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Raji, Exercises 3.2:
1. Give a reduced residue system modulo 12.
I suppose the best answer is 1; 5; 7; 11. Warning! A reduced residue system
modulo n is required to contain a representative for every number that is
relatively prime to n. So 1; 7; 11 is not a reduced residue system modulo
12. Raji, no doubt, meant to say that, although his De…nition 14 does not
include that requirement.
2. Give a complete residue system modulo 13 consisting of odd integers.
I would choose 1; 3; 5; 7; 9; 11; 13; 15; 17; 19; 21; 23; 25. Notice that these
numbers are congruent to 1; 3; 5; 7; 9; 11; 0; 2; 4; 6; 8; 10; 12. A more amusing choice is 13 11; 9; 7; 5; 3; 1; 1; 3; 5; 7; 9; 11 which are congruent to 0; 2; 4; 6; 7; 10; 12; 1; 3; 5; 7; 9; 11.
3. Find ' (8) and ' (101).
A reduced residue system modulo 8 is 1; 3; 5; 7, so ' (8) = 4. The number
101 is prime, so ' (101) = 100.
Raji, Exercises 3.3:
1. Find all solutions to 3x
6 (mod 9).
The numbers 2; 4; 8 are the solutions of this congruence that lie in the
complete residue system 0; 1; 2; 3; 4; 5; 6; 7; 8. Verify this by computing 2 0,
2 1, 2 2, . . . , 2 8. Notice that there are three solutions, in accordance
with Theorem 26, because 3 divides 6 and gcd (3; 9) is three.
2. Find all solutions to 3x
2 (mod 7).
Because gcd (3; 7) = 1, there is exactly one solution to this equation (Remark 2). You can …nd it by testing x = 0; 1; 2; 3; 4; 5; 6. In fact, x = 3
works because 3 3 = 9 2 (mod 7).
3. Find an inverse modulo 13 of 2 and of 11.
The inverse of 2 modulo 13 is 7 because 7 2 = 14
1 (mod 13). The
inverse of 11 modulo 13 is 6 because 6 11 = 66 1 (mod 13).
4. Show that if a is the inverse of a modulo m, and b is the inverse of b
modulo m, then ab is the inverse of ab modulo m.
Operating modulo m we see that abab
aabb
1 1 = 1.
Clark, Exercises
18.4 Determine whether or not each of the following is true. Give reasons in
each case.
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(1) x 3 (mod 7) ) gcd (x; 7) = 1.
True. As 7 is prime, gcd (x; 7) must either be 1 or 7. If 7 divides
x 3, then x 3 = 7n for some n, so 3 = x 7n. Thus gcd (x; 7)
must divide 3, so gcd (x; 7) 6= 7. Hence gcd (x; 7) = 1.
(2) gcd (68019; 3) = 3.
True: 68019 = 3 22673.
(3) 12x 15 (mod 35) ) 4x 5 (mod 7).
True. If 12x 15 is divisible by 35, then it is divisible by 7. So
3 (4x 5) is divisible by 7. But 7 doesn’t divide 3, and 7 is prime,
so 7 divides 4x 5 by Euclid’s lemma. Thus 4x 5 (mod 7).
(4) x 6 (mod 12) ) gcd (x; 12) = 6.
True. If x 6 is divisible by 12, then x = 6 + 12n = 6 (1 + 2n) so x
is divisible by 6. The only larger common divisor x could have with
12 would be 12 itself. But 6 = x 12n is not divisible by 12, so 12
does not divide x.
(5) 3x 3y (mod 17) ) x y (mod 17)
True. If 3x 3y (mod 17), then 3x 3y = 3 (x y) is divisible by
17. So, by Euclid’s lemma, x y is divisible by 17.
(6) 5x y (mod 6) ) 15x 3y (mod 18)
True. If 5x y (mod 6), then 5x y = 6n for some n, so 15x
18n whence 15x 3y (mod 18).
3y =
(7) 12x 12y (mod 15) ) x y (mod 5)
True. If 12x 12y (mod 15), then 12x 12y = 12 (x y) = 15n for
some n. So 4 (x y) = 5n. As 5 is a prime, Euclid’s lemma says that
5 divides x y, that is x y (mod 5).
(8) x 73 (mod 75) ) x mod 75 = 73
True. If x 73 (mod 75), then x 73 is divisible by 75, so x 73 =
75q for some q. Thus x = 75q + 73, and 0 73 < 75. By de…nition,
this says that x mod 75 = 73.
(9) x 73 (mod 75) and 0 x < 75 ) x = 73
True. If x 73 (mod 75), then x = 75q + 73 as above. But then 0
75q + 73 < 75, so q can’t be negative because of the …rst inequality,
and q can’t be positive because of the second. So q = 0, that is,
x = 73.
(10) There is no integer x such that 12x 7 (mod 33)
True. If 12x
7 (mod 33), then 12x 7 = 33n for some n, so
7 = 12x 33n = 3 (4x 11n). But 7 is not a multiple of 3.
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