#9: x ≡ 2 (mod 7) =⇒ x = 7k + 2. x ≡ 3 (mod 10) =⇒ 7k + 2 ≡ 3 (mod 10) =⇒ 7k ≡
1 (mod 10) =⇒ k ≡ 3 (mod 10) =⇒ x ≡ 7 · 3 + 2 ≡ 23 (mod 70).
#12(a): Finding the remainder of 210203 ÷ 101 is the same as solving 210203 mod 101. Since
101 is prime, 2100 ≡ 1 (mod 101) by Fermat’s Little Theorem. It follows that 210203 ≡
210200 · 23 ≡ (2100 )102 · 8 ≡ 1102 · 8 ≡ 8 (mod 101).
#14(a): 77 ≡ 3 (mod 4) can be seen easily (e.g. by noting that 72 ≡ 1 (mod 4)).
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(b): Let 77 = 4k + 3 (as must be true from part a). Then 77 = 74k+3 = (7k )4 · 73 ≡ (7k )4 .
Since 4 and 7 are relatively prime and φ(10) = 4, then by Euler’s Theorem, (7k )4 ≡ 1 (mod
7
7
10), and 77 ≡ 1 · 73 = 343 ≡ 3 (mod 10). So 3 is the last digit of 77 .
#15:
X
φ(d) = φ(1) + φ(2) + φ(5) + φ(10) = 1 + 1 + 4 + 4 = 10
d|10
X
φ(d) = φ(1) + φ(2) + φ(3) + φ(4) + φ(6) + φ(12) = 1 + 1 + 2 + 2 + 2 + 4 = 12
d|12
X
d|n
φ(d) =
XY
d|n pk ||d
φ(pk ) =
Y
Y k
φ(1) + φ(p) + . . . + φ(pk ) =
p =n
pk ||n
pk ||n
Exponent of 2
0 1 2 3 4 5 6
7
8
9
Evaluated
1 2 4 8 16 32 64 128 256 512
Congruence mod 11 1 2 4 8 5 10 9
7
3
6
3
(b): Let x be the inverse of 3 modulo 10. Then 2 ≡ 8 (mod 11) =⇒ (23 )x ≡ 8x (mod 11)
=⇒ 23x ≡ 21 ≡ 8x (mod 11), using the fact that 210 ≡ 1 (mod 11). The inverse of 3 mod 10
is 7, so 87 ≡ 2 (mod 11).
(c): 8 is 23 , and 3 is relatively prime to 10, which is the order of the cyclic group (Z/11Z)∗
with generator 2. Thus 8 is also a generator of (Z/11Z)∗ .
(d): p is prime, so φ(p) = p − 1. gcd(g, p) = 1 since g is a primitive root mod p. xy ≡
1 (mod p − 1) =⇒ xy ≡ 1 (mod φ(p)) =⇒ g xy ≡ g 1 = g (mod p), and therefore
hx ≡ (g y )x = g xy ≡ g (mod p).
(e): This is an abstraction of part (c). Because g is a primitive root mod p, g generates the
cyclic group (Z/pZ)∗ under multiplication modulo p. The order of said group is φ(p) = p − 1.
y is relatively prime to this order, and therefore h ≡ g y is also a generator.
#17(a):
#20(a): φ(n) satisfies aφ(n) ≡ 1 (mod n), and r is the smallest positive value that satisfies
the equivalence, so r ≤ φ(n).
(b): am = ark = (ar )k ≡ 1k = 1 (mod n)
(c): 1 ≡ at = aqr+s = (ar )q · as ≡ 1q · as = as (mod n) =⇒ 1 ≡ as (mod n)
(d): Assume s 6= 0. Then s is a positive integer which satisfies as ≡ 1 (mod n). But then
s < r is a contradiction, because r is the smallest such integer. It follows that s = 0.
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(e): Euler’s Theorem states aφ(n) ≡ 1 (mod n), which implies ordn (a)|φ(n) by the result just
obtained in part (d).
#21(a): r|600 =⇒ ∃ k : rk = 600. Then k|600, and therefore k|gcd(k, 600). r < 600 =⇒
k > 1 =⇒ gcd(k, 600) > 1. If k were relatively prime to 2, 3, and 5, then k would be
relatively prime to 600, i.e. gcd(k, 600) = 1, a contradiciton. Thus 2|k, 3|k, or 5|k. It follows
and r| 600
. Thus r| 600
, r| 600
, or r| 600
.
that if p|k, then rk = 600 =⇒ r · kp = 600
p
p
2
3
5
(b): This follows from part (a).
(c): Because ord601 (7)|n =⇒ 7n ≡ 1 (mod 601)
(d): We have shown that ord601 (7) ≮ 600. 7600 ≡ 1 (mod 601) by Fermat’s Little Theorem
(601 is prime). Thus the order of 7 mod 601 is 600 by definition, so 7 is a primitive root
mod 601.
(e): g is a primitive root mod p iff g (p−1)/qn 6≡ 1 (mod p) ∀ qn ∈ {q1 , . . . , qs }
#24: WLOG, let i = 1. Note that for j 6= 1, that zj = m1 . . . mj−1 mj+1 . . . mk , and therefore
m1 |zj It follows that m1 |aj yj zj . Then
x = a1 y1 z1 + a2 y2 z2 + . . . + ak yk zk ≡ a1 y1 z1 ≡ a1 (modm1 ),
where the last congruence holds because y1 ≡ z1−1 by definition.
#39(a): The first q −1 multiples m ≥ 1 of p are p, 2p, . . . , (q −1)p. The next multiple would
be m = pq, which does not satisfy m < pq. Thus there are exactly these q − 1 and no more.
The same argument shows that there are p − 1 multiples m of q such that 1 ≤ m < pq.
(Used later) Let p|x and q|x. p and q are distinct irreducibles, so gcd(p, q) = 1. Then by
some fundamental theorem, pq|x.
(b): Let d = gcd(m, pq). Then d|pq =⇒ ∃ f : df = pq. Assume p - d and q - d. Then
p|pq =⇒ p|df =⇒ p|f or p|d =⇒ p|f , because p is prime. Similarly, q|f . But then pq|f ,
and it follows from df = pq that d|1, and therefore d = 1, a contradiction. Thus p|d or q|d,
and because d|m, either p|m or q|m.
(c): p|m, q|m =⇒ pq|m. 1 ≤ m < pq =⇒ m 6≡ 0 (mod pq) =⇒ pq - m, a contradiction.
Thus p|m and q|m cannot both be true.
(d): There are pq − 1 integers m such that 1 ≤ m < pq. Of these, any m such that
gcd(m, pq) 6= 1 is either a multiple of p or q, but not both. There are exactly p − 1 multiples
of q and q − 1 multiples of p, so the number of m such that gcd(m, pq) = 1 is pq − 1 − (p −
1) − (q − 1).
#40(a): x ≡ 0 (mod 2) and x ≡ 1 (mod 4)
(b): x ≡ 0 (mod 2) and x ≡ 0 (mod 4)
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