Download Homework 9 Solutions

Homework 9 Solutions
Section 4.4
1b) To solve 5x ≡ 2 (mod 26) we solve 5x + 26y = 2 (but we only need x.) By Euclid’s
algorithm we get
26 = 5 · 5 + 1
5 = 5·1
So 26 − 5 · 5 = 1 (duh). Multiply by 2 to get that 2 · 26 − 10 · 5 = 2. We then see that
(−10)5 ≡ 2 (mod 26), thus 16 · 5 ≡ 2 (mod 26). Hence x = 16.
4a) This is just like Example 4.9 in the book. We get the same values for x1 , x2 and x3 .
Hence a solution is given by x = 1 · 35 · 2 + 2 · 21 · 1 + 3 · 15 · 1 = 157 ≡ 52 (mod 105).
4c) Here we have that n = 6 · 11 · 17 = 1122 We use the Chinese Remainder Theorem.
Now N6 = 11 · 17 = 187, N1 1 = 6 · 17 = 102, and N17 = 6 · 11 = 66.
We need to solve 187x6 ≡ 1 (mod 6), 102x11 ≡ 1 (mod 11), and 66x6 ≡ 1 (mod 17).
The first is just 1 · x6 ≡ 1 (mod 6), the second reduces to 3x11 ≡ 1 (mod 11), which
you can do in your head to get x11 ≡ 4 (mod 11). The third equation reduces to
66x6 ≡ 1 (mod 17) or 15x6 ≡ 1 (mod 17) or (−2)x6 ≡ 1 (mod 17), multiply by -1 to
get 2x6 ≡ −1 (mod 17), so we see that x6 ≡ 8 (mod 17). Hence we get that
x = 5 · 187 · 1 + 4 · 102 · 4 + 3 · 66 · 8 = 4151
(mod 1122)
Thus x ≡ 785 (mod 1122)
6) We see that we are required to find the smallest a > 2 such that a ≡ 0 (mod 2), a ≡ 2
(mod 3), a ≡ 2 (mod 4), a ≡ 2 (mod 5) and a ≡ 2 (mod 6). We see that obviously 2
is a solution, we seek to find the next smallest solution. But a ≡ 2 (mod 4) implies that
a ≡ 0 (mod 2). Also if a ≡ 0 (mod 2) and a ≡ 2 (mod 3), then this implies that a ≡ 2
(mod 6). So the set of 5 congruences reduces to just three congruences, namely a ≡ 2
(mod 3), a ≡ 2 (mod 4), and a ≡ 2 (mod 5). So by the Chinese Remainder Theorem
we know that all solutions will congruent modulo 60, hence since 2 is a solution, the
next smallest one will be 62.
9) We are required to find the smallest a such that a ≡ 1 (mod 2), a ≡ 1 (mod 3), a ≡ 1
(mod 4), a ≡ 1 (mod 5), a ≡ 1 (mod 6) and a ≡ 0 (mod 7). Now a ≡ 1 (mod 4)
implies that a ≡ 1 (mod 2) and a ≡ 1 (mod 6) implies that a ≡ 1 (mod 3). Hence we
see that the first 5 congruences reduce to the three congruences a ≡ 1 (mod 4), a ≡ 1
(mod 5), a ≡ 1 (mod 6) which has an obvious solution, namely a ≡ 1 (mod 120). So
a ≡ 0 (mod 7) and a ≡ 1 (mod 120) and hence the solution will be in mod 840. We
just list the numbers that are congruent to 1 (mod 120) until we get one that is a
multiple of 7. We find that 721 ≡ 0 (mod 7). Hence the solution is 721.
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18) We need to obtain two incongruent solutions to the three congruences 2x ≡ 3 (mod 5),
4x ≡ 2 (mod 6) and 3x ≡ 2 (mod 7). The solution to the first equation is x ≡ 4
(mod 5), the solution to the third equation is x ≡ 3 (mod 7). The reason there are
two incongruent solution is that the second equation above has two solutions, namely
x ≡ 2 (mod 6) and x ≡ 5 (mod 6). So we get two different systems of equations,
(1)
x≡4
(mod 5), x ≡ 3
(mod 7), and x ≡ 2
(mod 6)
(2)
x≡4
(mod 5), x ≡ 3
(mod 7), and x ≡ 5
(mod 6)
and
Solving the first equation we see that n = 5 · 6 · 7 = 210 with N5 = 42, N6 = 35, and
N7 = 30.
So we need to solve 42x5 ≡ 1 (mod 5), 35x6 ≡ 1 (mod 6), and 30x7 ≡ 1 (mod 7). We
can do each of these in our head to get that x5 ≡ 3 (mod 5), x6 ≡ 5 (mod 6) and
x7 ≡ 4 (mod 7).
We get that
x = 4 · 42 · 3 + 2 · 35 · 5 + 3 · 30 · 4 = 1214
(mod 210)
So x ≡ 164 (mod 210).
The only thing that changes for the second equation is the congruence modulo 6, but
the setup is all the same and we get the same x5 , x6 and x7 . So now we get that
x = 4 · 42 · 3 + 5 · 35 · 5 + 3 · 30 · 4 = 1529
So x ≡ 59 (mod 210).
So our two solutions are x = 59 and x = 164
2
(mod 210)