) and (mod 5) is congruent
to 3 (mod 20), the last two congruences are equivalent to x = 3 + 20t
and the first one becomes
7x ≡ x = 3 + 20t ≡ 2t ≡ 1
(mod 3);
t≡2
(mod 3).
Thus the smallest positive value for t is 2, giving x = 43 so that 7x =
301 is the number desired.